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il63 [147K]
3 years ago
5

A student is conducting an experiment to determine how far a ball will roll down a ramp based on the angle of incline.

Physics
2 answers:
Nataly_w [17]3 years ago
7 0

Answer:

Explanation:

There are several possible controls that can be made in order to control the experiment.

The angle of inclination

The size and mass of the ball.

The length of the ramp.

All these factors are controllable and very well define the experiment. More the weight and angle of inclination, faster and longer distance the ball will cover.

Sedbober [7]3 years ago
5 0
The size of the ball, the mass, and if the ball gets an extra push
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blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

Collect like terms

92 + 1 = x

93 = x

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239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0
3 years ago
In each cylinder, draw the piston and the gas particles based on the temperature shown.
tangare [24]
I need more information
8 0
3 years ago
A jet with mass m = 90000.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.
leva [86]
This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
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6 0
4 years ago
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially
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Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

3 0
3 years ago
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