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Vesnalui [34]
3 years ago
9

If a moon on jupiter has 1/8 the mass of the earth and 1/2 the earth's radius, what is the acceleration of gravity on the planet

's surface? the acceleration of gravity on earth's surface is 10 m/s 2 .
Physics
1 answer:
dusya [7]3 years ago
8 0

The formula we can use here is:

g = G m / r^2

where g is gravity, G is gravitational constant, m is mass, and r is radius

Since G is constant, therefore we can equate two situations:

g1 m1 / r1^2 = g2 m2 / r2^2

(10 m/s^2) r1^2 / m1 = g2 * (1/2 r1)^2 / (1/8 m1)

<span>g2 = 5 m/s^2</span>

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BigorU [14]

Answer:

The magnetic field through the wire must be changing

Explanation:

According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ  through it. This is stated mathematically as ε = dΦ/dt.

Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.

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dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ

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3 years ago
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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

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So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

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(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

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3 years ago
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