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The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced

Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

$K=\frac{1}{2}mv^2$ (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

$p=mv$ (2)

From eq.(1) we get

$v=\sqrt{\frac{2K}{m}}$

and substituting into (2),

$p=\sqrt{2mK}$

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

$K' = 8K$

- The mass is reduced by half:

$m'=\frac{m}{2}$

Substituting, we find the new momentum:

$p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p$

So, the momentum increases by a factor of 2.

Learn more about momentum and kinetic energy:

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4 0

3 years ago

Que es la virtud para aristoteles

neonofarm [45]

Can you write in English so I can help you

7 0

3 years ago

What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of

Murljashka [212]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

$A=\frac{\rho l}{R}\\ \frac{\pi d^2}{4}=\frac{\rho l}{R}\\ l=\frac{\pi d^2 R}{4 \rho}$

Here $d=0.15, R=15, \rho=1.68(10^{-8})$ .

Substituting numerical values,

$l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})} \\ l=15.78$

The length of the wire is $15.78 \;\;m$

5 0

4 years ago

How much work was done to accelerate a 2.2 kg object from rest to 26m/s?

Makovka662 [10]

Answer:

Explanation:

work done = change in kinetic energy, assuming there are no energy losses.

work done = 1/2 × m × change in v^2

= 1/2 × 2.2 × ( 26^2 - 0^2)

= 1/2 × 2.2 × 676

= 743.6 joules

4 0

2 years ago

A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. If the aver

Natasha_Volkova [10]

Answer:

$A_{area}=0.243m^{2}$

Explanation:

Given data

Average intensity of sunlight on one day I=700 W/m²

Power P=170 W

To find

Area A

Solution

As we know that:

$I_{intensity}=\frac{P_{power}}{A_{area}}\\ A_{area}=\frac{P_{power}}{I_{intensity}} \\ A_{area}=\frac{170W}{700W/m^{2} } \\ A_{area}=0.243m^{2}$

8 0

4 years ago

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