Please HELP me answer these three questions!! (show work)
Answer:
2) 42.6
3) 1500000 m
4) no he will not be late, it will take 3s
Explanation:
#2:
Formula for average velocity:
⇒ [tex]v=\frac{d}{t}[/tex]
Substitute
⇒ [tex]\frac{2560}{60}[/tex]
⇒ [tex]42.6[/tex]
#3:
1st convert 60km to m/s (correct si unit must always come first)
to convert kilometer to meter, multiply by 1000 (prefix 'kilo' = 1000)
60 km x 1000 = 60000 m
use the formula for speed, distance, & time
60000 = d/25
to find displacement just multiply the speed and time together
60000 x 25
1500000m is the displacement
#4:
s = d/t
given are displacement and speed/velocity
d = 25m
v = 8.5 m/s
to find the time just divide displacement by speed
8.5 = 25/t
25/8.5
≈ 3 seconds
so he will not be late for class.
When heal flows between systems Entropy
Answer: increases
Explanation:
Explain how a balloon is able to keep its shape?
Answer:
It depends on how they are made.
Explanation:
Rubber balloons are not always spherical in shape. When filled with air, the inflation forms a balance between the balloon material, including its shape and thickness and its elasticity and the pressure of the air. That’s what determines its shape. Although a sphere is often the shape, it could be tubular, offset, and most other shapes.
Two identical cars (m-1350 kg) are traveling at the same speed of 35.7 m/s. They are moving in the directions shown in the drawing
What is the magnitude of the total momentum of the two cars?
Car 1 - 60°
Car 2 - 30°
The magnitude of the total momentum of the two cars is 68,245.5 kg m/s.
What is the magnitude of the car's total momentum?To calculate the total momentum of the two cars, we need to first calculate the momentum of each car and then add them together.
The momentum of an object is given by the product of its mass and velocity. So, the momentum of each car can be calculated as:
p = m x v
where;
p is momentum, m is mass, and v is velocity.Since both cars have the same mass, their momenta will be equal if they have the same velocity. In this case, both cars are traveling at the same speed of 35.7 m/s.
The momentum of car 1 can be calculated by resolving its velocity into horizontal and vertical components:
vx1 = v1 cos(60°) = 0.5 x 35.7 = 17.85 m/s
vy1 = v1 sin(60°) = 0.866 x 35.7 = 30.97 m/s
The momentum of car 1 is then:
p₁ = m x v₁ = 1350 x √(vx₁² + vy₁²)
p₁ = 1350 x √(17.85² + 30.97²)
p₁ = 48,256.85 kg m/s
Similarly, the momentum of car 2 can be calculated by resolving its velocity into horizontal and vertical components:
vx2 = v2 cos(30°) = 0.866 x 35.7 = 30.97 m/s
vy2 = v2 sin(30°) = 0.5 x 35.7 = 17.85 m/s
The momentum of car 2 is then:
p₂ = m x v₂ = 1350 x √(vx₂² + vy₂²)
p₂ = 1350 x √(30.97² + 17.85²)
p₂ = 48,256.85 kg m/s
The total momentum of the two cars is the vector sum of their momenta, which can be calculated using the Pythagorean theorem:
ptotal = √(p₁² + p₂²)
= √((48,256.85 )² + (48,256.85²) = 68,245.5 kg m/s
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The rate of flow of heat through different materials of the same thickness is different. Plan and design an experiment to test this statement based on the rate of flow of heat
Based on the results of the experiment, it can be concluded that the rate of flow of heat through different materials of the same thickness is different.
What is the experiment plan?Here is an experimental plan to test the rate of flow of heat through different materials of the same thickness:
Materials:
Three blocks of different materials (e.g., metal, plastic, and wood)
Thermometer
Heat source (e.g., hot plate)
Stopwatch
Insulating material (e.g., foam)
Procedure:
Cut three blocks of the same thickness from each of the three materials.
Measure the initial temperature of each block using a thermometer.
Place the three blocks on a heat source (e.g., hot plate) with the same amount of heat and start the stopwatch.
Measure the temperature of each block every 30 seconds using the thermometer.
Record the temperature of each block at each time interval.
After 5 minutes, turn off the heat source and measure the final temperature of each block.
