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3 years ago

Find the acceleration due to gravity on Titan's surface, given that its mass is 1.35 x 102kg and its radius is 2570 km

Physics

1 answer:

Anon25 [30]3 years ago

5 0

Answer:

1.36 x $10^{-15}$ $ms^{-2}$

Explanation:

The equation $g = \frac{G*M}{R^2}$ would be used where:

g = acceleration due to gravity

G = universal gravitational constant (6.67408 × $10^{-11}$ $m^{3} kg^{-1} s^{-2}$ )

M = mass

R = radius.

assuming the value following 10 is its exponent, which is a little confusing, the values given would simply be substituted.

Anyone can correct me if I'm wrong.

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A coil of wire with 60 turns lies on the plane of the page, and a uniform magnetic field points into the page with a strength of

baherus [9]

Answer:

0.85V

Explanation:

The emf is calculated by using the Lenz's Law

$\epsilon=-\frac{\Delta\Phi_B}{\Delta t}$

But for this case we have that the magnetic field is constant. Hence we have

$\epsilon=-\frac{B(A_{2}-A_{1})}{t_2-t_1}=-\frac{(1.10T)(0.270m^{2}-0.89m^{2})}{0.8s}\\\\\epsilon=0.85V$

where we have taken that the intial time is t1=0

I hope this is useful for you

regards

4 0

3 years ago

What do we mean when we say that motion is relative? What is everyday motion usually related to

Luda [366]

It’s measured in a reference frame that is usually the earth’s surface

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3 years ago

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Romashka-Z-Leto [24]

Answer:

True

Explanation:

when the object is larger , the inertia of the object is larger so its tendency to change its state of motion is reduced is reduced.

5 0

3 years ago

A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000

34kurt

Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N

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3 years ago

Suppose a free-fall ride at an amusement park starts at rest and is in free fall. What is the velocity of the ride after 2.3 s?

horsena [70]

Answer:

V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s

S = 1/2 g t^2 since initial speed is zero

S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m

8 0

3 years ago