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ANEK [815]
3 years ago
8

Technician A says if a fuse burns out when a switch is turned on, a short circuit is present before the switch. Technician B say

s if a fuse burns out when a switch is turned on, a short circuit is present after the switch. Who is correct?
Physics
1 answer:
lord [1]3 years ago
7 0

Answer:

Technician B

Explanation:

This can be explained as a fuse is a protective device which blows in order to protect the circuit and the equipment from short circuiting.

Short circuit is a condition where a very heavy current flow through the line with no low line impedance which is not the intended path for the current to flow.

Thus when heavy current flows through a fuse it blows out thus disconnecting the path for the current flow.

Therefore, if a fuse burns out when switch is on, short circuit is present after the switch

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Does the mass of a parachute affect terminal velocity?
vovangra [49]

Answer:

The greater weight increases the terminal velocity by acting as an extra force against gravity and air resistance.

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2 years ago
Most of the life is spent in a cell development called?
tatuchka [14]
It's called cellular differentiation. I think.
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3 years ago
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A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel
larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\      (1)

R1 =  13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}            (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

R_{eq}=\frac{23V}{4A}=5.75\Omega

Now, you can calculate the unknown resistor R2 by using the equation (1):

\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega

hence, the resistance of the unknown resistor is 10.31Ω

8 0
3 years ago
Read 2 more answers
A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa
kotykmax [81]
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is v_i=180~m/s, the final velocity is v_i=0~m/s, and the total time of the motion is \Delta t=0.02~s, so the acceleration is given by
a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2 
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m
So, the bullet penetrates the sandbag 1.8 meters.
5 0
3 years ago
I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to
Vladimir79 [104]

Answer:

293k

Explanation:

In this question, we are asked to calculate the temperature to which the reaction must be heated to double the equilibrium constant.

To find this value, we will need to use the Van’t Hoff equation.

Please check attachment for complete solution

7 0
3 years ago
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