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kotegsom [21]
3 years ago
9

A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest poi

nt and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?(A) net force of zero(B) Net constant force up the ramp(C) net increasing force up the ramp(D) net increasing force down the ramp(E) net constant force down the ramp(F) net decreasing force down the ramp(G) net decreasing force up the ramp
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer: Option (E)

Explanation:

There are only two forces acting on the car: The force of gravity, which point down, and the normal force generated by the interaction of the car and the ramp, which points perpendicular to the ramp.

N=mgcos(\alpha ),

m:mass of the car ; \alpha:angle of the ramp.

g: gravity's acceleration.

N is constante since all parameters are constants.

So, since the force of gravity is also constant, the net force acting on the car is constant. Since velocity decreases over time and the car stars moving down, the force is pointing down the ramp.

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Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts
podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height =?

y_{0}: is the initial height = 1 m

v_{0_{y}: is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time    

First, we need to find the time by using the following equation:

t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s

Now, the height is:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m      

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}

1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!                                                                                  

4 0
3 years ago
Giving 20 points and brainiest <br> But please help me
monitta

Answer:

yeah, Do you want me to check your answers? Yes is correct

8 0
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GrogVix [38]

Answer:

"Longitudinal wave" is the appropriate answer.

Explanation:

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3 0
3 years ago
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A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)
Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

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