Whatare two diffent units that represent work

Answers

Answer 1

Answer: $W=Pt$
This means:
Work in Joules (J) = Power in Watts (W) × Time in Seconds (s)

$W=Fd$
This means:
Work in Joules (J) = Force in Newtons (N) × Distance in Meters (m)

Answer 2

Answer: two units for work :
~ first is Joule which is equal to Newton .meter ..
~ second is kilogram .meter square per second square i.e kgm²/s²
the second unit is derived from the first unit Newton .meter
Newton = kgm/s² and then multiply m .. it will be equal to kgm²/s²..

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Convert the following:a)2500 kilogram into gram.
b)335 hours into seconds.
c)1 day into seconds.

Answers

Explanation:

A) 1 kg = 1000g

2500kg = 2500000g

B 1 hour = 3600 sec

335 hour = 1206000sec

C 1day = 24 hours

1 hour = 3600 sec

24 hours = 86400sec

Answers:

a.2500000grams

B.1206000seconds

c.86400seconds.

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How far does light travel in 1 millisecond that is, in one hundredth of a scecond?

Answers

First of all, we have to convert the time into seconds:
$t=1 ms=0.001 s$
Light travels at speed of $v=3\cdot 10^8 m/s$ .
The distance covered by the light during this time can be found with the formula
$v= (S)/(t)$
where S is the distance covered and t the time. By re-arranging this formula and by plugging the numbers into it, we find how far the light went in 1 millisecond:
$S=vt=(3\cdot 10^8)(0.001 s)=3 \cdot 10^5 m.$

Final answer:

The speed of light is approximately 299,792 kilometers (186,282 miles) per second. Given a millisecond is a thousandth of a second, light will travel approximately 299.792 kilometers in one millisecond.

Explanation:

The speed of light in a vacuum, such as outer space, is a set constant in physics, approximately 299,792 kilometers per second, or about 186,282 miles per second. Thus, if we want to know how far light travels in 1 millisecond, that is, one hundredth of a scecond, we need to convert seconds into milliseconds.

There are 1,000 milliseconds in one second. Hence, in 1 millisecond, light will travel about 299.792 kilometers (or about 186.282 miles).

Learn more about Speed of Light here:

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Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are separated by 12 cm. (Why can you solve this problem without knowing the magnitudes of the charges

Answers

The force between the two point charges is "5 N".

Force and magnitudes:

The total of almost all of the forces operating on an item is referred to as the magnitude of force. When some of the forces operate in the same direction, this same magnitude rises. Whenever forces occur in opposite directions around an item, the amount of something like the force diminishes.

Distance between charges, d = 6 cm

Attractive force, F = 20 N

Separated, d' = 12 cm

We know the formula,

F = $k (q_1q_2)/(d^2)$

or,

F $\propto (1)/(d^2)$ ...(Equation 1)

New force will be:

F' $\propto (1)/(d'^2)$ ...(Equation 2)

From "Equation 1" and "Equation 2", we get

→ $(F)/(F') =(d'^2)/(d^2)$

By substituting the values,

$(20)/(F') =(12^2)/(6^2)$

$(20)/(F') =(144)/(36)$

F' = 5 N

Thus the approach above is appropriate.

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Answer:

The force between them when they are separated by 12 cm is 5 N.

Explanation:

Distance between two point charges, d = 6 cm

The attractive force between them is 20 N, F = 20 N

Let F' is the force between them when they are separated by 12 cm, d' = 12 cm

The force between point charges is given by the formula as :

$F=k(q_1q_2)/(d^2)$

$F\propto (1)/(d^2)$ ........(1)

New force,

$F'\propto (1)/(d'^2)$ ............(2)

From (1) and (2) :

$(F)/(F')=(d'^2)/(d^2)\n\n(20)/(F')=(12^2)/(6^2)\n\nF'=5\ N$

So, the force between them when they are separated by 12 cm is 5 N. Hence, this is the required solution.

Compute the resistance in ohms of a copper block 5.0 cm long and 0.10 cm2 in cross-sectional area. (ρ = 1.77 × 10-6 ohm-cm)

Answers

R=ρ l/a
l=5 cm
a=0.10 cm²
ρ=1.77×10∧-6 Ωcm
R=(1.77×10∧-6) ×5/0.10
=8.85 ×10∧-5 Ω

resistance is directly proportional to length and inversly proportional to area of cross section is given by above equation

A fast car (Car A) and a slow car (Car B) both travel the same distance. Which car takes more time to do this? Explain.

Answers

Answer:

Input work is the work done on a machine as the input force acts through the input distance. This is in contrast to output work which is a force that is applied by the body or system to something else. Output work is the work done by a machine as the output force acts through the output distance.Jan 15, 2020 whhdbd dhd dvd s bs sbs shs shs sh s

Explanation:

avsvsyxsgzvs zvx shs dhx xbx dhd dvd dj

Aspace walking astronaut has become detached from her spaceship! She’s floating in space with her handy tool belt attached to her waist, thinking about how she might get back to the ship, which she can see 50m from her current location. How can she get back to the ship? Use the language from the laws of motion in your answer.What can she do with the tool belt?

What will that action cause her body to do? Why?

Which of Newton’s Laws applies to this situation?

Write a paragraph which incorporates all of the above information.

Answers

Answer:

In order to get back to the ship, the astronaut must throw the belt in the opposite direction the spaceship.

From Newton's third law of motion:

For every action, there is equal and opposite reaction.

In order to throw the belt away, astronaut would apply some force which cause same amount of force acting on her body in the opposite direction which would take her towards the spaceship.

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