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seropon [69]
3 years ago
10

A 60-kilogram man holding a 20-kilogram box rides on a skateboard at a speed of positive 7 meters per second. He throws the box

behind him, giving it a velocity of negative 5 meters per second, with respect to the ground. What is his velocity, in meters per second, after throwing the object?
Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

v_1=11\ m/s

Explanation:

It is given that,

Mass of the man, m₁ = 60 kg

Mass of the box, m₂ = 20 kg

Speed of man+ box, V = +7 m/s

Speed of the box, when it was thrown, v₂ = -5 m/s

To find,

Velocity of the man after throwing the box

Solution,

Let v₁ be the velocity of the man after throwing the box. By using the conservation of momentum, it can be calculated.

(m_1+m_2)V=m_1v_1+m_2v_2

(60+20)\times 7=60\times v_1+20\times (-5)

v_1=11\ m/s

Therefore, the velocity of the man after throwing the object is 11 m/s towards his initial direction.

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Answer:

W = N!/(n0! * n1!)

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Let n0 = number of particles in the lowest energy state

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Using this, we can say that N = n0 + n1

From this we can then express the weight, W of the close system by finding the factorials of each particles

W = N!/(n0! * n1!)

Hence, the weight W is expressed as W = N!/(n0! * n1!)

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Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th
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Answer:

r_{cm} = 0.074 m from the position of the center of the Sun

Explanation:

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distance between Sun and Jupiter is given as

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r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}

r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}

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3 0
3 years ago
According to the first rule, if a force pulls on one end of a rope, the tension in the rope equals the magnitude of the pulling
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Answer:

F.

Explanation:

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magnitude of tension in rope 1 is

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Hence the tension T1 is rope 1 is F.

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3 years ago
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Define average velocity.Immersive Reader
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Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
slava [35]

Answer:

69.68 N

Explanation:

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W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

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ki = initial kinetic energy

d = distance traveled

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vf = 8m/s

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d = 25m

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inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

7 0
3 years ago
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