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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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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
2 years ago
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80
Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0
2 years ago
The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change
mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m

(b)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m

(c)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m

(d) The three lines belong to the ultraviolet region.

8 0
2 years ago
Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani
Vika [28.1K]

Answer:

d not joining FRA and joining HOSA INSTEAD

3 0
2 years ago
Read 2 more answers
A ball of mass 5.0kg is lifted off the floor a distance of 1.7m. 1. What is the change in the gravitational potential energy of
tangare [24]

Answer:

Explanation:

Change in gravitational energy of the ball = mgh

5 mutiply 10 multiply 1.7 = 85J

Potential energy at height = Kinetic energy at bottom

KE= 85J

Velocity

v=5.83m/s

4 0
1 year ago
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