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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:

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Describe a technology used in space exploration.
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Answer:

High speed optical communication technology

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3 years ago
2. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10-9 C and the oth
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Answer:

A. -2.16 * 10^(-5) N

B. 9 * 10^(-7) N

Explanation:

Parameters given:

Distance between their centres, r = 0.3 m

Charge in first sphere, Q1 = 12 * 10^(-9) C

Charge in second sphere, Q2 = -18 * 10^(-9) C

A. Electrostatic force exerted on one sphere by the other is:

F = (k * Q1 * Q2) / r²

F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²

F = -2.16 * 10^(-5) N

B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:

Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))

= - 6 * 10^(-9) C

Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C

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I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app
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The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

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  • final velocity of the sprinter, v = 27 km/h
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Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

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