Ask question

Ask question

All categories

1 year ago

5

A motor does a total of 480 J of work if 5 s to lift a 12 clock to the top of a ramp. The average power developed by the motor i

s

2 answers:

yawa3891 [41]1 year ago

7 0

Answer:

ok,lets go.

Explanation:

Average power =work/time

P=480j/5s

P=96W

marta [7]1 year ago

3 0

Answer:

Power = work / time

P = 480 / 5 = 96W

You might be interested in

Conduction is the transfer of heat in:<br> 1 gases<br> 2 liquids<br> 3 solids

kykrilka [37]

Answer:

Conduction heat transfer is the transfer of <em>heat by means of molecular excitement within a material without bulk motion</em> of the matter.

Explanation:

Conduction heat transfer in gases and liquids is due to the collisions and diffusion of the molecules during heir random motion.

4 0

2 years ago

A 180-ohm resistor has 0.10 A of current in it. what is the potential difference across the resistor

Firlakuza [10]

We know V=IR (Ohm's law).

We are given R=180Ω and I=0.1A, then V=(0.1AΩ)(180Ω). Therefore

V=18V

5 0

3 years ago

How do human break N2 into a useable form

Fiesta28 [93]

A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>

<span />

6 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

2 years ago

How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w

sergij07 [2.7K]

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.

4 0

2 years ago