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2 years ago

If the resistance remains constant and the voltage doubles, what effect will that have on the power?

1 answer:

6 0

One of the handy formulas for electrical power is

Power = (Voltage)² / (Resistance)

Since the voltage is squared, if you double the voltage and
the resistance doesn't change, the power will increase by
the factor of (2²), which is 4 times the original power.

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Calculate the acceleration of a bus whose speed changes from 7 m/s to 16 m/s over a period of 5 s.

Bezzdna [24]

Let’s do this together!

Okay so the acceleration formula is vf-vi over time .

So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s

So the time is 5s so 9m/s divided into 5s is 1.8m/s/2

So the answer is 1.8m\s2

7 0

3 years ago

The graph shows two runners participating in a race.

Olin [163]

Answer:

'Daniela had a 5-meter head start, and Leonard caught up to her at 25 meters.'

Explanation:

hope that helps :)

6 0

2 years ago

Read 2 more answers

WHAT are the highest frequency and the lowest frequency pets of the EM spectrum?

Rus_ich [418]

Highest frequency EM waves: cosmic rays and gama rays

Lowest frequency EM waves:
Radio and Tv waves

8 0

3 years ago

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

$V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA$

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0

2 years ago

If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o

larisa86 [58]

Answer:

$f_k = 310N$

the answer is A.

Explanation:

Using the laws of newton:

∑F = ma

where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.

Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:

∑F = m(0)

∑F = 0

It means that:

F - $f_k$ = 0

where F is the force applied and $f_k$ is the friction force. Replacing the value of F, we get:

310N - $f_k$ = 0

Finally, solving for $f_k$ :

$f_k = 310N$

8 0

3 years ago