Answer:
a) To find the acceleration at the bottom of the pipe, we can use the formula for centripetal acceleration: a = v^2 / r where v is the velocity, and r is the radius (half of the diameter) of the pipe. Since the velocity is constant and equal to 5 m/s, and the radius of the pipe is 4 m, the acceleration at the bottom is:
a = (5 m/s)^2 / 4 m a = 6.25 m/s^2
b) To find the force on the bike at an angle of 30° up from the bottom, we need to use the formula for centripetal force: F = m * a where m is the mass of the bike and rider (given as 400 kg) and a is the centripetal acceleration calculated in part (a). The force on the bike at an angle of 30° up from the bottom is:
F = 400 kg * 6.25 m/s^2 * cos(30°) F = 3,464 N
c) To find the minimum velocity at the top for the bike and rider to stay moving in a circle, we can use the same formula for centripetal acceleration and solve for velocity: a = v^2 / r v = sqrt(a * r) where r is the radius of the pipe (again, 4 m) and a is the centripetal acceleration required to keep the bike and rider moving in a circle, which is equal to the acceleration due to gravity at the top of the pipe:
a = g = 9.81 m/s^2 v = sqrt(9.81 m/s^2 * 4 m) v = 6.26 m/s
d) Comparing the minimum velocity calculated in part (c) to the constant speed of 5 m/s given in the question, we can see that the bike and rider do have sufficient velocity to stay moving on a circle at the top of the pipe.
ILL MARK BRAINLIEST!!!!!!
Which of the following statements regarding its orbit is true?
A. The sun and a planet are at the two foci of an orbit
B. The central body is located at one focus of the ellipse describing the orbit.
C. In an elliptical orbit, there is one focus and the satellite is located there.
D. In an orbit, the satellite and the central body are the two foci of the ellipse
Answer:
The answer to your problem is, D. In an orbit, the satellite and the central body are the two foci of the ellipse
Explanation:
Our solar system contains the sun and eight planets revolving around the sun. Planets like earth, mars, mercury etc. are revolving around sun in their own orbits. Every planet's orbit around the Sun is an ellipse, according to Kepler's First Law.
Two focal points, or foci, make up an ellipse. The overall distance of a planet from these 2 focus points is constant during its orbit. Additionally, an ellipse has two symmetry lines.
The orbital ellipse's central focus is always where the sun is positioned. The sun is centered. The planet's orbit is an ellipse, thus as it revolves around its axis, the distance from the sun changes continuously.
Thus the answer to your problem is, D. In an orbit, the satellite and the central body are the two foci of the ellipse
A ball is dropped from the height of 10m.at the same time, another ball is thrown vertically upward at an initial speed of 10m/s. How high above the ground will thr two balls collide?
Answer: H=5.1m
Explanation:
Given:
Ball 1 height= 10m
Ball 2 initial velocity=10m/s
use the kinematic equation:
S=(vi)t+12at2
I choose my sign convention to be up=positive, down=negative, so a=−9.81ms2 (a is taken as the value of gravity)
Ball 1 dropped from 10m :
−(10−H)=0+12(−9.81)t2
Note that (10-S) is negative because that displacement is *below* the starting point.
12(9.81)t2=10−H
——- equation (1)
Ball 2 thrown upward at 10 m/s :
H=(10)t+12(−9.81)t2
or
12(9.81)t2=10t−H
——- equation (2)
equation (1) minus equation (2):
0=(10−H)−(10t−H)
t=1 equation (1):
12(9.81)12=10−H
H=5.1m
Given Information:
Ball One:
[tex]\vec y_{0} = 10 \ m[/tex] (Indicating the initial position)
We also know the ball was dropped from rest. So, [tex]\vec v_{0_{1} } = 0 \ m/s[/tex].
Ball Two:
[tex]\vec v_{0} = 10 \ m/s[/tex] (Indicating the initial velocity)
We also know the ball was throw from the ground. So, [tex]\vec y_{0_{1} } = 0 \ m[/tex].
The Information we want to Find:
[tex]\vec y_{c} = ?? \ m[/tex] (Indicating the position the two projectiles collide)
Using the Following Kinematic Equation to Solve:
[tex]\Delta \vec x = \vec v_{0}t + \frac{1}{2} \vec at[/tex]
For ball one...
[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y} t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \vec v_{0_{1} }t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = (0)t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex] => Equation 1
For ball two...
[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0_{1} } = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t +\vec y_{0_{1} }[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t + 0[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex] => Equation 2
Set equations 1 and 2 equal to each other and solve for the time that they collide.
