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pentagon [3]
2 years ago
13

An astronaut is in an all-metal chamber outside the space station when a solar storm results in the deposit of a large positive

charge on the station. Which statement is correct?
a. The astronaut must abandon the chamber immediately to avoid being electrocuted.
b. The astronaut will be safe only if she is wearing a spacesuit made of non-conducting materials.
c. The astronaut does not need to worry: the charge will remain on the outside surface.
d. The astronaut must abandon the chamber if the electric field on the outside surface becomes greater than the breakdown field of air.
d. The astronaut must abandon the chamber immediately because the electric field inside the chamber is non-uniform.
Physics
1 answer:
ArbitrLikvidat [17]2 years ago
3 0

Answer:

<em>c. The astronaut does not need to worry: the charge will remain on the outside surface.</em>

<em></em>

Explanation:

The astronaut need not worry because <em>according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside.</em> This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.

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Someone who wants to sell you a Superball claims that it will bounce to a height greater than the height from which it is droppe
sergeinik [125]

Answer:

No

Explanation:

Unless there are other external forces, this will never be true. Because according to energy conservation, potential energy will be converted to kinetic energy as the ball falls down (so it loses height and gain speed). And vice versa, kinetic to potential when it bounces back. So the potential energy after must be the same (or smaller if losing heat to external environment), so it can only get the the same height or less, but not more.

7 0
1 year ago
To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.
lozanna [386]

Answer:

a)

Explanation:

  • A block sliding down an inclined plane, is subject to two external forces along the slide.
  • One is the component of gravity (the weight) parallel to the incline.
  • If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

        F_{gp} = m*g* sin \theta (1)

       (taking as positive the direction of the movement of the block)

  • The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
  • When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

       F_{f} = \mu_{k} * N  (2)

  • The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

       N = m*g* cos \theta (3)

  • In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

        m*g* sin \theta =  \mu_{k} * m*g* cos \theta

  • As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
  • In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

        \mu_{k}  = tg \theta

8 0
1 year ago
Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee
valina [46]

Answer: The frequency = 1714.3Hz

Explanation: The solution can be achieved by using doppler effect formula.

Since the source is moving toward the observer, the velocity of the observer will be positive.

Please find the attached file for the solution

3 0
1 year ago
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
statuscvo [17]

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required = \frac{(14.644 x 10^6)}{784}

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

3 0
1 year ago
A spaceship which is 50,000 kilometers from the center of Earth has a mass of 3,000 kilograms. What is the magnitude of the forc
Anna11 [10]
Fg=Gx(M1M2/r^2)
Fg=478N
6 0
1 year ago
Read 2 more answers
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