Calculate the temperature difference between the initial and final temperatures for each block.
Calculate the rate of heat flow for each block by dividing the temperature difference by the time interval.
Repeat the experiment at least three times for each block and take an average of the results.
Place each block on an insulating material (e.g., foam) and repeat the experiment to compare the effect of insulation on the rate of heat flow.
Data analysis:
Plot a graph of the rate of heat flow (y-axis) versus time (x-axis) for each block.
Compare the slopes of the graphs to determine the rate of heat flow for each block.
Compare the rates of heat flow for the three blocks to test the statement that the rate of flow of heat through different materials of the same thickness is different.
Compare the rates of heat flow for each block with and without insulation to determine the effect of insulation on the rate of heat flow.
Conclusion:
The rate of heat flow depends on the thermal conductivity of the material, which is a measure of how well a material conducts heat. Materials with higher thermal conductivity will have a higher rate of heat flow, while materials with lower thermal conductivity will have a lower rate of heat flow.
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how long will be required for a car to go from a speed of 20.0 m/s to a speed of 25.0 m/s if the acceleration is 3.0 m/s2?
Answer:
Explanation:
We can use the following kinematic equation to solve for the time required:
v_f = v_i + at
where:
v_f = final velocity = 25.0 m/s
v_i = initial velocity = 20.0 m/s
a = acceleration = 3.0 m/s^2
t = time
Rearranging the equation to solve for t, we get:
t = (v_f - v_i) / a
Substituting the given values, we get:
t = (25.0 m/s - 20.0 m/s) / 3.0 m/s^2
t = 1.67 s
Therefore, it will take 1.67 seconds for the car to go from a speed of 20.0 m/s to a speed of 25.0 m/s, assuming a constant acceleration of 3.0 m/s^2.
Work Energy Theorem Question: You apply 50 N to a 10 kg object to cause it to move from rest to 2.5 m/s. What distance was the object moved?
Answer:
0.625 meters
Explanation:
We can use the work-energy that the work done on an object is equal to the change in its kinetic energy:
Work = ΔK = Kf - Ki
Where:
Work is the work done on the object
ΔK is the change in kinetic energy of the object
Kf is the final kinetic energy of the object
Ki is the initial kinetic energy of the object (which is zero since the object is at rest)
The work done on the object is equal to the force applied to the object multiplied by the distance over which the force is applied:
Work = F × d
Where:
F is the force applied to the object (50 N)
d is the distance over which the force is applied (unknown)
So we can write:
F × d = Kf - Ki
Substituting the given values:
50 N × d = 1/2 × 10 kg × (2.5 m/s)^2 - 0
Simplifying:
50 N × d = 31.25 J
Solving for d:
d = 31.25 J / 50 N = 0.625 m
Therefore, the object was moved a distance of 0.625 meters.
please rate
Suppose the angles shown in Fig. 5.31 are 52° and 25°. If the left-hand mass is 2.5 kg, what should the right-hand mass be so that it accelerates (a) downslope at 0.64 m/s2 and (b) upslope at 0.76 m/s2?
Downslope at 0.64 m/s², m = 12.4 kg
Upslope at 0.76 m/s², m = 6.35 kg
Define Mass?In Physics, mass is the most basic property of matter, and it is one of the fundamental quantities. Mass is defined as the amount of matter present in a body. The SI unit of mass is the kilogram (kg). The formula of mass can be written as:
Mass = Density × VolumePart A)
The sum of forces on the left-hand mass
T - mgsin(angle) = ma
T - (2.6) (9.8) (sin 70) = 2.6(.64)
T = 25.6 N
m = left mass.........M = right mass
T - mg×sin70 = ma
Mg×sin16 - T = Ma
Mg×sin16 - mg×sin70 = a×(M+m)
M×g×sin16 - mg×sin70 = Ma + ma
M× (g×sin16 -a) = m× (a + gsin70)
M = m× (a + gsin70) / (g×sin16 -a)
a) a = 0.64
M = 10.98 Kg
b) M = 11.82 kg
For the right-hand mass, the sum of forces...
mgsin(angle) - T = ma
m (9.8) (sin 16) - 25.6 = m (.64)
2.7m - 25.6 = .64m
m = 12.4 kg
Part B)
For the left-hand mass
mgsin(70) - T = ma
(2.6) (9.8) (sin 70) - T = (2.6) (.76)
T = 21.97 N
Then for the right-hand mass
T - mgsin(16) = ma
21.97 - m (9.8) (sin 16) = m (.76)
21.97 - 2.7m = .76m
m = 6.35 kg
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The right-hand mass should be 3.3 kg to accelerate up the slope at 0.76 m/s². To solve this problem, we need to use the principles of Newton's laws of motion and trigonometry.