[tex]\left \{ {{\vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}} \atop { \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t } \right.[/tex]
[tex]\Longrightarrow \frac{1}{2} \vec at +\vec y_{0}= \vec v_{0}t + \frac{1}{2} \vec at[/tex]
[tex]\Longrightarrow \vec y_{0}= \vec v_{0}t[/tex]
[tex]\Longrightarrow t=\frac{\vec y_{0}}{\vec v_{0}}[/tex]
[tex]\Longrightarrow t=\frac{10}{10}[/tex]
[tex]\Longrightarrow t=1 \ s[/tex]
Thus, the balls collide at time, t=1 s. We can now use this time to plug into equation 1 or 2 to find the height at which they collide. I will use equation 1.
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex]
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} (-9.8)(1) +10[/tex]
[tex]\Longrightarrow \vec y_{c} = 5.1 \ m \ \therefore \ Sol.[/tex]
*Note* [tex]\vec a_{y}[/tex] is the acceleration of gravity ([tex]-9.8 \ m/s^2 \ or \ -32 \ ft/s^2[/tex])
Final Answer: The balls collide at the height 5.1 m.
A 70.0 kg
ice hockey goalie, originally at rest, has a 0.110 kg
hockey puck slapped at him at a velocity of 43.5 m/s.
Suppose the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. What would the final velocities goalie
and puck
of the goalie and the puck, respectively, be in this case? Assume that the collision is completely elastic.
The final velocity of the goalie is -0.613 m/s (indicating that he moves backwards), and the final velocity of the puck is 30.2 m/s (indicating that it moves forwards).
What does mechanics' collision mean?When particles, collections of particles, or solid bodies move in the same direction and get close enough to each other, they collide and generate mutual force.
This issue can be resolved by applying the laws of motion and kinetic energy conservation.
momentum conservation
Before the collision:
Total momentum = 0 (since the goalie is at rest)
After the collision:
Total momentum = m_goalie * v_goalie + m_puck * v_puck
Conservation of kinetic energy:
Before the collision:
Total kinetic energy = 0
After the collision:
Total kinetic energy = (1/2) * m_goalie * v_goalie² + (1/2) * m_puck * v_puck^2
We can use these two equations to solve for the final velocities of the goalie and the puck.
First, let's use the conservation of momentum equation to solve for v_goalie:
0 = m_goalie * v_goalie + m_puck * v_puck
v_goalie = - m_puck * v_puck / m_goalie
Now, we can substitute this expression for v_goalie into the conservation of kinetic energy equation:
(1/2) * m_puck * v_puck² = (1/2) * m_goalie * (- m_puck * v_puck / m_goalie)² + (1/2) * m_puck * 43.5²
Simplifying this equation and solving for v_puck, we get:
v_puck = 2 * (m_oalie / (m_goalie + m_puck)) * 43.5
v_puck = 30.2 m/s
Finally, we can substitute this value for v_puck back into the equation for v_goalie
v_goalie = - m_puck * v_puck / m_goalie
v_goalie = -0.613 m/s
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When a force F stretches a rope of mass per unit length r, the velocity of a wave in the rope is given by xxxx. You pull on a rope with a certain force, and a wave travels in the rope with a certain velocity. If you double your force, the velocity of a wave in the rope is now ____________ the original velocity.
A. 1/2
B. xxxxx times
C. the same as
D. xxxx times
E. 2 times
Velocity of the wave in the rope is now 1.4 times the original velocity that is option B.
What is force?An influence that causes motion of any object with mass to change its velocity is called as force.
The velocity of a wave in a rope is given by the following equation:
v = √(F/r)
F is the force applied to the rope and r is the mass per unit length of the rope.
If the force is doubled (2F), the velocity of the wave in the rope can be found as follows:
v' = √(2F/r)
v'/v = √(2F/r) / √(F/r) = √(2)
Therefore, velocity of the wave in the rope is now 1.4 times the original velocity, or option B.
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Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is W = 72.0 N, is sitting at distance L= 0.80 m to the left of the pivot, at what distance L1 should Marcel place Gilles, whose weight is w, to the right of the pivot to balance the seesaw? Keep 2 digits after the decimal point, in meters.
Marcel should place Gilles at about 0.828 m to the right of the pivot in order to balance the seesaw.
Moment of forcesFor the seesaw to be balanced, the clockwise moment caused by Gilles sitting on the right side of the pivot must be equal to the counterclockwise moment caused by Jacques sitting on the left side of the pivot. The moment (M) is given by the weight of the child multiplied by the distance from the pivot:
M = w × L1 = W × L
where w is the weight of Gilles.
