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3 years ago

PLEASE HELP ASAP!!! THANKS! how do stars and planets effect eachother?

Physics

2 answers:

Dvinal [7]3 years ago

7 0

Answer:

Stars and planets each have their own gravitational pull. Obviously a star would have a larger gravitational pull than a planet because it's larger. A star's gravitational pull keeps a planet in orbit around that star.

Explanation:

Serhud [2]3 years ago

6 0

Answer:

"Planets May Affect the Chemistry of Their Stars. Planets are, by and large, at the mercy of their stars. Not only do stars provide a ready energy source of radiated light and heat, but the mass and gravitational pull of stars flat-out dwarfs the summed masses and pulls of any orbiting companions."

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An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break

mamaluj [8]

Answer:

Value that the spring constant k = 12Mg / h

Explanation:

According to 2nd law of Newton:

upward force of the spring= F

The weight of the elevator W = mg

F = Mg = M(5g)

==> F =6Mg.

As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence

F =ks = 6Mg

==> s = 6Mg/k

We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:

Mg(h+s) = 1/2ks2

And plugging in our expression for s:

Mg(h+6Mg/k)= 1/2k(6Mg / k)2

gh + 6M2g2/k = 1/2k(36M2g2 /k2)

Mgh +6M2g2/k = 1/2k(36M2g2 /k2)

gh + 6Mg2/k = 18Mg2 / k

gh = 12Mg2 / k

h = 12Mg / k

k = 12Mg / h

7 0

4 years ago

Read 2 more answers

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

4 years ago

When Danny Diver who weighs 500 N steps off a diving board 10 m above the water, he hits the water with kinetic energy of

Andreas93 [3]

Answer:

5000 J

Explanation:

Given that,

Weight of Danny Diver, F = 500 N

He steps off a diving board 10 m above the water

We need to find the kinetic energy of the water with which it hits. Let its is K. It is given by :

E = mgh

Where g is acceleration due to gravity

$E=500\ N\times 10\ m\\\\=5000\ J$

So, he hits the water with kinetic energy of 5000 J.

8 0

3 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

miv72 [106K]

Answer:

C. In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

I = Δp = m vf - m v₀

I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

I₁ = m (-0.3 - 0.3)

I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

I₂ = m (0 - 0.3)

I₂ = -0.3 m

Let's calculate the relationship between the two impulses

I₁ / I₂ = -0.6m / -0.3m

I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A. False The momentum changes

B. False. The momentum is less in the second collision