We know that the force of gravity acting on the mass is equal to its weight, which can be calculated using the formula Fg = mg, where Fg is the force of gravity, m is the mass, and g is the acceleration due to gravity (which is approximately 9.8 m/s²).
To find the force acting down the slope, we need to calculate the component of the weight that acts down the slope, which is given by Fg sin θ, where θ is the angle of the slope. Using the given angle of 52°, we can calculate the force acting down the slope for the left-hand mass as:
Fdown = Fg sin θ
Fdown = (2.5 kg)(9.8 m/s²) sin 52°
Fdown = 18.9 N
To find the required mass for the right-hand mass to accelerate at 0.64 m/s^2 down the slope, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass times its acceleration (F = ma). Therefore, we can calculate the required force for the right-hand mass as:
F = ma
F = (m)(0.64 m/s²)
Since the force acting down the slope is 18.9 N, we can set these two equations equal to each other and solve for the mass:
F = Fdown
(m)(0.64 m/s²) = 18.9 N
m = 29.5 kg
Therefore, the right-hand mass should be 29.5 kg to accelerate down the slope at 0.64 m/s².
To find the required mass for the right-hand mass to accelerate at 0.76 m/s² up the slope, we can use the same approach, but this time we need to use the component of the weight that acts up the slope, which is given by Fg cos θ, where θ is the angle of the slope. Using the given angle of 25°, we can calculate the force acting up the slope for the right-hand mass as:
Fup = Fg cos θ
Fup = (m)(9.8 m/s²) cos 25°
Setting this equal to the force required for the right-hand mass to accelerate up the slope, we get:
Fup = ma
(m)(0.76 m/s²) = (m)(9.8 m/s²) cos 25°
m = 3.3 kg
Therefore, the right-hand mass should be 3.3 kg to accelerate up the slope at 0.76 m/s².
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1. A 8.2 kg mass hanging from a spring scale is slowly lowered onto a vertical spring.
A) What does the spring scale read just before the mass touches the lower spring?
B) The scale reads 14 N when the lower spring has been compressed by 2.4 cm . What is the value of the spring constant for the lower spring?
C) At what compression length will the scale read zero?
The spring scale read just before the mass touches the lower spring is 80.36N, the spring constant for the lower spring is 2765N/m and at 2.9cm length the scale will read zero.
Given the mass of spring = 8.2kg
The force exerted for compressing of spring = 14N
The compression in spring = 2.4cm = 0.024m
(A.) Initially the spring scale reads only the weight of the mass = mg
W = 8.2 * 9.8 = 80.36N
(B) Let the value of spring constant = k
The net force exerted so that the scale reads(F') = 80.36N - 14 = 66.36N
We know that according to Hooke's law the force exerted on spring F = kx such that:
F' = kx then:
66.36 = k * 0.024
k = 66.36/0.024 = 2765N/m
(C) the compression where scale reads zero = x'
The scale reads zero when the restoring force equals to the weight of the mass then the scale reads zero such that:
x' = 80.36/2765 = 0.029m = 2.9cm
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There are two objects (A and B) each with its own mass and charge. The electrostatic force between them has a value of 23.2N and is attractive. The two objects are separated by a distance of 2.3cm with object A being to the left of B. Assume one charge has 4 times the charge of the other.
a. Do the two charges have the same signs or opposite signs?
b. IF the two charges have opposite signs, is charge A or B negative?
c. Which charge (A or B) has the higher value?
d. Find the value of each charge.
