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3 years ago

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?

Physics

2 answers:

Alex_Xolod [135]3 years ago

4 0

A.900 watts That would be your correct answer

Bess [88]3 years ago

3 0

Answer : Power = 900 watts

Explanation :

It is given that,

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. The power delivered is given by the following relation :

$P=VI$

$P=I^2R$

$P=\dfrac{V^2}{R}$

In this case, we use relation (1)

$P=12\ V\times 75\ A$

$P=900\ W$

The correct option is (A) " 900 Watts ".

Hence, this is the required solution.

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A 5kg ball is on top of the school building at a height of 40m above the ground.

mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0

2 years ago

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on i

Sergeu [11.5K]

Answer:

$a = 2.77~{\rm m/s^2}$

Explanation:

Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.

The moment of inertia of a ring is

$I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127$

The mass on the left is heavier, that is the pulley is rotating counterclockwise.

By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.

$\tau = I \alpha$

Here, the net torque is the sum of the weight on the left and the weight on the right.

$\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}$

Applying Newton's Second Law gives the angular acceleration

$\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}$

The relation between angular acceleration and linear acceleration is

$a = \alpha R$

Then, the linear acceleration of the masses is

$a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}$

5 0

3 years ago

Make the following conversion.

anastassius [24]

The answer is 0.00230 cm

4 0

3 years ago

Read 2 more answers

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creativ13 [48]

Answer:

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3 years ago

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Alexandra [31]

Answer:

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8 0

2 years ago