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Karolina [17]
2 years ago
12

Pls need an answer asap

Physics
1 answer:
Andrei [34K]2 years ago
3 0

c] Electrons

Electron gets transferred

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A 15 kg mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done
sashaice [31]
Kinetic energy = (1/2) (mass) x (speed)²

At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules

At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules

The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.

That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
8 0
3 years ago
Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller element
Zolol [24]

Answer:

<em>Element C will be best for a nuclear fission reaction</em>

Explanation:

<em>Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process.</em> For the reaction to be continuous in a chain reaction,<em> the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission</em>. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

5 0
3 years ago
Given a block of glass that is semi-circular, a laser pointer, protractor/ruler, and index card, design your own experiment to o
Charra [1.4K]

Answer:

we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.

Explanation:

To design the experiment of measuring the critical angle, we describe the phenomenon, when the light passes from a medium with a higher refractive index to one with a lower index, it separates from the normal one and the Critical Angle is defined as the Angle for which the refraction occurs at 90º

            n₂ sin θ₂ = n₁ sin 90

           n₁ / n₂ = sin θ₂

As we can see, we have to measure the angle with which the laser touches the exit surface of the glass block.

Design of the experiment:

    We place the glass block on the ramp and at the top we hit the conveyor for half the angle, we climb the block on the ramp and see that the angle of incidence of lightning on the exit face changes, part of the beam comes out of the glass , we see it by dispersion in the particles of dirty in the air; Maybe the conveyor or the laser should be moved slightly so that the beam touches the point of origin on the conveyor.

   

   When we go up the ramp there is a point where the beam is reflected inside the block, we carefully step back to the point where the beam is horizontal, we measure this angle which is our critical angle.

5 0
3 years ago
Read 2 more answers
Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca
Brut [27]

Let say the two train cars are of masses m_1 and m_2

now if the speed of two cars are v_1 and v_2

then we can say that the momentum of two cars before they collide is given by

P = m_1v_1 - m_2v_2

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

P = m_1v_1 - m_2v_2

5 0
3 years ago
Read 2 more answers
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
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