Answer:
b
Explanation:
when a ball falls from the sky, it falls down fast due to no friction.
Pest describes the act of using senses or tools to gather information?
ating a hypothesis
king an observation
mmarizing the results
ording the measurements
Answer:
Making an observation
Explanation:
The use of senses as a tool to gather information is described as making an observation.
While making an observation, the senses must be at alert.
Observation making is paramount to the scientific method. It is from observations that questions are asked and then hypothesis which can be tested are formulated. Observation can be carried out using the eyes, nose, feeling e.t.c. Nowadays, observation can also be carried out using some sophisticated equipment in the laboratory. This necessary for phenomenon the eludes our natural senses. A scientist must be a keen observer and their senses must be sharp.Therefore, the act of using senses or tools to gather information is called making an observation.
is a guideline to help an individual write and achieve well-specified goals.
Answer:
is a guideline to help an individual write and achieve well-specified goals.
Explanation:
An action plan is a guideline to help an individual write and achieve well-specified goals.
Answer:
yes, the guideline is to help an individual write and achieve well-specified goals.
Explanation:
A hockey player whacks a 162-g puck with her stick, applying a 171-N force that accelerates it to 42.3 m/s. A. If the puck was initially at rest, for how much time did the acceleration last? B. The puck then hits the curved corner boards, which exert a 151-N force on the puck to keep it in its circular path. What’s the radius of the curve?
Given parameters:
Mass of puck = 162g = 0.162kg (1000g = 1kg)
Force exerted on puck = 171N
Final velocity = 42.3m/s
Unknown
A. time of the acceleration
B. radius of the curve?
Solution:
A. time of the acceleration
the initial velocity of the puck = 0m/s
We know that;
Force = mass x acceleration
Acceleration = [tex]\frac{Final velocity - Initial velocity}{time taken}[/tex]
Acceleration = [tex]\frac{42.3 - 0}{t}[/tex]
So force = mass x [tex]\frac{42.3 }{t}[/tex]
Input the parameters and solve for time;
171 = 0.162 x [tex]\frac{42.3 }{t}[/tex]
171 = [tex]\frac{6.85}{t}[/tex]
t = [tex]\frac{6.85}{171}[/tex] = 0.04s
The time of acceleration is 0.04s
B. radius of the curve;
to solve this, we apply the centripetal force formula;
F = [tex]\frac{mv^{2} }{r}[/tex]
where;
F is the centripetal force
m is the mass
v is the velocity
r is the radius
Since the force exerted on the puck is 151;
input the parameters and solve for r²;
151 = [tex]\frac{0.162 x 42.3^{2} }{r}[/tex]
151r = 0.162 x 42.3²
r = 1.92m
The radius of the circular curve is 1.92m
procedure for determining the thermal conductivity of a solid involves embedding thermocouple in a thick slab of the material and measuring the response to a prescribed change in temperature at one surface. Consider an arrangement for which the thermocouple is embedded 10 mm from a surface that is suddenly brought to a temperature of 100degreeC by exposure to boiling water. If the initial temperature of the slab was 30degreeC and the thermocouple measures a temperature of 65degreeC, 2 minutes after the surface is brought to 100degreeC, what is the thermal conductivity. The density of the material is 2200 kg/m3 and the specific heat is 700 J/M- Find: What is the thermal conductivity of the material
Answer:
The thermal conductivity [tex]k = 1.4094 W/ m\cdot K[/tex]
Explanation:
From the question we are told that
The depth of the thermocouple from the surface is x = 10 mm = 0.01 m
The temperature is [tex]T_f = 100 ^o C[/tex]
The initial temperature is [tex]T_i = 30 ^o C[/tex]
The temperature of the thermocouple after t = 2 minutes( 2 * 60 = 120 \ seconds) is [tex]T_t = 65 ^o C[/tex]
The density of the material is [tex]\rho = 2200 kg/m^3[/tex]
The specific heat of the solid [tex]c_s = 700 J/kg \cdot K[/tex]
Generally the equation for semi -infinite medium is mathematically as
[tex]\frac{T_s - T }{T_i - T} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]\frac{65 - 100 }{30 - 100} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]0.5 = erf [\frac{0.01}{2 \sqrt{\alpha * 120} } ][/tex]
Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]
[tex]\frac{0.01}{ 2 (\sqrt{\alpha * 120 } )}[/tex] this is from the Gaussian function table
[tex]0.0 1 = 0.954 * (\sqrt{\alpha * 120 } )[/tex]
=> [tex]\sqrt{\alpha * 120 } = \frac{0.01 }{0.954 }[/tex]
=> [tex]\alpha = 9.1525 *10^{-7} \ m^2 /s[/tex]
Generally the thermal conductivity is mathematically represented as
[tex]k = \alpha * \rho * c_s[/tex]
[tex]k = 9.1525 *10^{-7} * 2200 * 700[/tex]
[tex]k = 1.4094 W/ m\cdot K[/tex]
What happens to the oceans tides if the ocean if the Earth spins slower?
Answer:
As the earth rotates, it tries to drag/bring the tidal bulges with it. When a large amount of friction is applied, the earth spin will gradually and slow down but not all the way down.
Answer:
If the Earth spun slower the rate of tides will be higher because the moon will start revolving faster tan the Earth, creating more tides as the moon will revolve more around the Earth in a month.
If my answer helped, kindly mark me as the brainliest !!
Thank You!!
A mass of 15 kg is resting on a horizontal, frictionless surface. Force 1 of 206 N is applied to it at some angle above the horizontal, force 2 has a magnitude of 144 N and is applied vertically downward, force 3 has a magnitude of 5 N and is applied vertically upwards, and force 4 has a magnitude of 42 N and is applied in the -x direction to the object. When these forces are applied to the object, the object is moving at 20 m/s in the x direction in a time of 3 seconds. What is the normal force acting on the mass in Newtons
Answer:
N = 136.77 N
Explanation:
This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.
Let's use trigonometry to decompose the force F1
cos θ = F₁ₓ / F₁
sin θ = F_{1y} / F₁
F₁ₓ = F₁ cos θ
F_{1y} = F₁ sin θ
now let's apply Newton's second law to each axis
X axis
F₁ₓ - F4 = m a
Y axis
N + F3 + F_{1y} -F₂ -W = 0
the acceleration can be calculated with kinematics
v = v₀ + a t
since the object starts from rest, the initial velocity is zero v₀ = 0
a = v / t
a = 20/3
a = 6.667 m / s²
we substitute in the equation
F₁ₓ = F₄ + m a
F₁ₓ = 42 + 15 6,667
F₁ₓ = 142 N
F₁ cos θ = 142
cos θ = 142/206 = 0.6893
θ = cos⁻¹ 0.6893
θ = 46.42º
now let's work the y axis
N = W + F₂ - F₃ - F_{1y}
N = 15 9.8 + 144 -5 - 206 sin 46.42
N = 286 - 149.23
N = 136.77 N
Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 90 m/s, what the magnitude of bigger vector?
Answer:
Vy = 80.5 [m/s]
Explanation:
In order to solve this problem we must use the Pythagorean theorem.
V = 90 [m/s]
The components are Vx and Vy:
Therefore:
[tex]v=\sqrt{v_{x}^{2} + v_{y}^{2} }[/tex]
where:
Vy = 2*Vx ; because one is twice of the other.
[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]
and the bigger vector is:
Vy = 40.25*2
Vy = 80.5 [m/s]
A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete ( k = 0.79 W/m⋅K, α = 5.94 × 10 −7 m2/s, rho = 1600kg/ m 3 , and c p = 0.84kJ/kg⋅K ) cooled to 14° C during a cold night is heated again during the day by being exposed to ambient air at an average temperature of 28° C with an aver-age heat transfer coefficient of 14 W/ m 2 ⋅K. Using the analyti-cal one-term approximation method, determine (a) how long it will take for the column surface temperature to rise to 27° C, (b) the amount of heat transfer until the center temperature reaches to 28° C, and (c) the amount of heat transfer until the surface temperature reaches 27° C.
Answer:
a) Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b) Amount of heat transfer is 5320 kJ
c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
Explanation:
Given that;
Diameter D = 30 cm
Height H = 4m
heat transfer coeff h = 14 W/m².°C
thermal conductivity k = 0.79 W/m.°C
thermal diffusivity α = 5.94 × 10⁻⁷ m²/s
Density p = 1600 kh/m³
specific heat Cp = 0.84 Kj/kg.°C
a)
the Biot number is
Bi = hr₀ / k
we substitute
Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C
Bi = 2.658
From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,
λ₁ = 1.7240
A₁ = 1.3915
Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁
the Fourier number is determined to be
[ T(r₀, t) -T∞ ] / [ Ti - T∞] = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)
(27 - 28) / (14 - 28) = (1.3915)e^-(17240)²t (0.3841)
t' = 0.6771
Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes
t = t'r₀² / ₐ
= (0.6771 × 0.15 m)² / (5.94 x 10⁻⁷ m²/s)
= 23,650 s
= 7.1 hours
Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b)
The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.
Maximum heat transfer between the ambient air and the column is
m = pV
= pπr₀²L
= (1600 kg/m³ × π × (0.15 m)² × (4 m)
= 452.389 kg
Qin = mCp [T∞ - Ti ]
= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C
= 5320 kJ
Amount of heat transfer is 5320 kJ
(c)
the amount of heat transfer until the surface temperature reaches to 27°C is
(T(0,t) - T∞) / Ti - T∞ = A₁e^(-λ₁²t')
= (1.3915)e^-(1.7240)² (0.6771)
= 0.1860
Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes
(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)
= 1 - 2 × 0.1860 × (0.5787 / 1.7240)
= 0.875
Q = 0.875Qmax
Q = 0.875(5320 kJ)
Q = 4660 kJ
Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
Sand is made of different types of rocks. The shapes and sizes are not all the same. Sand is a ----?
Answer:
A soil
Explanation:
Sand is a granular material composed of finely divided rock and mineral particles. It is defined by size, being finer than gravel and coarser than silt. Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.
Answer:
mixture
sand is a mixture
A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.81 g, what is the tension in the web?
N=
Answer:
The tension in the web is 0.017738 N
Explanation:
Net Force
The net force exerted on an object is the sum of the vectors of each individual force applied to an object.
If the net force equals 0, then the object is at rest or moving at a constant speed.
The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.
A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:
T=W=m.g
The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:
[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]
[tex]T=0.017738\ N[/tex]
The tension in the web is 0.017738 N
Which of the following elements are most likely to have similar chemical properties?
Answer:
transition metal, and inner transition metals groups are numbered 1-18 from left to right
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r
(A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E
Answer:
E ’= E / 8
therefore the correct answer is A
Explanation:
Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface
Фi = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
E (4πr²) = q_{int} / ε₀
density is
ρ = q_{int} / V
q_{int} = ρ V = ρ 4/3 π r³
we substitute
E (4π r²) = ρ 4/3 π r³ /ε₀
E = 1 /3ε₀ ρ r
let's change the density by
ρ = Q / V = Q / (4/3 π R³)
E = 1 / 4πε₀ Q r / R³
if we now distribute the same charge on a sphere of radius R' = 2R
E ’= 1 / 4pieo Q r / (2R)³
E ’= 1 / 4ft Qr / R³ ⅛
E ’= E / 8
therefore the correct answer is A
Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of this circle with the symbol for “initial position”. Move the object 5.0cm to the right and stop. Label this circle with the correct symbol for “final position.”

(A) What was the initial position of the object?
(B) What is the final position of the object?
(C) What is the distance traveled by the object?
(D) What is the displacement of the object?
(E) Of the three underlined quantities, which are numerically equal?
What is the displacement from the forest to the doctor’s office?
Look up "Everything You Need To Know About Math In One Big Fat Notebook pdf." It's the best thing I've ever been given, I have it with me in math class all the time and I've aced every test. I have it with me right now and it has everything I've ever been taught about math in it so it might help you.
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside.
(a) Under this assumption, show that the cost of operating an air con- ditioner is proportional to the square of the temperature difference.
(b) Give a numerical example for a typical house and discuss implications for your electric bill.
(c) Suppose instead that heat enters the building at a rate proportional to the square-root of the temperature difference between inside and outside. How would the operating cost now depend on the temperature difference?
Answer:
Considering first question
Generally the coefficient of performance of the air condition is mathematically represented as
[tex]COP = \frac{T_i}{T_o - T_i}[/tex]
Here [tex]T_i[/tex] is the inside temperature
while [tex]T_o[/tex] is the outside temperature
What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity
So it implies that the air condition removes [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity
Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as
[tex]Q \ \alpha \ (T_o - T_i)[/tex]
=> [tex]Q= k (T_o - T_i)[/tex]
Here k is the constant of proportionality
So
since 1 unit of electricity removes [tex] \frac{T_i}{T_o - T_i}[/tex] amount of heat
E unit of electricity will remove [tex]Q= k (T_o - T_i)[/tex]
So
[tex]E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]
=> [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]
given that [tex]\frac{k}{T_i}[/tex] is constant
=> [tex]E \ \alpha \ (T_o - T_i)^2[/tex]
From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.
Considering the second question
Assuming that [tex]T_i = 30 ^oC[/tex]
and [tex]T_o = 40 ^oC[/tex]
Hence
[tex]E = K (T_o - T_i)^2[/tex]
Here K stand for a constant
So
[tex]E = K (40 - 30)^2[/tex]
=> [tex]E = 100K [/tex]
Now if the [tex]T_i = 20 ^oC[/tex]
Then
[tex]E = K (40 - 20)^2[/tex]
=> [tex]E = 400 \ K[/tex]
So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher
Considering the third question
Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside
We have that
[tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]
So
[tex]E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]
=> [tex]E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]
Assuming [tex]\frac{k}{T_i}[/tex] is a constant
Then
[tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]
From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.
What is the acceleration of a ball with a mass of 0.40 kg is hit with a force of 16N?
Answer:
40 m/s^2
Explanation:
Mass= 0.40 kg
Force= 16 N
Therefore the acceleration can be calculated as follows
F = ma
16= 0.40 × a
16= 0.40 a
a= 16/0.40
a= 40 m/s^2
Hence the acceleration is 40 m/s^2
The perception of an image first, followed by noticing individual pieces of the
image, can be described as:
A. sensation.
B. perceptual processing.
C. top-down processing.
D. bottom-up processing.
SUBMIT
Answer:
The answer is Top-Down processing
Explanation:
I had this question on a apex quiz and i got it correct.
What is the momentum of a 20.0 kg scooter traveling at 5.00 m/s?
Answer:
The answer is 100 kgm/sExplanation:
To find the momentum of an object given it's mass and velocity we use the formula
momentum = mass × velocityFrom the question
mass = 20 kg
velocity = 5 m/s
We have
momentum = 20 × 5
We have the final answer as
100 kgm/sHope this helps you
a torque of 100Nm is required to open a door. WHAT IS the minimum distance of the handle fromt he hinge. if the door is to be pulled open wth a force at handle not greater than 50N?
Answer:
At least [tex]2\; \rm m[/tex].
Explanation:
The torque [tex]\tau[/tex] that a force exerts on a lever is equal the product of the following:
[tex]F[/tex], the size of that force,[tex]r[/tex], the distance between the fulcrum and the point where that force is applied, and[tex]\sin\theta[/tex], the sine of the angle between the force and the lever.[tex]\tau = F\cdot r \cdot \sin\theta[/tex].
The force in this question is (at most) [tex]50\; \rm N[/tex]. That is: [tex]F = 50\; \rm N[/tex].
[tex]\sin \theta[/tex] is maximized when [tex]\theta = 90^\circ[/tex]. In other words, the force on the door gives the largest-possible torque when that force is applied perpendicular to the door. When [tex]\theta = 90^\circ\![/tex], [tex]\sin \theta =1[/tex].
If the force here is applied at a distance of [tex]r[/tex] meters away from the hinge (the fulcrum of this door,) the torque generated would be:
[tex]\begin{aligned}\tau &= F \cdot r \cdot \sin \theta \\ &= (50\, r)\; \rm N \cdot m\end{aligned}[/tex].
That torque is supposed to be at least [tex]100\; \rm N\cdot m[/tex]. That is:
[tex]50\, r \ge 100[/tex].
[tex]r \ge 2[/tex].
In other words, the force needs to be applied at a point a minimum distance of [tex]2\; \rm m[/tex] away from the hinge of this door.
This glass of lemonade is sitting in the hot summer sun. As time passes, in which direction will heat transfer take place?
The heat transfer takes place from the Ice to lemonade (ice → lemonade).
What is heat transfer?The term “heat transfer” refers to the movement of heat. The flow of heat across a system's boundary is due to a temperature differential between the system and its surroundings.
When a temperature difference exists between states of matter, heat transfer happens solely in the direction of decreasing temperature, that is, from a hot object to a cold item.
The temperature of ice is increasing while the lemonade is decreasing. Heat transfer happens solely in the direction of decreasing temperature,
Hence, the heat transfer takes place from the Ice to lemonade (ice → lemonade.
To learn more about the heat transfer, refer to the link;
https://brainly.com/question/13433948
#SPJ2
Answer: C
Explanation:
When you shock your self because of a door knob this is an example of
A.the law of conservation of charge
B. Force field
C.static discharge
D.torture lol
Answer:
C
Explanation:
Static electricity is what makes your hair stand up when you rub a balloon against it or gives you a shock from your doorknob. In static electricity, electrons are moved around mechanically (i.e. by someone rubbing two things together).
HOPE THIS HELPED AND LETTER D MADE ME LAUGH
A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff 132 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.) J (b) Suppose the projectile is traveling 89.3 m/s at its maximum height of y = 297 m. How much work has been done on the projectile by air friction? J (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
Answer:
a) The initial total mechanical energy of the projectile is 498556.296 joules.
b) The work done on the projectile by air friction is 125960.4 joules.
c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
Explanation:
a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:
[tex]E = U_{g} + K[/tex] (Eq. 1)
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.
[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.
If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:
[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 498556.296\,J[/tex]
The initial total mechanical energy of the projectile is 498556.296 joules.
b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:
[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)
Where:
[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.
[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]
[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.
[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.
If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:
[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]
[tex]W_{loss} = 125960.4\,J[/tex]
The work done on the projectile by air friction is 125960.4 joules.
c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:
[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)
[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]
Where:
[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.
[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.
[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)
[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]
[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]
[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]
If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:
[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]
[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]
The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
Find the velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south.
The velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south would be 10.35 meters / seconds.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
Note that these equations are only valid for a uniform acceleration.
As given in the problem we have to find the velocity of the car we have to find the velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south,
The acceleration of the car = 1.5 m/s²
The time taken by the car = 6.9 seconds
By using the first equation of the motion,
v = u + at
v = 0 + 1.5*6.9
v = 10.35 meters / seconds
Thus, the velocity of the car after 6.9 s, if its acceleration is 1.5 m/s² due south, would be 10.35 meters / seconds.
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Chen is testing the friction of three surfaces. He pushes the same ball across three different surfaces with the same force and measures the distance the ball rolls over each surface. The ball moved ten inches across Surface 1, six inches across Surface 2, and fifteen inches across Surface 3. Which could most likely describe the three surfaces?
Answer:Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.
Explanation:
Answer: its C
Explanation:
100 POINTS HELP!!!!!!!!!!!!!
Answer:
it is whatever the temperature is at 5(I cant seem to see it clearly)
Explanation:
Ray runs 78 feet north, then 61 feet west. Calculate the total displacement traveled by in feet
Answer:
hdhshsisjsbrtheisebvrtctvsjusyevevrvrg eggs haushehegehs
11. What is the momentum of a truck with a mass of 3000 kg traveling north at a
speed of 25 m/s?
Answer:
We are given:
mass of the truck (m) = 3000 kg
velocity of the truck (v) = 25 m/s
Calculating the Momentum:
We know that the momentum of an object is equal to its mass times it's velocity
P = mv (where P is the momentum of the truck)
Momentum of the truck:
replacing with the values of the truck
P = mv
P = 3000 * 25
P = 75000 kg m/s
Therefore, the momentum of the truck is 75000 kg m/s
Why after a certain time bouncing a ball does it stop?
Answer:
If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. ... Some energy is dissipated in the compression and decompression phases.
Explanation:
what’s the best answer , a , b , c or d
Answer:
The Basic Concept
When we look at an object at rest, we say that it is in equilibrium since all the forces being applied on it, cancel out
now, if one of the force was slightly more than the other one in the same case. The object will start to move and NOT be in equilibrium
BUT
if an object is in a vacuum and moving on a frictionless surface, the object will attain equilibrium after some seconds since it will be moving with constant speed and all the forces acting on it will be equal
Hence, if the object is accelerating. we can say with surety that the object is not is equilibrium since from the second law of motion,
F = ma ; when a is a non-zero value, there is definitely some net force being applied on the object
Looking at the given case
in the question, we are given that the object is 'accelerating' upwards
we proved above that if an object is accelerating, there is some net force on that object and hence the object is NOT in equilibrium
Since the object is accelerating, from the second law of motion:
F = ma; m cannot be zero and if a is a non-zero value as given in the question, there is definitely some net force on the object
Since there is some force being applied on the object, the object is NOT in equilibrium
Conclusion
Since we found that the object is NOT in equilibrium and that there is some net force on the object,
The first option is correct
1. What happens to the current in a series circuit as it moves through each component? a. The current stays the same throughout the circuit.
b. The current will increase or decrease depending on the resistance.
c. The current decreases with each component it goes through.
d. The current increases with each component it goes through.
2. A 10-volt power supply is placed in series with two 5-ohm resistors. What is the current in the circuit after it passes through each of the two resistors?(1 point)
a. The current will stay the same at 1 amp after passing through both resistors.
b. The current will drop to 2 amps after the first resistor and then to 1 amp after the second resistor.
c. The current will stay the same at 2 amps after passing through both resistors.
d. The current will drop to 1 amp after the first resistor and then to 0 amps after the second resistor.
3. What is the voltage that passes through R1 and R2?
a. R1: 12 V, R2: 24 V
b. R1: 8 V, R2: 4 V
c. R1: 12 V, R2: 12 V
d. R1: 6 V, R2: 6 V
4. Which of the following correctly describes the magnitude of currents I1 and I2 ?
a. I1 is equal to I2
b. I1 and I2 approach zero
c. I1 is greater than I2
d. I1 is less than I2
5. If the energy of an electric charge flowing in a circuit is conserved, which of the following obeys the Kirchhoff junction rule?
a. The sum of the current flowing in is greater than the sum of current flowing out.
b. The sum of the current flowing in is less than the sum of the current flowing out.
c. The sum of the current flowing in is equal to the sum of current flowing out.
d. The sum of the current flowing in is zero and the sum of the current flowing out is greater than zero.
Answer: sorry here’s the answers, I didn’t feel like typing it all
Explanation:
The correct answer to the 5 questions are;
1) Option A; The current stays the same throughout the circuit.
2) Option A; The current will stay the same at 1 amp after passing through both resistors.
3) Option C; R1: 12 V, R2: 12 V
4) Option A; I1 is equal to I2
5) Option C; The sum of the current flowing in is equal to the sum of current flowing out.
1) In an electrical circuit, usually as current moves through each component, it stays the same. Thus, option A is correct.
2) Formula for current is;
I = V/R
We are told that there are two resistors in series each having a resistance of 5Ω. Thus; Total resistance = 5 + 5 = 10 Ω.
Thus; Current = 10/10 = 1 A.
The current will stay same at 1 A after passing through both resistors.
3) From the circuit we are given, we see that the Voltage is 12 V. Now, the same voltage would be transmitted through both resistor R1 and R2.
Option C is correct
4) The current splits upon passing resistor 1 and as such it means the current I2 going through the second resistor would be the same. Thus; I1 = I2.
5) Kirchoff's junction rule states that all the incoming currents to a particular junction must be equal to sum of all currents going out of that same junction. Thus, option C is correct.
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