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3 years ago

CHAPTER 6: KINETICS OF A PARTICLE

Physics

1 answer:

vekshin13 years ago

8 0

Explanation:

Work done by winch = kinetic energy of car

∫ T ds = ½ mv²

∫ 225s ds = ½ mv²

225/2 s² = ½ mv²

225 s² = mv²

v = 15s / √m

Given s = 10 m and m = 2500 kg:

v = 15 (10) / √2500

v = 3 m/s

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t

dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

$P=m*v$

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

$(m_{1}*v_{1})=(m_{2}*v_{2})$

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

$(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]$

8 0

3 years ago

A circular test track for cars has a circumference of 3.5 km . A car travels around the track from the southernmost point to the

Marta_Voda [28]

<span>The car would have traveled exactly one-half of the circumference of the track, since it would have gone from one extreme point to its opposite extreme point. This would be equal to (3.5 / 2), or 1.75 km. The northernmost point would be 1.75km away from the southernmost point.</span>

8 0

4 years ago

If an object has a mass of 20 kg, what is the force of gravity acting on it on earth?

stealth61 [152]

That force is what most people would call the object's "weight".

Wherever the object is, its weight is

(mass) x (acceleration due to local gravity) .

On Earth, the acceleration due to gravity is 9.8 m/s² . (rounded)

The object's weight is (20 kg) x (9.8 m/s²) = 196 newtons .

(about 44.1 pounds)

7 0

3 years ago

Read 2 more answers

A runner starts from rest and in 2 s reaches a speed of 7 m/s. If we assume that the speed changed at a constant rate (constant

PtichkaEL [24]

Answer:

$v=3.5\frac{m}{s}$