How much power does a machine have that does 15204 J of work in 40 seconds?

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

A student uses an electronic force sensor to study how much force the student’s finger can apply to a specific location. The stu

melisa1 [442]

Answer:

B. Trial 2

Explanation:

Trial 2, because the student’s finger applied the largest force to the sensor.

Because the trial 2 student finger applied to largest force.

7 0

2 years ago

Read 2 more answers

The thrust of a certain boat’s engine generates a power of 10kW as the boat moves at constant speed 10ms through the water of a

Lunna [17]

Answer:

The change in power is 4400 W.

Explanation:

Given that,

Power = 10 kW

Speed = 10 m/s

Increases speed = 12 m/s

Given equation is,

$F=kv$

We know that,

The power is,

$P=Fv$

Put the value of F into the formula

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

We need to calculate the new power

Using formula for power

$\dfrac{P}{P'}=\dfrac{v^2}{v'^2}$

Put the value into the formula

$\dfrac{10}{P'}=(\dfrac{10}{12})^2$

$P'=(\dfrac{12}{10})^2\times10$

$P'=14.4\ kW$

We need to calculate the change in power

Using formula of change in power

$\Delta P=P'-P$

Put the value into the formula

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Hence, The change in power is 4400 W.

6 0

3 years ago

Astronomers were at first surprised to find complicated molecules in the interstellar medium. They thought ultra-violet light fr

jeka57 [31]

Answer:

The dust present in the clouds.

Explanation:

The complicated composition molecules that can be found in space are generally associated with clouds of dust. The significant amount of dust in these clouds provides protection not only for these molecules, but for any body that makes up or is associated with dust clouds.

It is exactly this dust that protects the molecules against the action of ultraviolet rays.

8 0

2 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

2 years ago