Answer:
Explanation:
a. Since the electrostatic force is attractive, the charges must have opposite signs.
b. Let's assume that charge A is positive and charge B is negative. Then, the electrostatic force would be given by:
F = (k * |qA| * |qB|) / r^2
where k is Coulomb's constant, r is the distance between the charges, and |qA| and |qB| are the magnitudes of the charges. Since the force is attractive, |qA| > |qB|.
c. From the given information, we know that:
F = 23.2 N
r = 2.3 cm = 0.023 m
Let |qB| = q, then |qA| = 4q. Substituting these values into the equation for the electrostatic force, we get:
23.2 = (k * 4q * q) / (0.023)^2
Solving for q, we get:
q = 3.38 x 10^-7 C
Then, |qA| = 4q = 1.35 x 10^-6 C.
d. Charge A has a value of 1.35 x 10^-6 C and charge B has a value of -3.38 x 10^-7 C.
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A 4.0-kg mass is moving to the right at 3.0 m/s. An 8.0 kg mass is moving to the left at 2.0 m/s. If after collision the two
masses join together, what is their velocity after collision?
O-0.33 m/s
O-0.20 m/s
O +1.4 m/s
O +2.3 m/s
Answer:
- 0.33 m/s
Explanation:
An illustration is shown above,
In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,
m1u1 - m2u2 = (m1 + m2)v
Where,
m1 = mass of the first object
u1 = initial velocity of the first object
v1 = final velocity of the first object
m2 = mass of the second object
u2 = initial velocity of the second object
v2 = final velocity of the second object
Therefore,
(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v
12 - 16 = 12v
-4 = 12v
Divide both sides by 12,
-4 / 12 = 12v / 12
-1 / 3 = v
v = -0.33 m/s
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Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kilometers to the supermarket, which is 5.0 kilometers from her own home. She walked the total distance in 1.5 hours. What were her average speed and average velocity?
A.
Her average speed was about 4.6 km/hr, and her average velocity was about 3.3 km/hr.
B.
Her average speed was about 3.3 km/hr, and her average velocity was about 4.6 km/hr.
C.
Her average speed was about 3.3 km/hr, and her average velocity was 0 km/hr.
D.
Her average speed was 0 km/hr, and her average velocity was about 4.6 km/hr.
Her average speed was about 4.6 km/hr, and her average velocity was about 3.3 km/hr.
The entire distance travelled divided by the total time taken is the definition of average speed. In this case, the total distance travelled was 7.0 km, and the total time taken was 1.5 hours. Hence, the average speed can be determined as follows:
Average Speed = [tex]\frac{7.0 km }{ 1.5 \ hours }= 4.6 km/hr[/tex]
The displacement divided by the whole time travelled is the average velocity. In this case, the displacement was 3.0 km (from Mary's home to Sheila's home), and the total time taken was 1.5 hours.The average velocity can therefore be determined as follows:
Average Velocity = [tex]\frac{3.0 km }{1.5 \ hours} = 3.3 km/hr[/tex]
Therefore,Her average velocity was roughly 3.3 km/hr, and her average speed was roughly 4.6 km/hr.
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A 65 kg-mass person stands at the end of a diving board, 1.5 m from the board's pivot point. Determine the torque the person is exerting on the board with respect to the pivot point. Show your work.
Answer:
Explanation:
The torque is given by the formula:
τ = F × r × sin(θ)
where τ is the torque, F is the force applied, r is the distance between the force and the pivot point, and θ is the angle between the force and the lever arm.
In this case, the person's weight is the force being applied, and it can be calculated as:
F = m × g
where m is the mass of the person and g is the acceleration due to gravity (9.81 m/s^2).
F = 65 kg × 9.81 m/s^2 = 637.65 N
The distance between the person and the pivot point is 1.5 m, so r = 1.5 m.
The angle between the person's weight and the lever arm is 90 degrees, so sin(θ) = 1.
Therefore, the torque the person is exerting on the board is:
τ = F × r × sin(θ) = 637.65 N × 1.5 m × 1 = 956.475 N·m
So the person is exerting a torque of 956.475 N·m on the diving board with respect to the pivot point.
A girl with a mass of 32 kg is playing on a swing. There are three main forces
acting on her at any time: gravity, force due to centripetal acceleration, and
the tension in the swing's chain (ignore the effects of air resistance). At the
instant shown in the image below, she is at the bottom of the swing and is
traveling at a constant speed of 4 m/s. What is the tension in the swing's
chain at this time? (Recall that g = 9.8 m/s²)
Tension
Weight
A. 333.6 N
OB. 817.8 N
C. 562.8 N
D. 441.6 N
4 m/s
The tension in the swing's chain at the instant shown in the image is 441.6 N.
option D
What is the tension at bottom swing?At the bottom of the swing, the girl is traveling at a constant speed, so her acceleration is zero. Therefore, the net force acting on her is also zero.
