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4 years ago

Do amps in circuit change if something is removed from the circuit for example:

Physics

1 answer:

AlekseyPX4 years ago

8 0

Answer:

Yes, the current will change if one light bulb is removed from the circuit.

Explanation:

Recall that each light bulb possess in fact resistance, and electric energy is converted into light and heat as current circulates through each. therefore, removing one light bulb will change the total resistance of the circuit , thus affecting the current through it according to Ohm's Law.

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Two stationary carts are tied together by a thread with a compressed spring held between them. When the thread is cut, the two c

Fantom [35]

Answer:

1.45 m/s

Explanation:

The spring exerts equal and opposite forces on the carts.

Therefore, the impulses on the carts are equal and opposite.

J₁ = -J₂

m₁Δv₁ = -m₂Δv₂

(3.00 kg) (0.82 m/s − 0 m/s) = -(1.70 kg) (v − 0 m/s)

v = -1.45 m/s

The magnitude of the second cart's velocity is 1.45 m/s.

8 0

4 years ago

A driver without a seat belt getting thrown from the car in a collision is an example of Newton’s?

kupik [55]

Newton’s first law. An object stays in motion unless acted on by an external force or in this case the driver was in motion but the sudden crash stoped the car but not the driver

5 0

3 years ago

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

6 0

3 years ago

A block of mass 8 m can move without friction

nekit [7.7K]

Answer:

Let M1 = 8 kg and M2 = 34 kg

F = M a = (M1 + M2) a

F = M2 g the net force accelerating the system

M2 g = (M1 + M2) a

a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2

5 0

2 years ago

A circular conducting loop with a radius of 0.10 m and a small gap filled with a 10.0 ȍresistor is oriented in the xy-plane. If

lidiya [134]

Answer:

The magnitude of the current is 5.45 mA.

Explanation:

Given that,

Resistance = 10.0 ohm

Radius = 0.10 m

Magnetic field = 1.0 T

Angle = 30°

Increase magnetic field = 7.0 T

Time t = 3.0 s

Number of turns = 1

We need to calculate the initial flux

Using formula of flux

$\phi=NB_{1}A\cos\theta$

Put the value into the formula

$\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}$

$\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.027\ wb$

We need to calculate the final flux

$\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.1904\ wb$

We need to calculate the induced emf

Using formula of emf

$\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}$

Put the value into the formula

$\epsilon=\dfrac{0.1904-0.027}{3.0}$

$\epsilon=0.0545\ V$

We need to calculate the current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.0545}{10.0}$

$I=5.45\ mA$

Hence, The magnitude of the current is 5.45 mA.

4 0

3 years ago