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pishuonlain [190]
2 years ago
14

Using the formula for work, Match the Force × Distance problems with their correct "Work Performed" answer. The first example ha

s been completed for you.
Force × Distance = Work Performed

10 newtons 6 meters =60 Nm

1. 62 Nm
40 pounds 100 feet

2. 4000ft.-lbs.
2.5 pounds 24 inches

3. 5 ft.-lbs
62 newtons 1 meter

4. 200ft.-lbs.
1 ton 20 feet

5. 40,000 ft.-lbs.
400 pounds 0.5 feet
Physics
1 answer:
Alenkinab [10]2 years ago
3 0
1. 62NM = 62 newtons 1 meter
2. 4000ft.-lbs = 40 pounds 100 feet
3. 5ft.-lbs = 2.5 pounds 24 inches
4. 200ft.-lbs = 400 pounds 0.5 feet
5. 40,000 ft.-lbs = 1 ton 20 feet
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Considering how the parts of a system work together and affect one another, other systems, and the environment is called _______
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Answer:

Hi

before I answer a question I think very deeply and try my best, hope it helps...

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Mechanical energy is the sum of ________ energy and potential energy.apex
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Can someone help me with this question
Mademuasel [1]

Answer:

Net force: 20 N to the right

mass of the bag: 20.489 kg

acceleration:  0.976  m/s^2

Explanation:

Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:

195 N - 175 N = 20 N

So net force on the bag is 20 N to the right.

The mass of the bag can be found using the value of the weight force: 201 N:

mass = Weight/g = 201 / 9.81 = 20.489 kg

and the acceleration of the bag can be found as the net force divided by the mass we just found:

acceleration = 20 N / 20.489 kg = 0.976  m/s^2

8 0
3 years ago
Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r
IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance

V is the potential difference

Calling C_1 the capacitance of capacitor 1 and V_1 its potential difference, the energy stored in capacitor 1 is

U=\frac{1}{2}C_1 V_1^2

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: C_2 = \frac{C_1}{2}

- The voltage is twice the voltage of capacitor 1: V_2 = 2 V_1

so the energy stored in capacitor 2 is

U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2

So the ratio between the two energies is

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3 years ago
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