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balandron [24]
3 years ago
5

A bulb is rated as 25Watt -250 Volt. On connecting this bulb to a mains of 250 volts, what will be the value of current (I) flow

ing in it, and how much resistance (R) will the bulb provide to this flow of current?
Physics
1 answer:
dangina [55]3 years ago
7 0

Answer:

I=0.1          R=2500

Explanation:

Resistance

R=\frac{V^{2} x}{P} \\=\frac{250^{2} }{25} \\=2500

Current

I=\frac{250}{2500} \\=0.1

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The drawing below shows two different types of pulley systems designed to lift a weight. In pulley system A, the end of the rope
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What is Biomass?<br> What is Biofuel?<br> Please help me science students
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Biomass-Total of mass of organisms in a given area/volume

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2 years ago
A 50 N object is on Earth. What is its mass?
faust18 [17]

Answer:   5.10 KG

Explanation:  W = 50 N

                       BY FORMULA

                         W = MG

                          M = W/G

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7 0
2 years ago
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
2 years ago
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