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Gennadij [26K]
3 years ago
15

Explain the mathematical equation for KVL (The voltage of a circuit through each component in that circuit is proportional to th

e resistance).
Physics
1 answer:
deff fn [24]3 years ago
6 0

Explanation:

Kirchhoffs Voltage Rule, or KVL, explains that "the overall voltage throughout the circuit in any closed loop network is equal to the sum of all voltage decreases inside the same loop" which is equivalent to zero. In other words, the mathematical sum of all within the loop voltages must be equal to zero.

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What type of speed looks at a particular point in time?
bearhunter [10]
The answer is D because instantaneous means at a particular point in time
4 0
3 years ago
HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
Pls help I’m being timed!!
Firdavs [7]

Answer:

B

Explanation:

Heat increase molecular motion

6 0
3 years ago
Read 2 more answers
Ngan has a weight of 314.5 N on Mars and a weight 833.0 N on Earth. What is Ngan's mass on Earth?
mr_godi [17]

Ngan's mass on earth is 85kg.

Ngan has a weight on Mars = 14.5 N

Ngan’s weight on Earth = 833.0 N

Ngan’s mass on Earth = ?

<span>Fg,earth = mg(earth)</span>

<span>M = Fg,earth </span><span>/ g(earth)</span>

<span>M = 833.0 N / 9.8 m/s2</span>

<span>M = 85 kg</span>

4 0
3 years ago
Read 2 more answers
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
2 years ago
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