Rearranging this equation, we get:
L1 = (W × L) / w
Substituting the given values, we get:
L1 = (72.0 N × 0.80 m) / w
We don't know the weight of Gilles, so we cannot solve for L1. However, we can set up an equation to find the weight w needed to balance the seesaw:
W × L = w × L1
Substituting the given values, we get:
72.0 N × 0.80 m = w × L1
Solving for w, we get:
w = (72.0 N × 0.80 m) / L1
Now we can substitute this expression for w into the earlier equation for L1, giving:
L1 = (W × L) / [(72.0 N × 0.80 m) / L1]
Simplifying, we get:
L1^2 = (W × L × L1) / (72.0 N × 0.80 m)
L1^2 = (W × L) / (72.0 N × 0.80 m)
Substituting the given values and solving, we get:
L1 = sqrt[(72.0 N × 0.80 m) / (76.8 N)]
L1 ≈ 0.828 m
Therefore, Marcel should place Gilles at a distance of approximately 0.828 m to the right of the pivot to balance the seesaw.
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What is the force of attraction between a balloon with a charge of +4.0 x 10^-6 C is held a distance of 0.41 m from a second balloon having the same charge?
Explanation:
The force of attraction between two charged objects is given by Coulomb's law:
F = k * (q1 * q2) / r^2
where:
F is the force of attraction
k is Coulomb's constant, which has a value of 9.0 x 10^9 N*m^2/C^2
q1 and q2 are the charges of the two objects
r is the distance between the two objects
In this case, we have two balloons with the same charge of +4.0 x 10^-6 C each, and they are held at a distance of 0.41 m from each other. Plugging these values into Coulomb's law, we get:
F = 9.0 x 10^9 N*m^2/C^2 * [(+4.0 x 10^-6 C)^2] / (0.41 m)^2
F = 1.92 x 10^-3 N
Therefore, the force of attraction between the two balloons is 1.92 x 10^-3 N.
A block is attached to two identical ideal springs. It is initially held so the whole setup is horizontal and the springs are not extended. The unextended length of each spring is 0.9 m with a spring constant of 473 N/m. The block is slowly lowered until the mass is in static equillibrium. The springs now make an angle θ = 20o with the horizontal as shown. What is the mass of the block? You may assume the block is a point mass (no volume).
Answer:
Approximately [tex]1.9\; {\rm kg}[/tex] (assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)
Explanation:
Let [tex]L = 0.9\; {\rm m}[/tex] denote the unextended length of each spring.
The length of each spring is now [tex](L / \cos(\theta))[/tex]. The displacement of each spring would be [tex]x = L - (L / \cos(\theta)) = (1 - (1 / \cos(\theta)))\, L[/tex].
The tension in each spring would be [tex]T = k\, x[/tex], where [tex]k[/tex] is the spring constant.
Decompose the tension that each spring exerts on the block into two components:
Horizontal: [tex]T\, \cos(\theta)[/tex].Vertical: [tex]T\, \sin(\theta)[/tex].The two horizontal components balance each other since they are equal in magnitude. The two vertical components add on to each other to exert a total upward force of [tex]2\, T\, \sin(\theta)[/tex] on the block.
Since the block is in equilibrium, the resultant force on the block will be [tex]0[/tex]. The sum of these two (upward) vertical components of tension should balance the (downward) weight of the block:
[tex]2\, T\, \sin(\theta) = m\, g[/tex], where [tex]m[/tex] is the mass of the block.
Rearrange this equation to find the mass of the block:
[tex]\begin{aligned} m &= \frac{2\, T\, \sin(\theta)}{g} \\ &= \frac{2\, k\, x\, \sin(\theta)}{g} \\ &= \frac{2\, k\, L\, (1 - (1 / \cos(\theta))\, \sin(\theta)}{g} \\ &= \frac{2\, (473)\, (0.9)\, (1 - (1 / \cos(20^{\circ})))\, \sin(20^{\circ})}{(9.81)}\; {\rm kg} \\ &\approx 1.9\; {\rm kg}\end{aligned}[/tex].
[tex]\blue{\huge {\mathrm{MASS \; OF \; THE \; BLOCK}}}[/tex]
[tex]\\[/tex]
[tex]{===========================================}[/tex]
[tex]{\underline{\huge \mathbb{Q} {\large \mathrm {UESTION : }}}}[/tex]
A block is attached to two identical ideal springs. It is initially held so the whole setup is horizontal and the springs are not extended. The unextended length of each spring is 0.9 m with a spring constant of 473 N/m. The block is slowly lowered until the mass is in static equillibrium. The springs now make an angle θ = 20° with the horizontal as shown. What is the mass of the block? You may assume the block is a point mass (no volume).[tex]{===========================================}[/tex]
[tex] {\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} [/tex]
The mass of the block is 1.9 kg.[tex]{===========================================}[/tex]
[tex]{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}}[/tex]
To solve for the mass of the block, we can use the forces acting on the block at equilibrium. We know that the force of gravity pulling down on the block is equal to the force of the springs pulling up.
