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3 years ago

What is meant by 'wasted energy'?

2 answers:

7 0

So it means that it wastes electricity e.g if you leave your computer on standby or TV on when your not using it it can waste alot of electricity

schepotkina [342]3 years ago

5 0

Wasted energy is an energy which is transformed without significant use.

You might be interested in

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

iren2701 [21]

Answer:

4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

$V(r) = \frac{4}{3} \ \pi \ r^3$

If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

$\Delta V(r) = 4 \ \pi \ r^2 \Delta r$

The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

$\frac{ 3 \Delta r }{ r}$

Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %

4 0

3 years ago

Among the alkali earth metals, the tendency to react with other substances

padilas [110]

Answer D
In alkali earth metals reacrivity increases from top to bottom (opposite of b)
This is because as you go down, the electron shells increase by 1 shell. The farther away a shell is from the nucleus, the higher its tendency to react.
D is true because the more reactive an alkali metal is, the more vigorous the reaction will be with water.

4 0

3 years ago

Read 2 more answers

. A ball is thrown downward at a speed of 20 m/s. Choosing

a_sh-v [17]

The final velocity is $v = 20 - gt$

The distance traveled by the ball at time t is $y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is $0 = (20)^2 - 2g(y-y_0)$

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

$v = 20 - gt$

The distance traveled by the ball at time t;

$y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is calculated as;

$v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)$

Learn more here: brainly.com/question/16878713

3 0

2 years ago

A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti

pochemuha

B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.

4 0

3 years ago

Read 2 more answers

The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star

Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

$PV=nRT$