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3 years ago

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Saturn has a radius of about 9.0 earth radii, and a mass 95 times the Earth’s mass. Estimate the gravitational field on the surf

ace of Saturn compared to that on the Earth. Show your work.

1 answer:

olchik [2.2K]3 years ago

6 0

Answer:The gravitational field on Saturn can be calculated by the following formula;

Explanation:

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You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

Readme [11.4K]

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

$d_1 = v*t_1$

$d_1 = 10*20 = 200 m$

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

$d_2 = v*t = 20*20 = 400 m$

so the net displacement is given as

$\vec d = \vec d_2 - \vec d_1$

$\vec d = 400 - 200 = 200 m$

so it will be displaced by total displacement 200 m

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3 years ago

What factors determine the magnitude of the electric force between two particles ?

svp [43]

The electric force between the two particles are calculated through the equation,

F = kQ₁Q₂ / d²

where F is the force, k is a constant called Coulomb's law constant, Q₁ and Q₂ are the charges, and d is the distance. This equation is called the Coulomb's law.

It can be seen from the equation above that the electric forces between the objects are majorly affected by the substance's charges and distance.

The answer to this item is therefore letter A.

7 0

3 years ago

Read 2 more answers

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

3 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago

The potential difference between points A and B in an electric

Flauer [41]

Answer:

i hope the answear is D becuase went over this long time ago when i was like you

Explanation:

3 0

3 years ago