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3 years ago

Help please !!!! LIKE ASAP!

Physics

1 answer:

serg [7]3 years ago

3 0

The first diagram is Nuclear Fission which is the splitting of a heavy, unstable nucleus into 2 lighter nuclei.

The second diagram is Nuclear Fusion which is the process where 2 light nuclei combine together releasing a lot of energy.

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A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati

zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0

3 years ago

Read 2 more answers

Which statement is true for particles of the medium of an earth quake p wave

katen-ka-za [31]

I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.

8 0

3 years ago

A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance? Show

melamori03 [73]

3 seconds my dude i think

5 0

3 years ago

An airplane is flying in the direction 10° east of south at 701 km/hr. Find the component form ofthe velocity of the airplane, a

solniwko [45]

Answer:

The component form will be;

In the x-axis = 121.73 due west

In the y-axis = 690.35 due south

Explanation:

An image of the calculation has been attached

7 0

3 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago