Answer:
20 * .3 = 6N
Explanation:
The impulse on the ball kicked with a force of 20 N force 0.3 s is 6 Ns
From the question given above, the following data were obtained:
Force = 20 N
Time = 0.3 s
Impulse = ?The impulse on the ball can be obtained as follow:
[tex]impulse \: = force \: \times time \\ impulse \: = 20 \: \times \: 0.3 \\[/tex]
Impulse = 6 NsTherefore, the impulse on the ball is 6 Ns
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The spacecraft was moved closer to the launcher by the same amount Wednesday as it was on Tuesday, and yet its speed went up much more. Claim 3 suggests that the magnetic force was much stronger on Wednesday than on Tuesday.
Consider the two subclaims for Claim 3 and answer the question below.
Claim 3.A: The magnetic force was much stronger on Wednesday because the magnet was stronger.
Claim 3.B: The magnetic force was much stronger on Wednesday because the magnetic force is stronger closer to the magnets.
Which claim do you think is more convincing, and why?
Answer:
The magnetic force was much stronger on Wednesday because the magnetic force is stronger closer to the magnets.
Explanation:
The magnetic force was much stronger on Wednesday than on Tuesday because the magnetic force is stronger closer to the magnets. Therefore claim 3B is more convincing.
What is magnetic force?Magnetic force can be described as a consequence of electromagnetic force which is caused due to the motion of charges. A moving charge surrounds itself with a magnetic field and the force that arises due to interacting magnetic fields.
The magnetic force between two moving charges is the effect exerted upon either charge by a magnetic field generated by the other. The magnetic force depends on the charge, the motion of each of the objects, and the separation between them.
The magnitude of the force is determined by the cross product of velocity and the magnetic field is equal to q.[v × B]. The resultant force can be described as perpendicular to the direction of the velocity and the magnetic field.
Therefore, the magnetic force was much stronger closer to the magnets.
Learn more about magnetic force, here:
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In a thunderstorm at 32.0°C, Reginald sees a bolt of lightning and hears the thunderclap 2.00s later. How far from Reginald did the lightning strike? The speed of sound through air at 32.0°C is 350.2 m/s. Show your work.
PLEASE HELP, THANKS!
Answer:
d = 700.4 m
Explanation:
Given that,
The speed of sound through air at 32.0°C is 350.2 m/s.
Reginald sees a bolt of lightning and hears the thunderclap 2.00s later.
We need to find how far from Reginald did the lightning strike. Let the distance be d. So,
Speed = distance/time
d = vt
So,
d = 350.2× 2
d = 700.4 m
So, the required distance is 700.4 m.
HELP ME HOW ARE BLACK HOLES FROMED
Answer:
Stellar black holes form when the center of a very massive star collapses in upon itself.
Answer:
A black hole can be formed by the death of a massive star. When such a star has exhausted the internal thermonuclear fuels in its core at the end of its life, the core becomes unstable and gravitationally collapses inward upon itself, and the star's outer layers are blown away.
Explanation:
HELP PLEASE
60 POINTS
HAVE A GREAT REST OF YOUR DAY PEOPLE :>
Answer:
The sun and the stars
Explanation:
I hope this helps!
Answer:
Northstar & North Pole
Explanation:
yyggggggggggggggg
An instructor wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition with two tiny slits separated by 0.180 mm. An interference pattern appears on a screen that lies 5.30 m from the slit pair. The instructor's measurements show that two adjacent bright interference fringes lie 1.60 cm apart on the screen. What is the laser's wavelength (in nm) ?
Answer:
λ = 5.434 x 10⁻⁷ m = 543.4 nm
Explanation:
To solve this problem we can use the formula provided by Young's Double Slit experiment:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta xd}{L}[/tex]
where,
λ = wavelength of light = ?
Δx = distance between adjacent bright fringes = 1.6 cm = 0.016 m
d = slit separation = 0.18 mm = 0.00018 m
L = Distance between slits and screen = 5.3 m
Therefore,
[tex]\lambda = \frac{(0.016\ m)(0.00018\ m)}{5.3\ m}[/tex]
λ = 5.434 x 10⁻⁷ m = 543.4 nm
A roller-coaster car has a mass of 1240 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 22 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.7 m/s? (b) What is FN if v = 20 m/s? Use g=9.81 m/s2.
Answer:
a) N = 7.90 10³ N, b) N = -1.04 10⁴ N
Explanation:
a) For this exercise we can use Newton's second law
N -W = m (-a)
The relationship is centripetal, the negative sign of the acceleration is because it points towards the center of the circle
a = v² / r
we substitute
N = mg -m v² /r
N = m (g - v² /r)
let's calculate
v = 8.7 m / s
N = 1240 (9.81 - 8.7²/22)
N = 7.90 10³ N
b) v = 20 m / s
N = 1240 (9.81 - 20²/22)
N = -1.04 10⁴ N
When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?
Answer:
1) v= 90km/h d = 70 m, 2) x₁ = v t_r, x₁ = 6.25 m, 3) x₁=6.25 no change
4) x = 22 m
Explanation:
1) for the first part, you are asked to find the minimum safety distance with the vehicle in front
The internet is searched for the stopping distance for two typical speeds on the highway
v (km/ h) v (m/s) d (m)
90 25 70
100 27.78 84
the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement
v = x / t_r
x₁ = v t_r
the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus
therefore the safe distance is
x_total = x₁ + d
2) The distance is the sum of the distance traveled in the reaction
x₁ = v t_r
for v = 90 km / h
x₁ = 25 0.25
x₁ = 6.25 m
for v = 100 km / h
x₁ = 27.78 0.25
x₁ = 6.95 m
the total distance is
x_total = x₁ + d
for v = 90 km / h
x_total = 25 0.25 + 70
x_total = 76.25 m
this is the distance until the cars stop and do not collide
3) the stopping distance of a truck is
v = 90 km / h d = 100 m
in this case we see that the braking distance is much higher,
the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change
4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s
for the car case at v = 90km / h = 25 m / s
according to this rule we must this to
x = 25 / 4.47 = 5.6 cars
each modern car is about 4 m long so the distance is
x = 22 m
we see that this distance is much greater than the reaction distance so it does not make much sense
a guitar string transmits waves at 509 m/s. to create a 491 Hz note, to what length should the guitar player shorten the string?
Answer:
Wavelength = 1.04 meters
Explanation:
Given the following data;
Speed = 509m/s
Frequency = 491Hz
To find the wavelength;
Wavelength = speed/frequency
Wavelength = 509/491
Wavelength = 1.04 meters
Therefore, the guitar player should shorten the length of the string to 1.04 meters.
Answer: 0.52
Explanation:
A box with mass 1.10 kg is located on a horizontal tabletop with friction. The coefficient of kinetic friction is 0.500. The tabletop is square and measures 1.00 m on its side. The box starts at one corner and finishes at the diagonal edge. The path it follows is by first traveling one edge, turning and traveling to the final location. You push the box by exerting a force on it that makes an angle of 30.0o with the horizontal. How much work does the friction force do
Answer:
W = 6.5 W
Explanation:
Work is defined by
W = F . d
W = f d cos tea
where the point represents the scaled product and the bold letters indicate vectors
they ask us the work of the friction force
we write the translational equilibrium equation
y Axis
N -W = 0
N = mg
x axis
F - fr = 0
F = fr
the formula for the friction force is
fr = μ N
we substitute
fr = μ m g
we substitute in the equation of work
W = fr d cos θ
W = μ m g d cos θ
let's calculate
W = 0.500 1.10 9.8 Σ d_i cos θ_i
W = 5.39 d cos tea
we have two displacement
the first on one side of the box, suppose that side is on the y-axis, therefore the angle between the displacement and the friction force is 70º
and there is a second displacement in the x axis, in this case the angle between the friction force and the displacement is 30º
therefore the total workload is the sum of those work
W = 5.39 (1 cos 70 + 1 cos 30)
W = 5.39 (0.342 + 0.866)
W = 6.5 W
Two girls are estimating each other's power. One runs up some step
ng each other's power. One runs up some steps, and the other times her. Here are their
results:
height of one step = 20 cm
number of steps = 36
mass of runner = 45 kg
time taken = 4.2 s
a .Calculate the runner's weight. (Acceleration due to gravity g=10m
b .Calculate the increase in the girl's gravitational potential energy as she runs up the steps.
c. Calculate her power. Give your answer in kilowatts (kW).
Answer:
A. 450 N
B. 3240 J
C. 0.77 KW
Explanation:
From the question given above, the following data were obtained:
Height of one step = 20 cm
Number of steps = 36
Mass of runner = 45 kg
Time taken = 4.2 s
Next, we shall convert 20 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
20 cm = 20 cm × 1 m /100 cm
20 cm = 0.2 m
Next, we shall determine the total height. This can be obtained as follow:
Height of one step = 0.2 m
Number of steps = 36
Total height =?
Total height = 36 × 0.2
Total height = 7.2 m
A. Determination of the runner's weight.
Mass of runner (m) = 45 kg
Acceleration due to gravity (g) = 10 m/s²
Weight (W) =?
W = m × g
W = 45 × 10
W = 450 N
B. Determination of the increase in the potential energy.
At the ground level, the potential energy (PE₁) is 0 J.
Next, we shall determine the potential energy at 7.2 m. This can be obtained as follow:
Mass of runner (m) = 45 kg
Acceleration due to gravity (g) = 10 m/s²
Total height (h) = 7.2 m
Potential energy at height 7.2 m (PE₂) = ?
PE₂ = mgh
PE₂ = 45 × 10 × 7.2
PE₂ = 3240 J
Final, we shall determine the increase in potential energy. This can be obtained as follow:
Potential energy at ground (PE₁) = 0 J
Potential energy at height 7.2 m (PE₂) = 3240 J
Increase in potential energy =?
Increase in potential energy = PE₂ – PE₁
Increase in potential energy = 3240 – 0
Increase in potential energy = 3240 J
C. Determination of the power.
Energy (E) = 3240 J
Time (t) = 4.2 s
Power (P) =?
P = E/t
P = 3240 / 4.2
P = 771.43 W
Finally, we shall convert 771.43 W to kilowatt (KW). This can be obtained as follow:
1000 W = 1 KW
Therefore,
771.43 W = 771.43 W × 1 KW / 1000 W
771.43 W = 0.77 KW
Therefore, her power is 0.77 KW
PLEASE HELP WITH THIS ONE QUESTION
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
The period is less, but not halved.
The period is more, but not doubled.
The period is halved.
The period is doubled.
Answer:
The period is more, but not doubled.
Explanation:
Recall that the period of a mass on a spring is T=2πmk.
How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?
thank you
Answer:
Newton's theory identified mass as the factor that causes gravity. On the other hand, Einstein's theory identified the curvature of space-time as the factor that causes gravity.
Answer:
Hey mate...
Explanation:
This is ur answer....
In the 17th century Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's interpretation was that these objects do not fall. According to Einstein, these objects and Earth just freely move in a curved spacetime and this curvature is induced by mass and energy of these objects.
Hope it helps you,
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An object is experiencing an acceleration of 0.4 m/s^2 while traveling in a circle of 35 m. What is it’s velocity?
Answer:
v = 3.74 m/s
Explanation:
Given that,
The acceleration of the object in circular path, a = 0.4 m/s²
The radius of the circle, r = 35 m
We need to find the velocity of the object. The acceleration of an object on the circular path is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{0.4\times35}\\\\v=3.74\ m/s[/tex]
So, the velocity of the object is equal to 3.74 m/s.
The part of the eye that gives its shape is the:
How to calculate this operation? m=10kg and L=2500J/kg? What is the Energy?
I hope you will give me the correct answer because the last answer was too rude. His known by SOD!
Answer:
25000J
Explanation:
Formula : Q = m×c×Δt
Q=Heat energy
m= mass
c=specific heat capacity
ΔT = change in temperature.
Q=M x C for this question.
Specific heat capacity = 2500 J/Kg
Mass = 10kg
2500 x 10 = 25000 J
While sitting in a boat, a fisherman observes that 2 complete waves pass by his position every 4 seconds. What is the period of these waves?A)2s B)8s C) 0.5 s D) 4s
Answer:
2
Explanation:
Because 2 waves 4 secs means 1 in 2s
The period of these waves is 2s. and The right option is A)2s.
The period of a wave is the time taken by a wave to complete one cycle.
The formula of period from the question is given below.
⇒ Formula:
T = t/2.................... Equation 1⇒ Where:
T = Period of the wavet = time taken for two complete oscillationFrom the question,
⇒ Given:
t = 4 seconds.⇒ Substitute these value above into equation 1
T = 4/2T = 2 seconds.Hence, The period of these waves is 2s.
The right option is A)2s
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If your water heat has an efficiency of 95 percent, how much energy would it take to heat 45kg of water from 23 C to 60 C. (The specific heat of water is 4.18 J/g/C.) Please show your work
Answer: 6611.715 joules
Explanation:
Q = MxCxdeltaT = 6959.7 which is 100%
95% = 6611.715
What is the earliest time at which the oscillator shown below is stationary?
Answer:
0.0s
Explanation:
I got it right in acellus
Answer: 0.0
Explanation:
What is the average kinetic energy of particles in a gas at a temperature of 245 Kelvins?
Answer:
2)
Explanation:
Is TV light is converging rays, divergent rays or parallel rays?
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume that the load is not tied down to the truck, but has a coefficient of friction of 0.500 with the flatbed of the truck.
A) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.
B) Is any piece of data unnecessary for the solution?
a) mass of the load.
b) mass of the truck.
c) velocity.
d) coefficient of static friction.
e) all are necessary.
Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load [tex]m_{LS}[/tex] = 10000 kg
mass of flat bed [tex]m_{FB}[/tex] = 20000 kg
initial speed of truck [tex]v_{0}[/tex] = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs [tex]F_{N}[/tex] -------------let this be equation 1
where [tex]F_{N}[/tex] = normal force = mg
so
Fs,max = μs mg
ma[tex]_{max}[/tex] = μs mg
divide through by mass
a[tex]_{max}[/tex] = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²
a[tex]_{max}[/tex] = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx
where [tex]v_{f}[/tex] is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load
A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?
Answer:
1) the required horizontal force F is 1095.6 N
2) W = 0 J { work done by rope will be 0 since tension perpendicular }
3) work is done by the worker is 1029.4 J
Explanation:
Given that;
mass of bag m = 125 kg
length of rope [tex]l[/tex] = 3.3 m
displacement of bag d = 2.2 m
1) What horizontal force is necessary to hold the bag in the new position?
from the figure below; ( triangle )
SOH CAH TOA
sin = opp / hyp
sin[tex]\theta[/tex] = d / [tex]l[/tex]
sin[tex]\theta[/tex] = 2.2/ 3.3
sin[tex]\theta[/tex] = 0.6666
[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )
[tex]\theta[/tex] = 41.81°
Now, tension in the string is resolved into components as illustrated in the image below;
Tsin[tex]\theta[/tex] = F
Tcos[tex]\theta[/tex] = mg
so
Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg
sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg
we know that; tangent = sine/cosine
so
tan[tex]\theta[/tex] = F / mg
F = mg tan[tex]\theta[/tex]
we substitute
Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )
F = 1225 × 0.8944
F = 1095.6 N
Therefore, the required horizontal force F is 1095.6 N
2) As the bag is moved to this position, how much work is done by the rope?
Tension in the rope and displacement of mass are perpendicular,
so, work done will be;
W = Tdcos90°
W = Td × 0
W = 0 J { work done by rope will be 0 since tension perpendicular }
3) As the bag is moved to this position, how much work is done by the worker
from the diagram in the image below;
SOH CAH TOA
cos = adj / hyp
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex]
we substitute
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex] = 1 - h/[tex]l[/tex]
cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]
h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]
h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
now, work done by the worker against gravity will be;
W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
we substitute
W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )
W = 4042.5 × ( 1 - 0.745359 )
W = 4042.5 × 0.254641
W = 1029.4 J
Therefore, work is done by the worker is 1029.4 J
Two teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 64 kg and exerts an average force of 1350 N horizontally. Each of the second team's members has an average mass of 69 kg and exerts an average force of 1367 N horizontally. (a) What is the acceleration (in m/s2 in the direction the heavy team is pulling) of the two teams
Answer:
[tex]a=0.13m/s^2[/tex]
Explanation:
From the question we are told that
Mass of first team man [tex]m_1=64kg[/tex]
Force of man first team man [tex]F_1=1350[/tex]
Mass of second team man [tex]m_2=69kg[/tex]
Force of man second team man [tex]F_2=1367N[/tex]
Generally the equation for net force F_n is mathematically given by
[tex]F_n=9(m_1+m_2)a[/tex]
[tex]9(m_1+m_2)a=9(f_2-f_1)[/tex]
[tex]9(64+69)a=9(1367-1350)[/tex]
[tex]a=\frac{9(1367-1350)}{9(64+69)}[/tex]
[tex]a=0.127819m/s^2[/tex]
Therefore the acceleration is given by
[tex]a=0.13m/s^2[/tex]
An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.
a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?
Answer:
(a) The change in momentum is 12.04 kg-m/s
(b) The force exerted by the bat is 1003.33 N
Explanation:
Given that,
The mass of a ball, m = 0.14 kg
Initial speed of the ball, u = 40 m/s
Final speed of the ball, v = -46 m/s
(a) The change in momentum of the ball during the collision with the bat is given by :
[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]
(b) Time for collision, t = 0.012 s
Now the force can be calculated as follows :
[tex]F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N[/tex]
Hence, this is the required solution.
Answer:
a. = 12.04 kg*m/s
b. = 1,003.3N
Explanation:
The answer above is correct.
a car moved 120km to the north. what is its displacement?
examples of buildup of static energy?
Answer:
Materials that can lose or gain electrons in this way are called triboelectric, according to Northwestern University. One common example of this would be shuffling your feet across carpet, particularly in low humidity which makes the air less conductive and increases the effect.
Explanation:
Hope that helps :)
Explain why the same side of the moon is always facing Earth.
Answer:
The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.
a race car goes around a circular track of radius 150 m at speed of 10.0 m/s. How long does it take to complete one lap?
Answer:
94.25 seconds
Explanation:
Solve for period (T) using: v=(2*pi*r)/T
rearrange: vT=2*pi*r
rearrange: T=(2*pi*r)/v
Plug in values.
T=(2*pi*150)/10
T=94.25 seconds
If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.
vT = 2 × π × r
T = (2 × π × r)/ v
T = (2 × π× 150)/10
T = 94.25 seconds
Thus, the time taken to complete the one lap would be 94.25 seconds.
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Which of the following is an example of an object with kinetic energy?
a. a plane lifting off of the runway
b. a bobsled perched at the top of a run
c. a snowball tumbling down a hill
d. both a and c
Answer:
A and C
Explanation:
Both have mass and are in motion