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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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Answer:

Let No be initial no of atoms

N = N0 / 2      after 1 half-life

N = N0 / 4     after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

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3 years ago
A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.​
Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
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lisov135 [29]

C.There is an invisible magnetic field around the magnet where it exerts magnetic force

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3 years ago
Waves that hit a fixed boundary return to the starting point on the____
Jet001 [13]

Answer:

A: Opposite

Explanation:

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