The force of each spring can be found using Hooke's Law:
[tex]\sf F = kx[/tex]where:
F is the force,k is the spring constant, andx is the displacement from the equilibrium position.In this case, the displacement is equal to the extension of the spring, which is given by:
[tex]\sf x = L(1-\cos\theta)[/tex]where:
L is the unextended length of the spring, andθ is the angle it makes with the horizontal.So the force of each spring is:
[tex]\sf F_{spring} = kx = kL(1-\cos\theta)[/tex]At equilibrium, the forces in the vertical direction must balance, so we have:
[tex]\sf 2F_{spring} = mg[/tex]where
m is the mass of the block andg is the acceleration due to gravity.Substituting in the expression for [tex]\sf F_{spring}[/tex] and simplifying, we get:
[tex]\sf\qquad\implies 2kL(1-\cos\theta) = mg[/tex]
Solving for m, we obtain:
[tex]\sf\qquad\implies m = \dfrac{2kL(1-\cos\theta)}{g}[/tex]
Plugging in the given values, we get: [tex]\\\begin{aligned}\sf m&=\sf \dfrac{2(473\: N/m)(0.9\: m)[1-\cos(20^{\circ})][\sin(20^{\circ})]}{(9.81 m/s^2)}\\&=\boxed{\bold{\:1.9\: kg\:}}\end{aligned}[/tex]
Therefore, the mass of the block is 1.9 kg.
[tex]{===========================================}[/tex]
[tex]- \large\sf\copyright \: \large\tt{AriesLaveau}\large\qquad\qquad\qquad\qquad\qquad\qquad\tt 04/02/2023[/tex]
Your ability to process language in the left hemisphere of the brain and spatial in the right hemisphere of the brain is called
Answer: Parietal lobe
Explanation: The parietal lobe controls the ability to read, write, and understand spatial concepts. Therefore, you gain the ability to process language through the left hemisphere of your brain.
What happens to the Sun's energy as it passes through the atmosphere to Earth's surface?
Some of the Sun's energy is absorbed, scattered, or reflected as it travels through the atmosphere to the surface of the Earth by a variety of atmospheric constituents, including gases, aerosols, clouds, and the Earth's surface.
What uses does the atmosphere make of solar energy?50% of the heat energy from the Sun can reach Earth's surface thanks to the atmosphere. The Sun's energy is reflected back into space at a rate of 30%. The atmosphere greenhouse gases, such as carbon dioxide, water vapour, and methane, absorb 20% of the remaining solar energy.
When solar energy travels through the stratosphere, what happens to it?Ozone (O3) in the upper atmosphere absorbs a substantial amount of the Sun's ultraviolet (high-energy, shortwave) light (the stratosphere). The Earth system does not become hotter as a result of solar radiation that Earth's surface or atmosphere reflect back into space. Heat is produced as a result of absorbed radiation.
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A ring of Aluminum bronze alloy has internal diameter 300 mm and 50 mm wide. The coefficient of cubic expansion of alloy is 51 x 10-6/°C. For a temperature rise of 600°C, find the following in mm: a) The final internal diameter. b) The change in width of the ring.
As a result, the ultimate internal diameter is D = 300 mm + D,D = 309.18 mm, and the ring's change in breadth is 1.53 mm.
Why does thermal expansion occur? What is it?Thermal expansion is the process through which an item enlarges and expands as a result of a change in temperature. The molecules take up more space because they move more quickly on average at higher temperatures. As a result, when anything is heated up, it gets bigger.
We must apply the thermal expansion formula to this issue in order to find a solution:
ΔL = α L ΔT
where L is the length change, is the thermal expansion coefficient, L is the starting length, and T is the temperature change.
a) The final internal diameter:
ΔD = α D ΔT
Substituting the values given, we get:
ΔD = (51 x 10^-6/°C) x 300 mm x 600°C
ΔD = 9.18 mm
The final internal diameter is therefore:
D = 300 mm + ΔD
D = 309.18 mm
b) The change in width of the ring:
The original width of the ring is 50 mm. We can use the same formula to find the change in width:
ΔW = α W ΔT
Substituting the values given, we get:
ΔW = (51 x 10^-6/°C) x 50 mm x 600°C
ΔW = 1.53 mm
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1. Given that the atmospheric pressure and density at sea level is 100 kPa and 1.3 kg/m respectively. Calculate the height of the atmosphere if all air in the atmosphere has this density (g = 10 N/kg)
Answer:
P = ρ g H pkgressure due to liquid (gas) of height H
H = 1.00E5 N/m^2 / (1.3 kg / m^3 * 10 N/kg) = 7,700 m
The towing lines of two tugboats pulling horizontally on a barge are at an angle of 30° to each other. The tensions in the towing lines of the first and second tugboats are 3 kN and 4 kN respectively. Calculate the magnitude of the resultant force which the tugboats exert on the barge.