Thus, we have:
0 = T - mg - mv²/r
where;
T is the tension in the swing's chain, m is the girl's mass, g is the acceleration due to gravity, v is her speed, and r is the radius of the swing.At the bottom of the swing, the radius is equal to the length of the chain, so we have:
r = L = 4.0 m
Substituting the values we have:
T = (32 kg)(9.8 m/s²) + (32 kg)(4 m/s)²/4.0 m
Solving for T, we get:
T = (32 kg)(9.8 m/s²) + (32 kg)(4 m/s)²/4.0 m
T = 441.6 N.
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A metal filament Lamp rated at 750w, 100v into be connected in series with a capacitor across a 230v, 60hz supply. Calculate the capacitance required
Answer : 0.00885 farads or 8.85 microfarads
Explanation: To calculate the capacitance required, we can use the following formula:
C = 1 / [2 * pi * f * ((V^2 - Vlamp^2)/P)]
where:
C = capacitance in farads (F)
pi = 3.14159...
f = frequency in hertz (Hz)
V = voltage in volts (V)
P = power in watts (W)
Vlamp = voltage of the lamp in volts (V)
Using the given values, we have:
C = 1 / [2 * pi * 60 * ((230^2 - 100^2)/750)]
C = 1 / [2 * 3.14159 * 60 * ((230^2 - 100^2)/750)]
C = 1 / [113.09724]
C = 0.00885 farads (F)
Therefore, the capacitance required is approximately 0.00885 farads or 8.85 microfarads.
Select in the ticker-timer a frequency of 25 Hz or 50 Hz. Determine the period of the ticker-timer.
Answer:
The period of a ticker-timer is the time interval between two consecutive dots made by the ticker.
If the frequency of the ticker-timer is 25 Hz, then it makes 25 dots in one second. Therefore, the period of the ticker-timer can be calculated as:
Period = 1/frequency = 1/25 Hz = 0.04 seconds
If the frequency of the ticker-timer is 50 Hz, then it makes 50 dots in one second. Therefore, the period of the ticker-timer can be calculated as:
Period = 1/frequency = 1/50 Hz = 0.02 seconds
So, the period of the ticker-timer is 0.04 seconds for a frequency of 25 Hz and 0.02 seconds for a frequency of 50 Hz
Explanation:
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Liam pulls a box with a horizontal force of 17 N over a distance of 19 m. What is the change in energy of the box after the 19 m?
So the change in energy of the box after being pulled by Liam over 19 m is 323 J.
What are some illustrations of energy change?Energy is capable of changing its forms. For instance, electrical energy transforms into thermal and light energy when a lightbulb is turned on. A automobile transforms the gasoline's molecular bonds' stored energy into a variety of other forms. Chemical energy is converted to light in the engine through a chemical reaction.
The work done on the box by Liam's force is:
work = force x distance x cos(theta)
where theta is the angle between the force and the direction of motion. Since the force is horizontal and the motion is also horizontal, theta is 0 degrees and cos(theta) is 1. Therefore, the work done on the box is:
work = 17 N x 19 m x 1 = 323 J
The change in energy of the box is equal to the work done on it, according to the work-energy principle. Therefore, the change in energy of the box is:
change in energy = work = 323 J
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A block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . determine the displacement of the velocityA block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . Determine how far has block 1 moved during the 1.2-s interval? A) 13.4 m B) 2.1m C) 28.2m D) 7.6mA block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . determine the displacement of the velocityA block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . Determine how far has block 1 moved during the 1.2-s interval?
To solve this problem, we can use the conservation of mechanical energy principle. When the blocks are released from rest, the potential energy of the system is converted to kinetic energy. Since the surface is frictionless, the mechanical energy of the system is conserved.
Using the principle of mechanical energy conservation, we can write:
m1*g*h = (m1+m2)*v^2/2
where m1 is the mass of the first block, m2 is the mass of the second block, g is the acceleration due to gravity, h is the height that the second block falls, and v is the velocity of the system after the blocks have moved a distance x.
