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algol [13]
3 years ago
8

The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part

icle if the force has a component parallel to the motion.The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion.
Physics
1 answer:
Masja [62]3 years ago
8 0

Answer:

The work-energy theorem states that a force acting on a particle as it moves over a <u>distance</u> changes the <u>kinetic</u> energy of the particle if the force has a component parallel to the motion.

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In the following description, QUOTE the portion that addresses weather in the area.
olga nikolaevna [1]

Answer:

Portland normally had rainy, but mild winters. Yesterday it rained 0.75 inches in Portland.

Explanation:

This talks about the weather

4 0
3 years ago
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
olya-2409 [2.1K]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

3 0
3 years ago
What ecosystem do killer whales and seals belong to?
Anestetic [448]

Answer:

can be found in many waters, but the Antarctic ecosystem is where the population is highly condensed.

Explanation:

4 0
3 years ago
If we increase the speed of rotation of wheel then the sound produced by cardboard touching the spokes of the wheel will become
Anarel [89]

it will become shrill

5 0
3 years ago
Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc
vivado [14]

Answer:

ΔR_{e} = 84   Ω,     R_{e} = (40 ± 8) 10¹   Ω

Explanation:

The formula for parallel equivalent resistance is

          1 / R_{e} = ∑ 1 / Ri

In our case we use a resistance of each

           R₁ = 500 ± 50  Ω

          R₂ = 2000 ± 5%

This percentage equals

        0.05 = ΔR₂ / R₂

        ΔR₂ = 0.05 R₂

        ΔR₂ = 0.05 2000 = 100   Ω

We write the resistance

        R₂ = 2000 ± 100    Ω

We apply the initial formula

        1 / R_{e} = 1 / R₁ + 1 / R₂

        1 / R_{e} = 1/500 + 1/2000 = 0.0025

        R_{e}  = 400    Ω

Let's look for the error  (uncertainly) of Re

      R_{e} = R₁R₂ / (R₁ + R₂)

       R’= R₁ + R₂

       R_{e} = R₁R₂ / R’

Let's look for the uncertainty of this equation

      ΔR_{e} / R_{e} = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

      ΔR’= ΔR₁ + ΔR₂

We substitute the values

     ΔR_{e} / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

     ΔR_{e} / 400 = 0.1 + 0.05 + 0.06

     ΔR_{e} = 0.21 400

     ΔR_{e} = 84   Ω

Let's write the resistance value with the correct significant figures

    R_{e} = (40 ± 8) 10¹   Ω

6 0
3 years ago
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