The magnitude of the resultant force exerted on the barge is approximately 3.6 kN.
Step by step explanationTo calculate the magnitude of the resultant force exerted on the barge, we can use the law of cosines:
c^2 = a^2 + b^2 - 2ab cos(C)
where c is the magnitude of the resultant force, a and b are the magnitudes of T1 and T2, respectively, and C is the angle between T1 and T2 (which is 30° in this case).
Substituting the given values, we get:
c^2 = (3 kN)^2 + (4 kN)^2 - 2(3 kN)(4 kN) cos(30°)
c^2 = 9 kN^2 + 16 kN^2 - 24 kN^2 cos(30°)
c^2 = 25 kN^2 - 24 kN^2 cos(30°)
c^2 = 25 kN^2 - 12 kN^2
c^2 = 13 kN^2
c = sqrt(13) kN
c ≈ 3.6 kN
Therefore, the magnitude of the resultant force exerted on the barge is approximately 3.6 kN.
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What is the momentum of an object with a mass of 4 kg traveling at 7 m/s?
Answer:
The answer to your problem is, 28 kg x m/s
Explanation:
m = 4
v = 7
P = 4 x 7
= 28
There is no other way to do it.
Thus the answer to your problem is, 28 kg x m/s
An average froghopper insect has a mass of 12.8 mg and jumps to a maximum height of 293 mm when its takeoff angle is 62.0∘ above the horizontal.
a) Find the takeoff speed of the froghopper.
b) How much kinetic energy did the froghopper generate for this jump? Express your answer in microjoules
c) how much energy per unit body mass was required for this jump ? Express your answer in joules per kilogram of body mass.
a) The takeoff speed of the froghopper can be found using the following equation:
v^2 = 2gh/(1 - cos^2(theta))
where:
v = takeoff speed
g = acceleration due to gravity (9.81 m/s^2)
h = maximum height (293 mm = 0.293 m)
theta = takeoff angle (62.0 degrees)
Substituting the given values into the equation, we get:
v^2 = 2(9.81)(0.293)/(1 - cos^2(62.0))
v^2 = 0.571
v = sqrt(0.571)
v ≈ 0.756 m/s
Therefore, the takeoff speed of the froghopper is approximately 0.756 m/s.
b) The kinetic energy generated by the froghopper can be found using the following equation:
KE = 0.5mv^2
where:
m = mass (12.8 mg = 0.0128 g)
v = takeoff speed (0.756 m/s)
Substituting the given values into the equation, we get:
KE = 0.5(0.0128)(0.756)^2
KE ≈ 0.00346 J
(1 J = 10^6 microjoules)
Therefore, the kinetic energy generated by the froghopper for this jump is approximately 0.00346 microjoules.
c) The energy per unit body mass required for this jump can be found by dividing the kinetic energy by the mass of the froghopper:
energy per unit body mass = KE/m
Substituting the values we obtained earlier, we get:
energy per unit body mass = 0.00346/0.0128
energy per unit body mass ≈ 0.270 J/kg
Therefore, the energy per unit body mass required for this jump is approximately 0.270 joules per kilogram of body mass.
Spring constant, Say if we have KEi + PEi + WNC= KEf + PEf , where does the spring potential energy go, before or after?
The spring potential energy is included in the initial potential energy term in the equation KEi + PEi + WNC= KEf + PEf.
What is the formula for calculating the potential energy of a spring?The formula for calculating the potential energy of a spring is PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position.
How does the spring constant affect the potential energy of a spring?The spring constant (k) determines the stiffness of the spring and how much force is required to stretch or compress it. The greater the spring constant, the more potential energy is stored in the spring for a given displacement from its equilibrium position. This means that a spring with a higher spring constant will have a greater potential energy than a spring with a lower spring constant when stretched or compressed by the same amount.
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Suppose that humans have created a colony outside of our solar system on a planet called Webb13. Webb13 has a mass of 2.75×1025 kg
and a day that lasts 22.9 h
(which defines the rotational period of the planet). The colony is located on the planet's equator.
The colonists set up a communications satellite which orbits Webb13. The satellite has a circular orbit that keeps it positioned directly above the colony.
Calculate the radius
of the satellite's orbit in kilometers.
