What is the angular acceleration experienced by a uniform solid disc of mass 5-kg and radius 20 cm when a net

How much power is needed to lift the 200-N object to a height of 4 m in 4 s?

Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

3 0

3 years ago

A proton is held at rest in a uniform electric field. When it is released, the proton will lose?

nydimaria [60]

A proton is held at rest in a uniform electric field. When it is released, the proton will lose its kinetic energy.

Kinetic energy

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.

To learn more about kinetic energy refer here:

brainly.com/question/11301578

#SPJ4

5 0

1 year ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago

A lamp has a power of 120 W and is left on for 24 hours . A television has a power of 350 W and is left on for three hours . com

Paladinen [302]

This is a power problem which requires the rearranging of a formula. The lamps energy used is 5 N, and the TV’s usage is 116.7 N (rounded from 116.6666repeating). Here my work:

7 0

3 years ago

Read 2 more answers

Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$