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alexandr1967 [171]

3 years ago

11

Why is radiation often used to destroy cancer cells ?

1 answer:

zimovet [89]3 years ago

8 0

Answer:

C

Explanation:

Radiation affects both cancer cells and healthy cells, but it affects cancer cells more.

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A major process by which the Sun's energy is transported around the Earth system?

Setler79 [48]

Is it Conduction. Here's my theory e<span>nergy is transferred between the earth's surface and the atmosphere by Conduction, Convection and Radiation.</span>

3 0

3 years ago

A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa

NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

$mV_1+3mV_2=(m+3m)V_f=4mV_f$

$V_1+3V_2=4V_f$ and making $V_f$ the subject then

$V_f=\frac {V_1+3V_2}{4}$ and since $V_1$ is initial velocity of car, value given as 4 m/s, $V_2$ is the initial velocity of the three cars stuck together, value given as 2 m/s and $v_f$ is the final velocity which is unknown. By substitution

$V_f=\frac {4+(3\times2)}{4}=2.5 m/s$

(b)

Initial kinetic energy is given by

$\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ$

Final kinetic energy is given by

$\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ$

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

6 0

4 years ago

Read 2 more answers

What is the GPE of a 15,000 kg airplane sitting on the ground?

Vesna [10]

Answer:

C, it is not moving

it has no potential

7 0

3 years ago

A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a

gizmo_the_mogwai [7]

Answer:

Acceleration, $767.08\ m/s^2$

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

$v^2=u^2+2ah$

u = initial velocity=0 m/s

a = acceleration=g

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s$

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2$

So, the average acceleration of the ball during the time it is in contact is $767.08\ m/s^2$ .

4 0

3 years ago

Which of these objects has kinetic energy?

Damm [24]

Answer:

A ball moving through the air.

Explanation:

The ball has momentum which is a form of kinetic energy.

I don't know if that is correct, but I hope it helps!!!!

3 0

3 years ago

Read 2 more answers