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frozen [14]
3 years ago
15

If you perform 40 J of work lifting a 10-N box from the floor to a shelf, how high is the shelf?

Physics
1 answer:
Kaylis [27]3 years ago
4 0
Before anything, if you are lifting the Work value have to be negative, reason is because the formula to the work of the weight is:

w = p \times d \times cos \:  \theta
*Photo to help you to visualize. (P = Weight Force, F = Force)

If you are lifting, you are exercising a force that opposes the Weight Force, Force up (lifting) and Weight Force down (gravity pulling).
This forms a 180 angle and the cos 180 = -1.

Well let's do the question:

Work (W) = -40 J (Joule = Kg x m^2/s^2)
(negative, explanation given)

Weight Force (P) = 10 N (Newton = kg x m/s^2)

Distance (D) = ?

W = P x d x cos theta

-> -40 = 10 x d x -1
-> -40 = -10 x d
-> d = -40/-10 (J/N)
-> d = 4 meters

d =j  \div n \\ d = ((kg \times  {m}^{2} ) \div  {s}^{2}) \div ((kg \times m) \div  {s}^{2}) \\ d = m \: (meters)

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A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?
Crazy boy [7]

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU  is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU  - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

6 0
3 years ago
The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t
Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension T_{da} is

T_{DA}=-4.433i-3.49j+2.735k

From the Question we are told that

M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft

\vec {DA}=-4.7i-3.7j+2.9k)ft

\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft

Generally the equation for T_{DA}  is mathematically given as

T_{DA}=\phi_{DA}* M_{da}

Where

\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}

Therefore

T_{DA}=\phi_{DA}* M_{da}

T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27

T_{DA}=-4.433i-3.49j+2.735k

For more information on this visit

brainly.com/question/20746649?referrer=searchResults

5 0
3 years ago
If a distant galaxy has a substantial redshift (as viewed from our galaxy), then anyone living in that galaxy would see a substa
mario62 [17]

Answer:

Option A

Explanation:

The statement makes sense since it's already explained that the galaxy is moving away from us and unlike option C which depicts that the galaxy is moving to us.

This statement makes sense. The redshift means that we see the galaxy moving away from us, so observers in that galaxy must also see us moving away from them—which means they see us redshifted as well

3 0
2 years ago
5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?
GenaCL600 [577]

Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

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3 0
2 years ago
Which pair of graphs represent the same motion?
GaryK [48]
Show me the graphs and i would be glad to help u
7 0
3 years ago
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