A motor exerts a force of 12,00N to lift an elevator 8.0m in 7.0secs what is the power produced by the motor?

At a certain elevation, the ________ , the air becomes saturated and water-vapor molecules ________.

andriy [413]

At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.

As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.

4 0

3 years ago

The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?

pishuonlain [190]

The average dissipated power in a resistor in a ac circuit is:
$P=I_{rms}^2 R$
where R is the resistance, and $I_{rms}$ is the root mean square current, defined as
$I_{rms} = \frac{I_0}{\sqrt{2}}$
where $I_0$ is the peak value of the current. Substituting the second formula into the first one, we find
$P=( \frac{I_0}{\sqrt{2} } )^2 R = \frac{1}{2} I_0^2 R$
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
$I_0 = \sqrt{ \frac{2 P}{R} }= \sqrt{ \frac{2 \cdot 2.0 W}{47 \Omega} }=0.29 A$

4 0

2 years ago

Who was the carthaginian general who used elephants to cross the alps in the second punic war

goldfiish [28.3K]

<span>the answer is Hannibal</span>

3 0

3 years ago

Read 2 more answers

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

2 years ago

A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is

Rudik [331]

Answer:

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 × 10² N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg(h+x) = 0.5 × k × x² ...................1

here x is compression in spring

so put here value in equation 1 we get

1.8 × 9.8 × (2.25+x) = 0.5 × 2× 10² × x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m

3 0

3 years ago