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kozerog [31]
3 years ago
9

A piston has an area of 4 m2 and needs 20 N of force to push it down. What is the pressure?P=F/A *

Physics
1 answer:
Mashutka [201]3 years ago
3 0
I believe the answer is C in this problem
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400 N

The weight of the man is evenly distributed on each scale

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Electrons are (2 points)
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<span>electrons are able to move from one atom to another; the atom that loses an electron becomes positively charged and the atom that gains an electron becomes negatively charged.</span>
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Read 2 more answers
I need help pleaseee
Ludmilka [50]

Answer:

8) 709.8875 J

9) The object is at 7.24375 m from the ground

10) Kinetic energy increases as the object falls.

Explanation:

We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:

h(t) = hi - (g/2) t^2

hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):

h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m

This is the position of the 10 kg object after 0.75 seconds (answer for part 9)

Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:

U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)

Part 10)

the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.

3 0
3 years ago
You measure its length as it passes. by how many millimeters do you determine the rod has contracted?
LUCKY_DIMON [66]
It sounds like a special relativity question but I need more info for a total answer. But remember it's length in the lab frame is
L•sqrt(1-(v/c)^2) where L is the rest length, v is its velocity magnitude and c is the speed of light. Sqrt is the square root (I'm on a phone so I can't see the math equation editor)
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4 years ago
Find the heat given off if 1 kg of 125 degree C steam is cooled to 50 degree C water.
Gnom [1K]
  • T_i=125°C=398K
  • T_f=50°C=323K
  • c=4184J/kg°K
  • m=1kg
  • ∆T=T_f-T_i=398-323=75K

\\ \sf{:}\Rrightarrow Q=mc\Delta T

\\ \sf{:}\Rrightarrow Q=1(4184)(75)

\\ \sf{:}\Rrightarrow Q=313800J

\\ \sf{:}\Rrightarrow Q=313.8KJ

4 0
3 years ago
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