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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 pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

3 0
2 years ago
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.
Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0
1 year ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

6 0
1 year ago
Read 2 more answers
Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.
creativ13 [48]

Answer: a) 456.66 s ; b) 564.3 m

Explanation: The time spend to cover any distance a constant velocity is given by:

v= distance/time so t=distance/v

The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

replacing in the above expression we have

time 1= 627 s

time2 = 297 s

The distance traveled is 564,3 m

8 0
2 years ago
A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e
m_a_m_a [10]

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

10 years = \frac{2d}{v}

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }

0.64 = 1-\frac{v^{2} }{186282^{2} }

solving for v, gives = 35 288 miles/s

4 0
2 years ago
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