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daser333 [38]
2 years ago
14

Jose is loading his luggage into his car so that he can go to visit his grandmother. He lifts his suitcase up a 10 m staircase i

n his basement to the entrance of his garage with an average force of 60 N. How much work is performed by Jose in moving the suit case?
Physics
1 answer:
maxonik [38]2 years ago
3 0

Answer:

The work done on the suitcase is, W = 600 J

Explanation:

Given,

The average force exerted by Jose on his suitcase,  F = 60 N

Jose carried the suitcase to a distance, S = 10 m

The work done on the suitcase is given by the relation

                           <em>W = F x S</em>

Substituting the given values in the above equation,

                            W = 60 N x 10 m

                                 = 600 J

Hence, the work done on the suitcase is, W = 600 J

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The answer is below

Explanation:

Well, a hurricane is a storm with violent winds (more less with a force of 12 in the scale that measures winds) and it happens especially in the Caribbean sea. While a typhoon is also a strong storm but it happens especially in the Indian region and the western pacific oceans.

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2 years ago
Which of the following involve an increase in the entropy of the system under consideration? a. melting of a solid b. evaporatio
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Explanation:

Since, entropy is the measure of degree of randomness. So, more randomly the molecules of a substance are moving more will be its entropy.

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  • Evaporation of a liquid will also cause the liquid to change its state from liquid to gas. This means molecules will go far away from each other leading to an increase in the entropy.
  • Sublimation is a process of conversion of a solid into gaseous phase without going through liquid phase. So, in this case also entropy will increase due to gain in energy by the molecules of a solid.
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8 0
3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
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So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
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