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3 years ago

Which resistors in the circuit must have the same amount of energy per unit charge across them?

Physics

2 answers:

egoroff_w [7]3 years ago

6 0

The correct answer is D): resistors B and C. In fact, the energy per unit charge is equal to the voltage across the resistors. The electrical potential energy is in fact: $U=qV$ , where q is the charge and V is the voltage; by rearranging the equation, we have $V= \frac{U}{q}$ , therefore the voltage is the energy per unit charge. In the circuit in the figure, the resistors B and C are connected to the same points of the circuit (they are connected in parallel), therefore they have the same voltage, so they have the same energy per unit charge.

Oksi-84 [34.3K]3 years ago

6 0

Answer:

D. B and C

Explanation:

We have to find that which of the given is having same amount of energy per unit charge.

So we know that energy per unit charge is defined as potential.

$V = \frac{U}{q}$

so here we have to find in this circuit that which of the given resistors are at same potential.

So here we can see that resistance B and C are parallel to each other in this circuit and all the resistors which are in parallel to each other are always having same potential.

So B and C are having same potential energy per unit charge

You might be interested in

A pendulum has 895 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

LuckyWell [14K]

Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.

Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

Where m is the mass of the pendulum.

g is the acceleration due to gravity

h is the height from the bottom z the ground.

At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

Where m is the mass of the pendulum

v is the velocity of the pendulum

At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.

Also as it has been mentioned energy can neither be created nor destroyed hence the entire potential energy is converted to kinetic energy at the bottom and would be equivalent to 895 J.

7 0

4 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0

3 years ago

1. A 20m steel wire is stretched to 20.0m by

Eduardwww [97]

Answer:

Force to stretch the wire is 250 N

Explanation:

As we know that modulus of elasticity will remain the same for the wire if the applied stretch to the wire is within elastic limit

So we will have

$\frac{F L}{\Delta L A} = constant$

now we have

$\frac{F_1 L}{\Delta L_1 A} = \frac{F_2 L}{\Delta L_2 A}$

so we can write it as

$F_2 = \frac{F_1 L_2}{L_1}$

$F_2 = \frac{50 (0.05)}{0.01}$

$F_2 = 250 N$

7 0

3 years ago

As rotational speed increases, thrust____?

never [62]

Increases exponentially is your correct answer

6 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

Read 2 more answers