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3 years ago

True or false?nuclear fusion happens when uranium combines

Physics

2 answers:

Leni [432]3 years ago

6 0

The correct answer is True

Arturiano [62]3 years ago

5 0

Answer: false

Explanation:

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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 test charge is introduced into the electric field of a charge. It feels a force of 2F where the electric field is 2E. What wou

Firdavs [7]

Surrounding every charge (or group of charges) is a thing, called an electric field. (it is a vector thing). E. Definition: The electric field at a point. E in empty space is a vector quantity which can be measured by the following procedure: place a small test charge q at that point, measure the force on q due to all other charges.

3 0

4 years ago

Read 2 more answers

Joey drives his Skidoo 13 kilometres north. He stops for lunch and then drives 10kilometres south. What distance did he cover? W

olganol [36]

Answer:

Total distance covered (scalar quantity) = 23 km

Displacement (vector quantity) = 3 km north from the original starting point

Explanation:

Since he drove 13 km north and then 10 km south, the total distance he cover in his drive was: 13 km + 10 km = 23 km

On the other hand, his displacement was 3 km north from where he started.

7 0

3 years ago

Which statements describe the sun? Check all that apply

lesya692 [45]

Answer:

All but 4 I believe

Explanation:

8 0

3 years ago

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A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.841 rad/s.0.841 rad/s. You, with a mass

kvv77 [185]

Answer:

The total angular momentum of the system is $217.46\ kg-m^2/s$ .

Explanation:

Given that,

Angular speed = 0.841 rad/s

Mass of platform = 72.1 kg

Speed = 1.17 m/s

Mass of poodle = 20.3 kg

Mass of mutt = 18.5 kg

Distance =3/4 of the platform's radius from the center

Mass of disk = 92.5 kg

radius = 1.87 m

Angular momentum is the product of moment of inertia and angular speed.

We need to calculate the angular momentum of person

$L_{person} = I\omega$

$L_{person}=\dfrac{1}{2}mr^2\times\omega$

$L_{person}=\dfrac{1}{2}\times72.1\times(\dfrac{v}{\omega})^2\times\omega$

$L_{person}=\dfrac{1}{2}\times72.1\times(\dfrac{1.17}{0.841})^2\times0.841$

$L_{person}=58.679\ kg m^2/s$

We need to calculate the angular momentum of platform

$L_{platform}=\dfrac{1}{2}mr^2\times\omega$

Put the value into the formula

$L_{platform}=\dfrac{1}{2}\times92.5\times1.87^2\times0.841$

$L_{platform}=136.02\ kg m^2/s$

We need to calculate the angular momentum of poodle

$L_{poodle}=\dfrac{1}{2}mr^2\times\omega$

Put the value into the formula

$L_{poodle}=\dfrac{1}{2}\times20.3\times(\dfrac{1.87}{2})^2\times0.841$

$L_{poodle}=7.4625\ kg m^2/s$

We need to calculate the angular momentum of mutt

$L_{mutt}=\dfrac{1}{2}mr^2\times\omega$

$L_{mutt}=\dfrac{1}{2}\times18.5\times(\dfrac{3}{4}\times1.87)^2\times0.841$

$L_{mutt}=15.302\ kg m^2/s$

We need to calculate the total angular momentum

$L=L_{person}+L_{platform}+L_{poodle}+L_{mutt}$

$L=58.679+136.02+7.4625+15.302$

$L=217.46\ kg-m^2/s$

Hence, The total angular momentum of the system is $217.46\ kg-m^2/s$ .

3 0

3 years ago

True or false?nuclear fusion happens when uranium combines ​

True or false?nuclear fusion happens when uranium combines