The displacement of the first block can be found by using the time it takes the system to reach this velocity. The time t can be found using the formula:
x = (1/2) * a * t^2
where a is the acceleration of the first block.
The acceleration of the first block is equal to the acceleration of the system, which can be found by using the equation:
m1*a = m2*g - m1*g
Substituting the value of a in the previous formula, we get:
x = (1/2) * (m2*g - m1*g) * t^2 / m1
Substituting the values we get:
x = (1/2) * (2.0 kg * 9.81 m/s^2 - 3.0 kg * 9.81 m/s^2) * (1.2 s)^2 / 3.0 kg
x ≈ 7.6 m
Therefore, the correct answer is D) 7.6 m.
A stuntman of mass 55 kg is to be launched horizontally out of a spring- loaded cannon. The spring that will launch the stuntman has a spring coefficient of 266N / m and is compressed 5 m prior to launching the stuntman. If friction and air resistance can be ignored, what will be the approximate velocity of the stuntman once he has left the cannon?
The approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.
StepsWe can use the conservation of energy, where the potential energy stored in the compressed spring is converted into the kinetic energy of the stuntman as he is launched out of the cannon.
The potential energy stored in the spring is given by:
PE = (1/2)kx²
where k is the spring constant and x is the distance the spring is compressed.
PE = (1/2)(266 N/m)(5 m)² = 3325 J
This potential energy is then converted into kinetic energy:
KE = (1/2)mv²
where m is the mass of the stuntman and v is his velocity.
3325 J = (1/2)(55 kg)v²
v² = (2*3325 J) / 55 kg
v² = 121 m²/s²
v = √(121 m²/s²) = 11 m/s
Therefore, the approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.
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it the frequency of a wave is increased it’s
and
will decrease but its
will stay the same.
Answer:
Wavelength will decrease but frequency will stay the same.
Explanation:
The higher the frequency, the shorter the wavelength. The relationship between wavelength and frequency is called an inverse relationship, because as the frequency increases, the wavelength decreases.
However, the frequency usually remains the same because it is like a driven oscillation and maintains the frequency of the original source. If v changes and f remains the same, then the wavelength λ λ must change. Since v = f λ v = f λ , the higher the speed of a sound, the greater its wavelength for a given frequency.
Describe the change to the graph of Y= X +3 when Y=2X -3 is graphed 
Answer: a stretch of 2
Explanation: because it 2 (x) - 3
Two stars, Bucky and Badger, form in the same giant molecular cloud. Bucky has 5 solar mass and Badger has 1 solar mass. Which of the followings is correct?
A) The main-sequence life of Bucky is 5 times longer
B) Bucky has a longer time to become a protostar
C) We can detect Badger first when it becomes a pre-main- sequence star
D) They have the same heavy elements
Answer:
Most likely the answer is D;
Explanation:
Because they formed from the same molecular cloud.
Bucky definitely will live shorter. And we can detect Bucky faster due it's enormous rate of burring fuel.
Both of the stars will have the same heavy elements.
The two stars, Bucky and Badger are formed in the same giant molecular cloud. Among them Bucky has five solar mass, which is five times the solar mass of Badger.
As a result of the higher solar mass of Bucky, its fuel will burn up very faster than Badger. So, Bucky will have shorter life. Also it will spin faster and become a protostar in short time.
Since, both the stars, Bucky and Badger are formed in the same giant molecular cloud, both of them will have the same heavy elements.
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What might happen if people did not have the rights established in Miranda v. Arizona
Answer:
If people did not have the rights established in Miranda v. Arizona, they could be subjected to police coercion and forced confessions without a lawyer present. This could lead to wrongful convictions and denial of due process, which are essential components of the American justice system
A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal velocity of 7.2 m/s.] How far does it move horizontally in this time
Answer:
Explanation:
Assuming negligible air resistance, the horizontal velocity of the shingle will remain constant and the vertical motion will be influenced by gravity.
We can use the kinematic equations of motion to determine the horizontal distance traveled by the shingle. The relevant equation is:
d = v * t
where d is the distance, v is the initial horizontal velocity, and t is the time of flight.
