All categories

3 years ago

In what two instances does power output become greater

1 answer:

5 0

The two instances wherein power output becomes greater are: "<span>when more work is done in a given amount of and time when the time it takes to do a certain amount of work is decreased." In addition, power output is typically measured in horsepower, watts or foot-pounds per minute.</span>

You might be interested in

EARTH SCIENCE HELP FAST PLEASE

xxTIMURxx [149]

I think its C not sure tho

8 0

3 years ago

Describe how the total mass of the particles before the reaction occurs compares to the total mass of the particles produced by

zzz [600]

The law of conservation of mass says you don't create or destroy atoms in a reaction. So the total mass of reagents equals the total mass of the products.

4 0

3 years ago

If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo

vovangra [49]

Answer:

13 N

Explanation:

The Net Force of an object should be the difference between the forces applied to the object if the object is not in equilibrium. This object is not in equilibrium so therefore by finding the difference between the forces, you will find your answer. 20 N - 7 N = 13 N.

6 0

3 years ago

Read 2 more answers

The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

6 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

3 years ago