What energy is directly dependent upon velocity and mass?

An electric heater has a Nichrome heating element with a resistance of 9 Ω at 20oC. When 112 V are applied, the electric current

maw [93]

Answer:

975.28 W.

Explanation:

Using,

R' = R(1+αΔt)....................... Equation 1

Where R' = Resistance at the final temperature, R = Resistance at the initial temperature, α = temperature coefficient of resistivity of Nichorome, Δt = Temperature rise.

Given: R = 9 Ω, α = 0.0004/°C, Δt = 1090-20 = 1070 °C

Substitute into equation 1

R' = 9(1+0.0004×1070)

R' = 9(1.428)

R' = 12.862 Ω.

Note: Operating wattage of the heater means the operating power of the heater

The power of the heater is given as,

P = V²/R'...................... Equation 2

Where P = Operating wattage of the heater, V = Voltage, R' = Operating resistance.

Given: V = 112 V, R' = 12.862 Ω

Substitute into equation 2

P = 112²/12.862

P = 975.28 W.

3 0

3 years ago

Which is greater than one liter

masya89 [10]

Megaliter > kiloliter > liter >centiliter >mililiter > deciliter > nanoliter

Have a nice day !

3 0

3 years ago

An astronaut weighing 248 lbs on Earth is on a mission to the Moon and Mars. (a) What would he weigh in newtons when he is on th

lesantik [10]

Answer:

$W_m=183.495N$

Explanation:

From the question we are told that:

Weight $W=248Ibs$

Mass of Weight $m= 248*0.453$

$m=112.344kg$

Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.

Therefore

The equation for Weight on Moon is given as

$W_m=m*g/6$

$W_m=122.344*\frac{9.8}{6}$

$W_m=183.495N$

3 0

3 years ago

Read 2 more answers

The acceleration of the object may be zero and the velocity of the object may not be equal to zero correct or not

Zinaida [17]

Answer:

Acceleration is the change in velocity, not the velocity itself; therefore, an object can have zero velocity but not zero acceleration because the velocity will be changing.

8 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a

Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

$M=-\frac{f_o}{f_e}$

Here, M is the magnification of an astronomical telescope, $f_e$ is the focal length of the eyepiece lens and $f_o$ is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put $f_o=74 cm$ and $f_e=2.4 cm$ in the above expression.

$M=-\frac{74}{2.4}$

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0

4 years ago