To find the time of flight, we can use the equation for the vertical displacement of an object under constant acceleration:
y = v0t + (1/2)at^2
where y is the vertical displacement, v0 is the initial vertical velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight. Solving for t, we get:
t = sqrt((2y)/a)
where sqrt means square root.
Substituting the given values, we have:
y = 9.4 m
a = -9.8 m/s^2
t = sqrt((2*9.4 m) / -9.8 m/s^2) = 1.45 s (using the positive root since time cannot be negative)
Now, we can use the horizontal velocity to find the distance traveled in this time:
d = v * t = 7.2 m/s * 1.45 s = 10.44 m
Therefore, the shingle moves a horizontal distance of 10.44 meters in this time.
A loudspeaker of mass 15.0 kg is suspended a distance of h = 1.00 m below the ceiling by two cables that make equal angles with the ceiling. Each cable has a length of l = 2.70 m .
1. What is the tension T in each of the cables?
Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Answer:
To solve the problem, we use the equations of equilibrium to find the tension in each cable holding up the loudspeaker. Since the loudspeaker is in equilibrium, the sum of the forces acting on it is zero. The weight of the loudspeaker is calculated first, and then we use trigonometry to find the horizontal and vertical components of the tension in one of the cables. We then apply the equation of equilibrium in the y direction to find the tension in each cable. The final answer is that the tension in each cable is approximately 81.1 N, which balances the weight of the loudspeaker.
___
Given:
Mass of loudspeaker (m) = 15.0 kg
Distance from ceiling (h) = 1.00 m
Length of cable (l) = 2.70 m
Acceleration due to gravity (g) = 9.80 m/s^2
Weight of loudspeaker:
Fg = mg
Fg = (15.0 kg)(9.80 m/s^2)
Fg = 147 N
Horizontal and vertical components of tension in one cable:
sinθ = h/l
sinθ = 1.00 m / 2.70 m
θ = sin^(-1)(1.00/2.70)
θ ≈ 21.6°
T_x = T sinθ
T_y = T cosθ
where T is the tension in the cable.
Equation of equilibrium in y direction:
ΣF_y = 2T cosθ - Fg = 0
Solving for T:
2T cosθ = Fg
T = Fg / (2 cosθ)
Plugging in the values:
T = (147 N) / (2 cos(21.6°))
T ≈ 81.1 N
13 Which of these fitness events happened LAST? OA. PE becomes part of American school curriculum. OB. Title IX forbids gender discrimination in sports. O C. The YMCA opens a gym in America. O D. The adjustable plate-loaded barbell is invented.
The answer is A. PE becomes part of American school curriculum.
Title IX, which forbids gender discrimination in sports, was enacted in 1972.
The YMCA opened its first gym in America in Boston in 1851.
The adjustable plate-loaded barbell was invented in the late 1940s by a man named George Snyder.
Physical Education (PE) has been a part of American school curriculums for well over a century, with the first school gymnasiums being built in the 1800s.
What is an american school curriculum?
The American school curriculum is the set of educational standards and guidelines used by schools in the United States to structure their academic programs. The curriculum typically includes courses in core academic subjects such as English, math, science, and social studies, as well as elective courses in areas such as the arts, foreign languages, and physical education.
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A 0.5kg wooden block is placed on top of a 1.0kgwooden block. The coefficient static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20 wht is the maximum horizontal force that can be applied to the lower block
A block of 0.5 kg is placed on top of another wooden block which weighs 1.0 kg. The coefficient of static friction between the two blocks is 0.35, whereas the coefficient of kinetic friction between the lower block and the level table is 0.20.
To calculate the maximum horizontal force that can be applied to the lower block, we need to determine the limiting frictional force between the two blocks.
Since the upper block is not moving, the force of static friction is acting on it. We can calculate this force as follows:
`F_static = friction coefficient * normal force`
where, normal force = weight of upper block = 0.5 kg * 9.81 m/s^2 = 4.905 N
`F_static = 0.35 * 4.905 = 1.718 N`
Therefore, the static frictional force acting on the upper block is 1.718 N.
Now, we need to find the maximum force that can be applied to the lower block before it starts moving. This force is equal to the force of static friction acting on the lower block.
Since the upper block is not moving, the force of static friction acting on the lower block is equal to the force of static friction acting on the upper block.
