What is the acceleration of a car that is going at a steady speed of 60 mph?

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago

Two small children decide it would be fun to toss a couple of large cats at each other. Cat A (7kg) is thrown at 7m/s and cat B

Alex777 [14]

Answer:

V=4.7m/s

Explanations:

Let Ma mass of cat A=7kg

Va velocity of cat A=7m/s

Mb mass of cat b=6.1kg

VB velocity of cat b=2m/s

From conservation of linear momentum

MaVa+MbVb=(Ma+Mb)V

7*7+6.1*2=(7+6.1)V

61.2=13.1V

V=4.7m/s

3 0

3 years ago

A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant s

Monica [59]

Hey There,

Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"

Answer: C. Force Of Friction
B. Force

If This Helps May I Have Brainliest?</span>

7 0

3 years ago

Read 2 more answers

n an object. one force is 3N to the east and the other force is 9n to the west. what is the net force acting on the object

cricket20 [7]

Answer:

-6N

Explanation:

The force to the east is acting in the positive x-direction therefore it is positive. The force to the east is in the negative x-direction therefore it is negative. The net force is just the sum of the two so 3-9=-6

4 0

2 years ago

A cartoon shows two friends watching an unoccupied car in free fall after it has rolled off a diff. One friend says to the other

olga55 [171]

Answer:

The statement is not correct.

Explanation:

To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.

Data obtained from the question include:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 3 s

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 3)

v = 0 + 29.4

v = 29.4 m/s

Thus, the velocity of the car after 3 s is 29.4 m/s.

Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.

8 0

3 years ago