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2 years ago

True or false energy is always lost in a circuit

Physics

1 answer:

Llana [10]2 years ago

7 0

Explanation:

Electrons will gain energy as they are "pushed" from different points in the circuit. This energy is then lost when the electrons flow through circuit components such as a light bulb.

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The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a

bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = $\frac{1}{2} mv_1^{2}$ = $\frac{1}{2} *3*3.6^{2}$ = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = $\frac{1}{2} mv_2^{2}$ = $\frac{1}{2} *3*7.0^{2}$ = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

3 0

3 years ago

Read 2 more answers

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

3 years ago

Objects in free fall are weightless. true or false???

frez [133]

The answer is false

7 0

3 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

What do tripped circuit breakers and blown fuses indicate.

vladimir2022 [97]

Answer:

When a circuit breaker regularly trips or a fuse repeatedly blows, it is a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits. Or, it may indicate that your house has too few circuits and is in need of a service upgrade.

7 0

2 years ago