The orbital radius of the satellite above Webb13's equator is around 11,360 kilometres.
What does the term "rotational period" mean?There are several rotating periods (of an astronomic object) the length of time needed for it to revolve in relation to the nearby stars. (of an object revolving on Earth) the duration of its axis rotation in relation to the earth (assumed fixed).
Since 1 hour equals 60 minutes x 60 seconds, or 3600 seconds, the orbital period of the satellite is the same as the planet's rotational period, which is 22.9 hours, or 82,440 seconds. The following formula may be used to determine the radius of the satellite's orbit:
[tex]r = (G * M * T^2 / 4π^2)^(1/3)[/tex]
where r is the orbit's radius, G is the gravitational constant, M is the planet's mass, and T is the satellite's orbital period.
Using the specified values:
[tex]G = 6.67 × 10^-11 m^3 kg^-1 s^-2 (gravitational constant)[/tex]
[tex]M = 2.75 × 10^25 kg (mass of Webb13)[/tex]
T = 82,440 s (orbital period of satellite)
The units can be changed to kilometres and then entered into the formula as follows:
[tex]r = (6.67 × 10^-11 m^3 kg^-1 s^-2 * 2.75 × 10^25 kg * (82,440 s)^2 / (4π^2))^(1/3)[/tex]
[tex]r = 1.136 × 10^7 m[/tex]
[tex]r = 11,360 km[/tex]
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consider a system consisting of two point particles m1 and m2. the particles are 25 cm apart, the mass of particle m2 is four times the mass of m1, and mass m1 is 5 cm from the origin. calculate the position of the center of mass, xcm.
The position of the center of mass of the system is 21 cm from the origin.
The position of the center of mass of a two-particle system can be calculated using the formula,
xcm = (m1x1 + m2x2) / (m1 + m2)
where x1 and x2 are the positions of particles m1 and m2, respectively, and m1 and m2 are their masses.
Substituting values into the formula for xcm,
xcm = (m1x1 + m2x2) / (m1 + m2)
xcm = (m15 cm + m225 cm) / (m1 + m2)
xcm = (m15 cm + 4m125 cm) / (m1 + 4m1)
xcm = (5 cm + 4*25 cm) / (1 + 4)
xcm = 105 cm / 5
xcm = 21 cm
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If the mass of Jupiter is defined as 1 M_j = 1.90 ✕ 10^27 kg, what is the mass of Saturn (5.68 ✕ 10^26 kg) in units of M_j?
What is the mass of Earth (5.97 ✕ 10^24 kg) in M_j?
What is the mass of Neptune (1.02 ✕ 10^26 kg) in M_j?
Answer: 1. Mass of Saturn in terms of Jupiter mass:
Saturn's mass = 5.68 × 10²⁶ kg
Jupiter's mass, MJ = 1.90 × 10²⁷
Therefore, Saturn's mass in terms of MJ = Saturn's mass/1.90 × 10²⁷
= 0.299 MJ
Therefore, Mass of Saturn is smaller than and is equal to 0.299 times mass of Jupiter.
To convert masses to units of M_j, we need to divide the given mass by the mass of Jupiter:
1 M_j = 1.90 x 10^27 kg
(a) Mass of Saturn in M_j:
Mass of Saturn = 5.68 x 10^26 kg
Mass of Saturn in M_j = (5.68 x 10^26 kg) / (1.90 x 10^27 kg/M_j)
= 0.299 M_j
Therefore, the mass of Saturn in units of M_j is approximately 0.299 M_j.
(b) Mass of Earth in M_j:
Mass of Earth = 5.97 x 10^24 kg
Mass of Earth in M_j = (5.97 x 10^24 kg) / (1.90 x 10^27 kg/M_j)
= 0.00315 M_j
Therefore, the mass of Earth in units of M_j is approximately 0.00315 M_j.
(c) Mass of Neptune in M_j:
Mass of Neptune = 1.02 x 10^26 kg
Mass of Neptune in M_j = (1.02 x 10^26 kg) / (1.90 x 10^27 kg/M_j)
= 0.0537 M_j
Therefore, the mass of Neptune in units of M_j is approximately 0.0537 M_j.
A child sits on a merry‑go‑round that has a diameter of 5.00 m. The child uses her legs to push the merry‑go‑round, making it go from rest to an angular speed of 20.0 rpm in a time of 41.0 s.
What is the average angular acceleration avg of the merry‑go‑round in units of radians per second squared (rad/s2)?
What is the angular displacement Δ of the merry‑go‑round, in units of radians (rad),
during the time the child pushes the merry‑go‑round?
The child pushes the merry-go-round, which causes it to rotate by an angle of around 85.5 radians.
