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wolverine [178]

2 years ago

13

A crate is released from the highest point of a 12 m long ramp inclined at 60° with the horizontal. The motion is frictionless.

What is the speed of the crate at the base of the ramp?

1 answer:

Darya [45]2 years ago

3 0

law of conservation of mechanical energy

PE1 = Ek2

mgh=1/2mv²

h = 12 sin 60

$\tt v=\sqrt{2gh}=\sqrt{2\times 9.8\times 12\times sin~60}=14.27~m/s$

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A large capacitor in an air conditioner is typically connected to the potential difference provided to your house (120 V). Altho

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$A=7.72\times 10^{-18}\ m^2$

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4 years ago

Read 2 more answers

A round asteroid has a surface gravity of .0288 m/s^2 if the mass of the asteroid is 1.10 x10^18 kg what is its radius

Oduvanchick [21]

Answer:

r = 50.47 x 10³ m = 50.47 km

Explanation:

Using the formula for the acceleration due to gravity:

$g = \frac{Gm}{r^2}$

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g = acceleration due to gravity on the surface of asteroid = 0.0288 m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

m= mass of asteroid = 1.1 x 10¹⁸ kg

r = radius of asteroid = ?

Therefore,

$0.0288\ m/s^2 = \frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{r^2} \\\\r = \sqrt{\frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^2)(1.1\ x\ 10^{18}\ kg)}{0.0288\ m/s^2}}$

<u>r = 50.47 x 10³ m = 50.47 km</u>

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3 years ago