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iogann1982 [59]

2 years ago

5

In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire

ction of the frictional force?
A. 20 newtons in the direction of the applied force
B. 20 newtons opposite to the direction of the applied force
C. 20 newtons perpendicular to the direction of the applied force
D. 20 newtons in two directions, perpendicular and in the direction of the applied force
E. No friction is acting on the block.

2 answers:

ololo11 [35]2 years ago

8 0

The answer is B because it is true my teacher literally told me this last week! Good thing I remember

sladkih [1.3K]2 years ago

5 0

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

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A transformer with X turns in primary coil and Y turns in secondary coil is used to change the magnitude of voltage to 240 V. Ca

Vinil7 [7]

The input voltage is 120 V and the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.

<h3>Input voltage </h3>

The input voltage of the transformer is the voltage of the primary coil and it is calculated as follows;

Ns/Np = Es/Ep

where;

Ns is the number of turn in the secondary coil
Np is the number of turn in the primary coil
Es is the secondary voltage
Ep is the primary voltage

2X/X = 240/Ep

2 = 240/Ep

Ep = 240/2

Ep = 120 V

Thus, the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.

Learn more about transformer here: brainly.com/question/25886292

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2 years ago

What is the weight of a 15kg dog on earth?

Mariulka [41]

In physics the standard unit of weight is Newton, and the standard unit of mass is the kilogram. On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.

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3 years ago

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.

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bearhunter [10]

No petroleum is a liquid

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