What is the merry-go-moment round's of inertia?The sum of the moments of inertia of the toddler and the merry-go-round (both about the same axis) represents the total moment of inertia: I=(28.13kg⋅m2)+(56.25kg⋅m2)=84.38kg⋅m2. When known values are substituted into the equation.
ω1 = 0 rad/s (initially at rest)
ω2 = (20.0 rpm) * (2π rad/rev) / (60 s/min) ≈ 4.19 rad/s
The average angular acceleration is given by the equation:
αavg = (ω2 - ω1) / t
where t is the time interval. Substituting the given values, we get:
αavg = (4.19 rad/s - 0 rad/s) / 41.0 s
αavg ≈ 0.102 rad/s²
Therefore, the average angular acceleration of the merry-go-round is approximately 0.102 rad/s².
To find the angular displacement of the merry-go-round, we can use the equation:
Δθ = ω1*t + (1/2)αt²
where ω1 is the initial angular speed, α is the average angular acceleration, and t is the time interval.
Substituting the given values, we get:
Δθ = 0 rad/s * 41.0 s + (1/2) * (0.102 rad/s²) * (41.0 s)²
Δθ ≈ 85.5 rad
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Catastrophes, significant life changes and sleep were the three examples provided in
the video that lead to stress?
True
False
The statement is False that Catastrophes, significant life changes and sleep were the three examples provided in the video that lead to stress.
What does catastrophe stress look like?disastrous stress .Serious sickness in the child or a family member, natural disasters, and child maltreatment are a few instances of this level of stress. The child's risk is at its maximum at this level.
What three stages of the body's reaction to stress are there?The alarm reaction stage, the resistance stage, and the weariness stage are the several phases of this illness. The "fight or flight" response and the body's earliest signs of acute stress are referred to as the alarm reaction stage.
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What did Erikson believe about the developmental stages of adolescence and young adulthood? Using what you know about eriksons stages of development, do you agree or disagree with the statement that all adolescents and young, adults pass through these stages? Explain
Overall, it is important to consider the individual differences and cultural context when applying Erikson's theory to adolescence and young adulthood.
What is Erikson's theory?
Erikson's theory is a psychoanalytic theory that describes the development of the human personality across eight stages throughout the lifespan. Each stage is characterized by a particular crisis or conflict that must be resolved in order for the individual to develop a healthy personality.
Erik Erikson proposed a theory of psychosocial development that includes eight stages spanning from infancy to old age. In adolescence and young adulthood, the stage is identity versus role confusion. Erikson believed that during this stage, individuals explore and experiment with different identities and roles as they seek to establish a sense of self and their place in the world.
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A 60.0 kg wrecking ball hangs from a uniform, heavy-duty chain of mass of 26.0 kg. Find the maximum tension in the chain.
Answer: The maximum tension in the chain will occur when the wrecking ball is at its lowest point, and the chain is vertical.
The weight of the wrecking ball is:
w = m_ball * g
w = 60.0 kg * 9.81 m/s^2
w = 588.6 N
The weight of the chain is:
w_chain = m_chain * g
w_chain = 26.0 kg * 9.81 m/s^2
w_chain = 254.8 N
At the lowest point, the tension in the chain must be equal to the sum of the weight of the ball and the weight of the chain:
T = w + w_chain
T = 588.6 N + 254.8 N
T = 843.4 N
Therefore, the maximum tension in the chain is 843.4 N.
Explanation:
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Consider three main sequence stars: Star A has a mass of 2 solar masses, Star B has a mass of 0.5 solar masses, and Star C has a mass of 20 solar masses
Which star will become a giant first?
Which star will become a giant last?
Answer:
Star C (with a mass of 20 solar masses) will become a giant first, after about 14 million years.
Star A (with a mass of 2 solar masses) will become a giant next, after about 85 million years.
Star B (with a mass of 0.5 solar masses) will become a giant last, after about 12.5 billion years.
Explanation:
Two horizontal forces, F and F₂, act on a box, but only Ę appears in the drawing. Ę₂ can refer to either the right or the left. The square moves along the x axis only. There is no friction between the box and the surface. Suppose F₁ = +4.0 N and the mass of the box is 4.1 kg. Find the magnitude and direction of F₂ when the acceleration of the box is (a) +7.0 m/s², (b) -7.0 m/s², and (c) 0 m/s².
Answer:
Since there is no friction, the net force acting on the box is equal to the sum of the two horizontal forces. From Newton's second law, we know that the net force is equal to the mass of the box times its acceleration. Therefore:
ΣF = m * a
where ΣF is the net force, m is the mass of the box, and a is the acceleration of the box.
We can use this equation to find the magnitude of F₂ in each case.
