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vovangra [49]
2 years ago
9

A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

out 10 seconds, with an average speed of 10 m/s.) (a) What is the average horizontal component of the force that the ground exerts on the runner's shoes? (b) How much work is done on the point-particle system by this force?
Physics
1 answer:
Darya [45]2 years ago
6 0

Answer:

a. F=126N

b. E_K=1323J

Explanation:

Given:

m=54kg

v=7 m/s

t= 3s

The runner force average to find given the equations

a.

F=m*a

a=\frac{v}{t}

F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}

F=126N

b.

Work done by the system by this force so

W=F*d

W=E_K

E_K=\frac{1}{2}*m*v^2

E_K=\frac{1}{2}*54kg*(7m/s)^2

E_K=1323J

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Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

E_{m} = E_{k} + E_{p}

E_{m} = \frac{1}{2}v^{2} + gh

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0
3 years ago
Two forces are acting on a wheelbarrow. One force is pushing to the right and an equal force is pushing to the left. What can yo
andreev551 [17]

-- As far as we know, the forces on the wheelbarrow are balanced.

-- That tells us that the net force on the wheelbarrow is zero, just
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-- That tells us that the wheelbarrow's acceleration is zero ... its
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-- That tells us that the wheelbarrow is moving in a straight line
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6 0
2 years ago
A white-blue star is hotter than a red star.
Katena32 [7]
This is true!!

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4 0
3 years ago
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PLEASE HELP ME WITH THIS ONE QUESTION
sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But,  the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0
2 years ago
if an object 18mm high is placed 12mm from a diverging lens and the image is formed 4mm in front of the lens what is the height
brilliants [131]

To solve this problem we use an amplification formula for divergent lenses

M = \frac{i}{o} = \frac{h'}{h}

Where:

i: distance of the image to the lens

o: Distance from the object to the lens

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M = \frac{4mm}{12mm} = \frac{h'}{18mm}

h'= 18mm\frac{4mm}{12mm}

h '= 6 mm

The height is 6 mm

5 0
2 years ago
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