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guapka [62]
3 years ago
10

A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is

done?
Physics
1 answer:
Tpy6a [65]3 years ago
5 0

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

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A 30 kg box is placed on a skateboard and pushed across the room with a velocity of 4 m/s. The pusher applies more force, increa
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A stone tumbles into a mine shaft and strikes bottom after falling for 4.2 second how deep is the mine shaft
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Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent
vfiekz [6]

The potential difference comes out to be

10 \times 10 {}^{ - 3} m

Given:

σ = 8. 85 × 10-9 c/m2

we know,

E = \frac{σ}{2ε0}

E =  \frac{8.85 \times 10 {}^{ - 9} }{2ε0}

E =  \frac{v}{d}

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

=  \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }

Δ = 10 \times 10 {}^{ - 3} m

Thus the potential difference is

10 \times 10 {}^{ - 3} m

Learn more about potential difference from here: brainly.com/question/28165869

#SPJ4

5 0
1 year ago
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