(a) When the acceleration of the box is +7.0 m/s²:
ΣF = F₁ + F₂
m * a = F₁ + F₂
(4.1 kg) * (7.0 m/s²) = 4.0 N + F₂
F₂ = (4.1 kg) * (7.0 m/s²) - 4.0 N
F₂ = 25.7 N to the right
So, the magnitude of F₂ is 25.7 N, and it acts to the right.
(b) When the acceleration of the box is -7.0 m/s²:
ΣF = F₁ + F₂
m * a = F₁ + F₂
(4.1 kg) * (-7.0 m/s²) = 4.0 N + F₂
F₂ = (4.1 kg) * (-7.0 m/s²) - 4.0 N
F₂ = -32.6 N to the left
So, the magnitude of F₂ is 32.6 N, and it acts to the left.
(c) When the acceleration of the box is 0 m/s²:
ΣF = F₁ + F₂
m * a = F₁ + F₂
(4.1 kg) * (0 m/s²) = 4.0 N + F₂
F₂ = -4.0 N
So, the magnitude of F₂ is 4.0 N, and it acts to the left.
Explanation:
a proton enters a uniform magnetic field that is perpendicular to the proton's velocity (figure 1). what happens to the kinetic energy of the proton?
A. it increases.
B. it decreases. C. it stays the same.
D. it depends on the velocity direction.
E. it depends on the b field direction.
The correct answer is C. The kinetic energy of the proton will remain the same if it enters a uniform magnetic field that is perpendicular to its velocity . Therefore, it stays the same.
Kinetic energy is a fundamental concept in physics that refers to the energy of an object in motion. When an object moves, it possesses kinetic energy that is proportional to its mass and the square of its velocity. The formula for kinetic energy is K = 1/2mv^2, where m is the mass of the object and v is its velocity.
The concept of kinetic energy is important in many areas of physics, including mechanics, thermodynamics, and relativity. It is used to describe the motion of particles in a gas, the motion of planets in a solar system, and the motion of subatomic particles in a particle accelerator. The concept of kinetic energy is closely related to other concepts in physics, such as potential energy, work, and momentum. Understanding kinetic energy is essential for understanding the behavior of physical systems in motion.
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60 N left
20 N right
what is the net force
Answer:
the net force is 40 N to the left.
Explanation:
To find the net force, we need to subtract the force going to the right from the force going to the left:
Net force = 60 N - 20 N
Net force = 40 N to the left
Therefore, the net force is 40 N to the left.
a hot iron is turned off and cools down to room temperature. the oron cools because
Answer:
heat energy is transferred from the arm iron to the cooler room. BRAINLIEST ANSWER
which part of dc motor reverse the direction of current through the coil every half cycle
In a DC motor, the part that reverses the direction of current through the coil every half cycle is the commutator. The commutator is a cylindrical structure mounted on the rotor shaft and consists of multiple metal segments separated by insulating material. The segments are connected to the ends of the coils, which are wound on the rotor.
As the rotor turns, the brushes (which are in contact with the commutator) transfer current to the coils through the commutator segments. The current direction in the coil is determined by the orientation of the coil relative to the magnetic field of the stator. When the coil is in one half of its rotation, the current flows in one direction, and when it is in the other half, the current flows in the opposite direction. The commutator reverses the direction of the current through the coil every half cycle to maintain the direction of the torque produced by the motor.
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Answer the 3 questions in the picture please
The clock will run faster in Pluto
The force constant is 81 N/m
The period is 3.5 s
Does clock run faster in pluto or jupitar?Clocks on Jupiter, which has a stronger gravitational field than Pluto, would run slower than clocks on Pluto. This means that clocks on Pluto would appear to run faster when compared to clocks on Jupiter.
We know that;
F = Ke
K = F/e
K = 68.28738 N/84.81007 * 10^-2 m
K = 81 N/m
Then;
T = 2π√L/g
T = 2(3.14)√0.5/1.6
T = 3.5 s
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A cyclist is rounding a 20-m -radius curve at 13 m/s.
What is the minimum possible coefficient of static friction between the bike tires and the ground?
The minimum possible coefficient of static friction between the bike tires and the ground is 0.6.
This is calculated by dividing the centripetal force formula.Centripetal force = m*v2/r
Centripetal force = (m*132)/20
Normal force = mg
Normal force = m*9.8
Let us find the minimum coefficient of static friction
The minimum coefficient of static friction = Centripetal force/Normal force
= (m*132)/(20*m*9.8) = 0.6
The minimum coefficient of static friction between the bike tires and the ground is 0.6, which is calculated by dividing the centripetal force of the cyclist by the normal force of the cyclist.
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