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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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Explanation:

Depending on what you are hauling.

7 0
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In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

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3 years ago
A person makes a cup of coffee by first placing a 200 W electric immersion heater in 0.32 kg
alexdok [17]

Answer:

Answer is D. 8.04 x 10^4 J

Explanation:

1. D

2. A

3. D

4. B

5. C

6. B

7. D

8. C

9. B

10. D

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maks197457 [2]
Mass is the right answer
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What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms
kirza4 [7]

Answer:

F' = (3/2)F

Explanation:

the formula for the electric field strength is given as follows:

E = F/q

where,

E = Electric Field Strength

F = Force due to the electric field

q = magnitude of charge experiencing the force

Therefore,

F = E q   ---------------- equation (1)

Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:

F' = (E/2)(3 q)

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using equation (1)

<u>F' = (3/2)F</u>

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3 years ago
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