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Andrew [12]
3 years ago
9

A child carries a 3N book at a constant velocity 4 meters across a horizontal floor. What is the net work done?

Physics
1 answer:
Oxana [17]3 years ago
5 0
Note that 4  m (meters) has the units of distance, not velocity.

By definition,
Work = Force x Distance
Therefore if a 3 N book is carried over a distance of 4 m, the work done is
3 *3 4 = 12 J

Answer:  12 J
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A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the pla
disa [49]

Answer:

d.

Explanation:

Since, the capacitance( decreases )

therefore voltage between the plates(increases ).

Hence, option d is correct.

C =εA/d.

d is doubled, therefore  C decrease ( inverse relation).

4 0
3 years ago
HELP MEH QUICK PLEASE
lozanna [386]

Answer:

D) a Battery

Explanation:

The best real-life example of direct current is a battery. Batteries have positive and negative terminals on a battery, the electrons in the wires will begin to flow to produce a current.

6 0
2 years ago
Help a Girl Out and Answer This you will get Brainliest, Thanks, and more points if your answer is correct (:
melamori03 [73]

Answer:

this was 2 weeks ago, but im pretty sure the correct answers are:

B=c

C=d

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3 0
2 years ago
Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li
Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

L=L_0\sqrt{1-\frac{v^2}{c^2} }

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }

= 1.136×10^17 m

L=  = \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}m

so L= 12 light years

8 0
3 years ago
The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0
3 years ago
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