What is transferred to an object when is done A:energy B:force C:effort D:resistance

What energy output objects work with the turbine?

Sonbull [250]

Answer:

Energy output from Solar panel is Electric Energy so any object which require electric energy an input will run by turbine. for example Electric Bulb,Electric Water Heaters.

hope it helps you!!

6 0

2 years ago

An elevator has a mass of 1000 Kg. What force is needed to accelerate it upward at a rate of 2 m/s/s?

zubka84 [21]

The force needed to accelerate an elevator upward at a rate of $2 m / s^{2}$ is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as $2 m / s^{2}$ , the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

$\text {Force}=\text {Mass} \times \text {Acceleration}$

$\text { Force }=1000 \times 2=2000 \mathrm{N}=2 \text { kilo Newon }$

So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of $2 m / s^{2}$ .

3 0

2 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

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

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$