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3 years ago

What does the force of gravity between two objects depend on?

1 answer:

3 0

-- The equal, mutual forces of gravity between two objects depend on the product
of the two masses ... it doesn't matter whether the masses are equal or there's
one huge mass and one minuscule mass. It's the product that counts.

-- The force also depends on the distance between the centers of the two objects.

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A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

3 years ago

A 45.0 kg ice skater needs a 25 N horizontal force to get moving on a smooth ice surface. What is the coefficient of friction be

SVEN [57.7K]

The horizontal force : f = k*N
k- coefficient of friction
k = f /N
N = m * g = 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N = 0.057
Answer C) 0.057

7 0

3 years ago

the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change

defon

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=mC_s \Delta T$
where
m is the mass of the substance
$C_s$ the specific heat capacity
$\Delta T$ the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is $C_s = 4.2 J/g ^{\circ}C$ and the amount of heat supplied is $Q=31500 J$ , so if we re-arrange the previous formula we find the increase in temperature of the water:
$\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C$

7 0

3 years ago

a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin

Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0

4 years ago

Which of these is a good example of Newton's First Law of inertia

Andru [333]

Answer:

one body movement to the side when a car makes a sharp turn

3 0

3 years ago

Read 2 more answers