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3 years ago

As the space shuttle orbits the Earth, the shuttle and the astronauts

Physics

1 answer:

maxonik [38]3 years ago

3 0

Answer:

a=g.

the body is in one of these INGRAVIDITY, so the weight of the body would be ZERO

Explanation:

When the spacecraft revolves around the Earth, it is subjected to Newton's second law

F = m a

where the force is the universal attractive force, the mass of the astronauts, and the centripetal acceleration.

a = F / m

a = (G M m / r²) / m

a = G M / r²

the amount of is the acceleration of gravity at height r.

a = g

Therefore, we see that the acceleration of the body is equal to the acceleration of gravity, which is why the body is in one of these INGRAVIDITY, so the weight of the body would be ZERO

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A 3.00 m-long 6.00-kg ladder pivoted at the top hangs down from a platform at the circus. A 42.0-kg trapeze artist climbs to a p

statuscvo [17]

Answer:

The period of the system of ladder and woman, T = 2.5 seconds

Explanation:

Mass of the ladder, $m_1 = 6 kg$

Mass of the artiste, $m_2 = 42.0 kg$

Length of the ladder, L = 42.0 kg

The total moment of inertia can be calculated using the equation:

$I = \frac{1}{3} M_1 L^2 + m_2 (\frac{L}{2} )^2\\I = \frac{1}{3} *6*3^2 + 42* (\frac{3}{2} )^2\\I = 18 + 94.5\\I = 112.5 kg m^2$

D = L/2 = 3/2

D = 1.5 m

The frequency of the system of ladder and woman follows that of a physical pendulum which can be given by the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{mgD}{I} } \\f = \frac{1}{2\pi } \sqrt{\frac{48*9.8*1.5}{112.5} }\\f = 0.4$

The period of the system of ladder and woman is given by:

T = 1/f

T = 1/0.4

T = 2.5 seconds

5 0

3 years ago

3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is

miskamm [114]

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}$

Applying values we get

$\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)$

4 0

3 years ago

Read 2 more answers

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$