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3 years ago

How much work can a 22kw car engine do in 60 seconds

Physics

2 answers:

const2013 [10]3 years ago

8 0

Answer:

Work done = 1320000J or 1.32 × 10^6J

Explanation:

power =22kw = 22000w

Time = 60seconds

From: power = work done/ Time

work done = power × time

work done = 22000 × 60

work done = 1320000J or 1.32 ×10^6 J

Sindrei [870]3 years ago

6 0

Answer:

13.2 x 10^5 J

Explanation:

Power = work/time

Given

Power = 22kw

= 22 x 1000 = 22000w

Time = 60 secs

Therefore

22000 = work/60

Cross multiply

Work = 22000 x 60

= 1320000

= 13.2 x 10^5J

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Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion o

denis-greek [22]

The center of mass isn't affected by the explosion.

To find the answer, we need to know about the trajectory of motion at zero external force.

<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>

When there's no net external force acting on an object, its momentum doesn't change with time.
As its momentum doesn't change, so it continues with the original trajectory.

<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>

When a firework is exploded, its internal forces are changed, but there's no external force.
So, although the fragments follow different trajectories, but the trajectory of center of mass remains unchanged.

Thus, we can conclude that the center of mass isn't affected by the explosion.

Learn more about the trajectory of exploded firework here:

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2 years ago

A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci

Galina-37 [17]

Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf - vi ) / t

Solution

6731 = 942*(vf - 0)/21 Multiply both sides by 21

6731 * 21 = 942*vf

141351 = 942*vf Divide by 942

141351/942 = vf

vf = 151 m/s

6 0

3 years ago

In what two ways can we increase the potential difference?

vredina [299]

Yes I hope this helps

4 0

3 years ago

Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica

DochEvi [55]

Answer:

So the specific heat of the liquid B is greater than that of A.

Explanation:

Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

$Q=m.c.\Delta T$

where:

m = mass of the body

c = specific heat of the body

$\Delta T=$ change in temperature of the body

The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.

So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.

$\Delta T=\frac{Q}{m} \times \frac{1}{c}$

$\Delta T\propto\frac{1}{c}$

So the specific heat of the liquid B is greater than that of A.

5 0

3 years ago

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

3 years ago