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leonid [27]
3 years ago
12

9. Calculate the distance (in km) that Charlie runs if he maintains the average

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

<u>Correct Question:</u>

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

<u>Answer:</u>

The total distance covered by Charlie is 8 km in 1 hour.

<u>Explanation:</u>

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$t=1 hour

As we know the formula to evaluate the total distance d when the average velocity and time is given;

v=\frac{d}{t}

d=v \times t

d=8 \times 1

d=8 k m

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

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Answer:

False

Explanation:

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8 0
4 years ago
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A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
3 years ago
You carry a 5 kg sack of potatoes across the store for 5 minutes, and you wait in line holding it for 30 seconds before you get
ludmilkaskok [199]

Answer:

The only work done is when the person lifts the sack over a distance, W = 78.48 [N]

Explanation:

We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.

W = F * d

where:

F = force [N]

d = distance [m]

The force is given by the producto of the mass by the gravity.

F = 5 * 9.81 = 49.05 [N]

W = 49.05 * 1.6 = 78.48 [N]

3 0
4 years ago
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
Tpy6a [65]

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 .

5 0
3 years ago
When a Lunar Module landed on the Moon, it used thrusters to slow its descent to the surface. When other spacecraft are returned
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A parachute is a device designed to generate a LOT of air resistance.
Parachutes do that very well in places where there's any air to work with.
There is no air on the Moon.

7 0
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