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1 year ago

When in orbit, astronauts experience weightlessness. What is this caused by?

Physics

1 answer:

Ulleksa [173]1 year ago

7 0

Answer:

Earth-orbiting astronauts are weightless for the same reasons that riders of a free-falling amusement park ride or a free-falling elevator are weightless. They are weightless because there is no external contact force pushing or pulling upon their body. In each case, gravity is the only force acting upon their body.

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A rock falls from a cliff and hits the ground at a velocity of 28m/s. How long did it take to fall?

liberstina [14]

Answer:

2.8s

Explanation:

=>a = v-u/t

=> t = v - u / t

= 28-0 / 10

= 2.8 s

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2 years ago

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3 years ago

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.50.(a) What is the o

bezimeni [28]

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2 years ago

Why does the speed of sound vary depending on the medium through which it travels. in general?

avanturin [10]

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2 years ago

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c

Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

$\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}$

hence, the speed of the spheres is 37.45 m/s

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2 years ago