`F_static(lower block) = F_static(upper block) = 1.718 N`
This means that the maximum horizontal force that can be applied to the lower block is 1.718 N.
However, if the applied force exceeds this value, the lower block will start moving and the force of kinetic friction will be acting on it, which is equal to:
`F_kinetic = friction coefficient * normal force`
`F_kinetic = 0.20 * 4.905 = 0.981 N`
Hence, if the applied force exceeds 1.718 N, the lower block will start moving and the force of kinetic friction will act on it, which is 0.981 N.
Therefore, the maximum horizontal force that can be applied to the lower block is 1.718 N.
Answer:
Explanation:
To determine the maximum horizontal force that can be applied to the lower block without causing the blocks to move, we need to calculate the maximum static friction force between the two blocks. This force is given by:
F_friction = coefficient of static friction * normal force
where the normal force is the force perpendicular to the surface of contact between the blocks. Since the blocks are resting on a level table, the normal force acting on the lower block is equal to the weight of both blocks, which is:
N = (m1 + m2) * g
where m1 is the mass of the lower block, m2 is the mass of the upper block, and g is the acceleration due to gravity (9.81 m/s^2).
Plugging in the given values, we have:
N = (1.0 kg + 0.5 kg) * 9.81 m/s^2 = 14.715 N
The maximum static friction force is then:
F_friction = 0.35 * 14.715 N = 5.15025 N
Therefore, the maximum horizontal force that can be applied to the lower block without causing the blocks to move is 5.15025 N. If a greater force is applied, the blocks will start to move and the kinetic friction force will take effect, which is given by:
F_kinetic = coefficient of kinetic friction * normal force
where the coefficient of kinetic friction is 0.20 in this case.
When a disrupted part of a wetland ecosystem is left alone so that nature can help restore it to what it once was, what are people counting on occurring? explain..
Answer: When a disrupted part of the ecosystem is left alone so that nature can help restore itself what people are counting on happening is secondary succession
Explanation:
Is this statement is correct ?? According to Newton's 4rd law, Action and reaction never start from the same point.
help me...
Explanation:
I'm sorry, but the statement you provided is incorrect. There is no such thing as Newton's 4th law. Newton's laws of motion consist of three laws, which are:
An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
The acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass.
For every action, there is an equal and opposite reaction.
None of these laws state that action and reaction never start from the same point. However, it is true that the action and reaction forces act on different objects, not necessarily at the same point. This is because Newton's third law states that every action has an equal and opposite reaction, which means that when one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.
A rocket takes off from a space station, where there is no gravity other than the negligible gravity due to the space station, and reaches a speed of 110 m/s in 10.0 s. If the exhaust speed is 1,600 m/s and the mass of fuel burned is 118 kg, what was the initial mass (in kg) of the rocket?
The initial mass of the rocket was 106 kg.
What is the initial mass of the rocket?We can use the principle of conservation of momentum to solve this problem.
The momentum of the rocket before takeoff is zero, since it is at rest, and the momentum after takeoff is the product of the mass of the rocket and its velocity.
However, during the takeoff, the rocket ejects a mass of fuel at a certain velocity, which creates a backward force (thrust) that propels the rocket forward.
This thrust can be calculated using the equation:
Thrust = (mass flow rate) x (exhaust velocity)
mass flow rate = (mass of fuel burned) / (burn time)
The mass of the rocket at any given time can be calculated using the equation:
mass = (initial mass) - (mass of fuel burned)
Using these equations, we can solve for the initial mass of the rocket:
Calculate the thrust:
Thrust = (118 kg / 10.0 s) x 1600 m/s = 1,888 N
Calculate the mass of the rocket at the end of the burn:
mass(end) = (initial mass) - (mass of fuel burned) = (initial mass) - 118 kg
Use the principle of conservation of momentum to find the initial mass:
momentum before = momentum after
0 = (mass(end) + 118 kg) x 110 m/s
mass(end) = -118 kg / 110 m/s = -1.07 kg/s
mass(end) = (initial mass) - 118 kg
(initial mass) = mass(end) + 118 kg
(initial mass) = (-1.07 kg/s x 10.0 s) + 118 kg
(initial mass) = 106 kg
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