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3 years ago

How much power does an electric device use if the current is 36.0 amps and the resistance is 3.9 ohms?

1 answer:

dsp733 years ago

8 0

Question: How much power does an electric device use if the current is 36.0 amps and the resistance is 3.9 ohms? 

How?:

Equation: P = I^2 R

Meanings:
P = Power in Watts
I = Current in Ampere
R = Resistance in ohms.

Plugged in: P = 36^2 x 3.9 = 5054.4

Answer: P= 5100 watts.

HOPE THIS HELPS! ^_^

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Which statement correctly describes the differences between positive and negative acceleration?

IceJOKER [234]

Answer: c) Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.

Hey

On Earth, you can move or you can not move. if you are moving 50 mph that means relative to Earth (not the Andromeda galaxy). When you start moving (accelerating) you are now moving relative to Earth. when you start slowing down (decelerating [most scientists just say you have negative acceleration]) you are starting to match your velocity to Earth's velocity.

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3 years ago

A gas was compressed to 30.0 mL at 1.5 atm from 65 mL. What was the original pressure?

astra-53 [7]

A gas as 30.0 Ml because pile of 1.4 add 1 that equals. To 1.5 and remove 10 so thebpresssure is 1399.93

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3 years ago

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

For the course of upward acceleration:

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

2 years ago

Imagine how mercury might be different if it had the same mass as earth. Explain.

maxonik [38]

I believe if it were heavier with more mass, then the sun would pull it in and there would be no mercury. It might also be hotter.

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3 years ago