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2 years ago

13

A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work i

s done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules

1 answer:

podryga [215]2 years ago

6 0

The answer is 3) 480 joules

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Which vector shows the direction of the centripetal acceleration at this point

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It's B

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Answer:

please the answer below

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k=8.89*10^9

For both cases we have

$k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}$

(b) by replacing this values of r in the expression for V we obtain

$k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}$

hope this helps!!

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2 years ago

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ryan hypothesizes that darker colors heat up faster. he places a thermometer inside a red wool sock, a green cotton glove, and a

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Light are transfer through waves in the atmosphere and yes it true that the darker the color is the more heat it could absorb thus it is also explain that the lighter the color is the less heat or light its absorb its because the light is bounces back through other form of light and it lessens the amount of heat in a substance. In Ryan's procedure the possible wrong that he done is the present of a green cotton glove. Green color are one of the color that bounces light and could not support the hypothesis of Ryan and the possible temperature he could get is not the accurate one.

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3 years ago

If an automobile engine delivers a power of 50.0 hp, how much time will it take for the engine to do 6.40 x 10^4 j of work? (1 h

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Given data

Power (P) = 50 hp,

= 50 × 746, we know that 1 hp = 746 W.

= 37300 Watts (Watt = J/s)

Work = 6.40 ×10⁴ J

Power is defined as rate of doing work and the unit of power is<em> Watt.</em>

Mathematically,

Power = (Work / time) Watts

= 6.40 ×10⁴ / time

37300 W = 6.40 ×10⁴ J /time (Where time in seconds)

=> time = Work/Power

= 6.40 ×10⁴/37300

= <em>1.74 seconds </em>

<em> </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>

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Explanation:

Solution:-

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