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Tcecarenko [31]
3 years ago
11

the gravitational force between the sun and earth is larger than sun and the moon. what do you think the reason for this is?

Physics
2 answers:
Tems11 [23]3 years ago
7 0

Answer: The greater an object's mass, the more gravitational force it exerts.

Explanation: So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive. Sorry if I get this wrong. I am in 5th grade! ♥

ivann1987 [24]3 years ago
4 0

hi i need help with one of my question but i don’t know how else to contact you :/. anyway i posted the question and i added the text so i was wondering if u could help me !

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A small electric motor produces a force of 5 N that moves a remote-control car 5 m every second. How much power does the motor p
kati45 [8]

Answer:

Explanation:

Given:

Force, f = 5 N

Velocity, v = 5 m/s

Power, p = energy/time

Energy = mass × acceleration × distance

Poer, p = force × velocity

= 5 × 5

= 25 W.

Note 1 watt = 0.00134 horsepower

But 25 watt,

0.00134 hp/1 watt × 25 watt

= 0.0335 hp.

4 0
3 years ago
Read 2 more answers
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
NikAS [45]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

7 0
3 years ago
A particle is moving along the x-axis so that its position at any time t is greater than and equal to 0 is given by x(t)=2te^-t?
erastovalidia [21]
For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left
6 0
3 years ago
An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach
matrenka [14]

Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, \theta=60^{\circ}

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m

So, the maximum height reached by the object is 34.43 m.

6 0
3 years ago
The horn on a fire truck sounds at a pitch of 350 hz a. what is the perceived frequency when the fire truck is moving toward you
solong [7]

The perceived frequency when the fire truck is moving toward you and away from you will be 370 Hz and 329.59 Hz respectively.

<h3>What is the Doppler effect?</h3>

A sudden change in the frequency due to the distance between the objects and source is explained by the doppler effect.

As the source and observer travel toward each other, the frequency of sound, light, or other waves increases or decreases.

The perceived frequency when the fire truck is moving toward you;

\rm r' = (\frac{v+v_0}{v}) V \\\\ \rm r' = (\frac{343+20}{343}) 350 \\\\ r' =370.4  \  Hz

The perceived frequency when the fire truck is moving away from you;

\rm r' = (\frac{v-v_0}{v}) V \\\\ \rm r' = (\frac{343-20}{343}) 343 \\\\ r' =329.59  \  Hz

Hence, the perceived frequency when the fire truck in cases 1 and 2 will be 370 Hz and 329.59 Hz.

To learn more about the doppler effect refer to the link;

brainly.com/question/15318474

#SPJ1

5 0
2 years ago
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