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MA_775_DIABLO
3 days ago
8

The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad

A(t) = 415 - \sin\left(\dfrac{2\pi (t + 23.2)}{92.8}\right)A(t)=415−sin( 92.8 2π(t+23.2) ​ )space, A, left parenthesis, t, right parenthesis, equals, 415, minus, sine, left parenthesis, start fraction, 2, pi, left parenthesis, t, plus, 23, point, 2, right parenthesis, divided by, 92, point, 8, end fraction, right parenthesis. The International Space Station reaches its perigee once in every orbit. How long does the International Space Station take to orbit the earth? Give an exact answer.
Physics
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"a very light ideal spring stretches by 21.0 cm when it is used to hang a 135-n object. what is the weight of a piece of electro
Keith_Richards [3271]
In the first case, once equilibrium is established,
mg = kx
135 = k(0.21) [Converted to SI units]
k = 135/(0.21)

In the second case, similarly when reaching equilibrium,
mg = kx
mg = [(135)(.449)]/(0.21)
mg = weight of the electronic equipment = approximately 289 N.
6 0
2 months ago
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A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diame
kicyunya [3294]
For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

                         = 2 revolutions per 1 second

                         = 2 * 2π  radians for each second

                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

                                                         ≈13.817  m/s²

. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

                                   = 0.175*13.817

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5 0
2 months ago
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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Sav [3153]

Answer:F_{v} =\mu_{k} mg

The force required has a magnitude of 2601.9 N

Explanation:

m = 450 kg

Static friction coefficient μs = 0.73

Kinetic friction coefficient μk = 0.59

The force necessary to initiate movement of the crate is F_{s} =\mu_{s} mg.

Once the crate begins to move, the frictional force decreases to F_{v} =\mu_{k} mg.

To maintain the motion of the crate at a steady velocity, we must lower the pushing force to F_{v} =\mu_{k} mg.

Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.

<pMagnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N

4 0
2 months ago
A large insulating sheet has a surface charge density σ. what is the electric field strength near the insulating sheet?
Ostrovityanka [3204]
The only scenarios where one can derive dependable data from a noncontact measurement of an insulator's charge state are with free insulative sheets or those that have a grounded conductor backing. In each of these instances, the electric field generated by the charge plays a crucial role.
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