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LekaFEV [45]
2 years ago
8

A ball accelerates from rest down a ramp at 2.4 m/s^2. Write an equation that could be used to determine the balls finals positi

on, assuming the positive direction points down the ramp.
Physics
1 answer:
Alik [6]2 years ago
8 0

Answer:

x=2.4t+4.9t^2

Explanation:

This equation is one of the kinematic equations to solve for distance. The original equation is as follows:

X=Xo+Vt+1/2at^2

We know that the ball starts at rest meaning that its initial velocity and position is zero.

X=0+Vt+1/2at^2

Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.

X=Vt+4.9t^2

Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.

Now, substitute in your velocity value.

X=2.4t+4.9t^2

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A cube has a density of 1800 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experimenter in
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Answer:

4341.44763 kg/m³

Explanation:

\rho' = Actual density of cube = 1800 kg/m³

\rho = Density change due to motion

v = Velocity of cube = 0.91c

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Relativistic density is given by

\rho=\frac{\rho'}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-\frac{0.91^2c^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-0.91^2}}\\\Rightarrow \rho=4341.44763\ kg/m^3

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5 0
4 years ago
How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
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Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

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q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

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