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Veseljchak [2.6K]
3 years ago
14

2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the

person has consumed 850 calories. If it takes them 40 seconds to push the fridge the full length of the court, calculate the power output by the person and how many calories they burned during this activity.
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

P=39.2205\, watt

E=374.948 \,cal

Explanation:

Given that:

  • force applied, F=66\,N
  • displacement, s=23.77\,m (length of a tennis court)
  • time taken for pushing, t = 40 s

Since, work is given by:

W=F.s

W=66\times 23.77

W=1568.82\,J

Now, power is given as:

P=\frac{W}{t}

P=\frac{1568.82}{40}

P=39.2205 \,watt

Calories consumed is:

E= 1568.82\times 0.239

E=374.948\, cal

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VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given \Delta x=7\ m

\theta=45 \°

Also, \Delta y=(3.5-2)=1.5\ m

a_x=0\ and\ a_y=-9.81\ m/s^2

Let us say the velocity in the x-direction is v_x and in the y-direction is v_y. And acceleration in the x-direction is a_x and in the y-direction is a_y.

Also, \Delta x\ and\ \Delta y is distance covered in x and y direction respectively. And t is the time taken by the ball to hit the backboard.

We can write v_x=v_0cos(45)\ and\ v_y=v_0sin(45). Where v_0 is velocity of ball.

Now,

\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt

\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}

Also,

\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s.

Plugging this value in

t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}

t=1.06\ seconds

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4 years ago
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

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where;

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Answer:

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The ratio is obtained by multiplying each side by \frac{m_{c}}{m_{b}}:

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\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }

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Answer:

 

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3 years ago
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