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tekilochka [14]
2 years ago
5

You are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceilin

g on a string. The tension in the string is measured to be exactly equal to the force due to gravity on the load. No other forces are acting on the load. Which of the following statements about the elevator are correct?
i. The elevator is an inertial frame of reference.
ii. The elevator may be at rest.
iii. The elevator may be moving at a constant velocity upward.
iv. The elevator may be moving at a constant velocity downward.
v. The elevator may be accelerating.
Physics
1 answer:
mote1985 [20]2 years ago
3 0
Hi there is the one you used in my house in a while away and it looks great on my phone but I hash
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A man of mass 85 kg runs up a flight of stairs of height 4.6 m in a time period
seraphim [82]

Explanation:

a) Power = work / time = force × distance / time

P = Fd/t

P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)

P ≈ 319 W

b) P = Fd/t

0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)

m = 47.6 kg

7 0
3 years ago
A student lifts a physical science book off the table and above their head. Is there work being done?
DiKsa [7]

Answer: A force must cause a displacement in order for work to be done. A book falls off a table and free falls to the ground. Yes. This is an example of work.

Explanation:

7 0
2 years ago
You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re
melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0
2 years ago
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0
3 years ago
Math Focus
zhenek [66]

Answer: 4.

Explanation:

Use formula v = d / t, where v = speed, d = distance and t = time.

v = 10 / 2.5

v = 4.

3 0
3 years ago
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