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3 years ago

If the sun had twice the mace how would that affect the gravitational force of the sun

Physics

1 answer:

daser333 [38]3 years ago

5 0

Answer: Gravity is the force that keeps planets in orbit around the Sun. Gravity alone holds us to Earth's surface.

Planets have measurable properties, such as size, mass, density, and composition. A planet's size and mass determines its gravitational pull.

A planet's mass and size determines how strong its gravitational pull is.

Models can help us experiment with the motions of objects in space, which are determined by the gravitational pull between them.

Explanation:

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What is the period of a wave if 3 waves passes every 6 seconds

Naya [18.7K]

Time period = time/no. of waves = 6/3 = 2s

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3 years ago

Two Mississippi companies, Howard Industries and Kuhlman Electric Company, are responsible for most of the transformers and powe

yanalaym [24]

I think b but im not sure just helping

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3 years ago

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A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0

kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

$\large {\boxed {a_s = v^2 / R} }$

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$ → in rad/s

$\omega = \frac{v}{2 \pi R}$ → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Uniform Circular Motion : brainly.com/question/2562955
Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

6 0

3 years ago

Read 2 more answers

A surface receiving sound is moved from its original position to a position three times farther away from the source of the soun

777dan777 [17]

Intensity, E follows inverse square law.

E α 1/r²

r is the distance.

So if the distance r is increased by 3, the intensity would be reduced by 3²

3² = 9

So the answer is nines times as low. C.

3 0

4 years ago

An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle o

klio [65]

Answer:

$3.58\:\mathrm{s}$

Explanation:

We can use the kinematics equation $\Delta y=v_it+\frac{1}{2}at^2$ to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

$\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}$

Now we can substitute values in our kinematics equation:

$\Delta y=-100$
$a=-9.8\:\mathrm{m/s^2}$ (acceleration due to gravity)
$v_i=-10.3283743813\:\mathrm{m/s}$
Solving for $t$

$-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}$

6 0

3 years ago