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2 years ago

What are two things that the amount of gravitational force between two object depends on

2 answers:

4 0

The equation behind this question is:

g = (G*m*M) / (r^2)

g is the gravitational force
G is the Universal Gravitational Constant
M is the mass of one object
and m is the mass of the other constant
r is the distance between the two objects

Amiraneli [1.4K]2 years ago

3 0

1). the product of the two masses being gravitationally attracted to each other

2). the distance between their centers of mass

And that's IT. The gravitational force between them depends on
only those two things, nothing else.

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How do you write a letter to the editor of a newspaper in reference to bullying

suter [353]

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2 years ago

A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

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2 years ago

How long does light take to travel from the sun to earth? heres the exact question: light travels at 300,000 km.s. The sun is a

Artist 52 [7]

I've heard it takes around 8 min

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2 years ago

A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)

bezimeni [28]

Answer:

205 V

V $_{R}$ = 2.05 V

Explanation:

L = Inductance in Henries, (H) = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

$V_{L} = - IwLsin(wt)$

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

= 11.0 V / 500 rad/s (0.500 H)

= 11.0 / 250

I = 0.044 A

Now

V $_{R}$ = IR

= (0.044 A) (93 Ω)

V $_{R}$ = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V $_{R}$ = V $_{R}$ cos (wt)

Putting V $_{R}$ = 4.092 V and w = 500 rad/s

V $_{R}$ = V $_{R}$ cos (wt)

= (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V $_{R}$ = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

= (4.092 V) (cos (500 rads/s)(0.00209))

= (4.092 V) (cos(1.045))

= (4.092 V)(0.501902)

= 2.053783

V $_{R}$ = 2.05 V

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2 years ago

Which of the following is the best example of a primary circular reaction?

olga55 [171]

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2 years ago