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vaieri [72.5K]
3 years ago
9

A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of

the electrons collide with a target, but a large majority pass through the target and impact a beam dump at the end of the accelerator. In one experiment the beam dump measured charge accumulating at a rate of -2.0 nC/s. How many electrons traveled down the accelerator during the 2.0 h run
Physics
1 answer:
diamong [38]3 years ago
7 0

Answer:i

9E13 ELECTRONS

Explanation:

First we find the number of charges in two hrs

Which is - 2x 7200=- 14400NC

So no of electrons is now q/e

-144x10^-7/-1.6*10^-19

= 9*10^13

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1 A ball is rolling along the floor with a constant velocity of 8 m/s. How far will it have gone after
Dennis_Churaev [7]

Answer:

448 meters

Explanation:

every second it moves 8 meters, so all you have to do is multiply 56x8 or 8x56 either way it is the same thing and you will get the same answer

3 0
2 years ago
Explain the difference between the three types of friction
USPshnik [31]
Static Friction

It is the friction that exists between a stationary object and the surface on which it's resting.


Sliding friction

It is the resistance created by two objects sliding against each other.

Rolling friction:-

It is the force resisting the motion when a body rolls on a surface.


hope this helps x
8 0
3 years ago
Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d
BartSMP [9]

La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (L), en metros:

L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})] (1)

Donde:

  • L_{o} - Longitud inicial del puente, en metros.
  • \alpha - Coeficiente de dilatación, sin unidad.
  • T_{o} - Temperatura inicial, en grados Celsius.
  • T_{f} - Temperatura final, en grados Celsius.

Si tenemos que L_{o} = 100\,m, \alpha = 11.5\times 10^{-6}, T_{o} = 8\,^{\circ}C y T_{f} = 24\,^{\circ}C, entonces la longitud final del puente de acero es:

L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]

L = 100.018\,m

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0
2 years ago
Three bulbs are connected in series to a 4.5 V battery. The second bulb burns out. The
Oksi-84 [34.3K]

Answer:

0v

Explanation:

3 0
3 years ago
Suppose that you have a spring gun that you use to launch a small metal ball. You try the first two settings of the gun. The fir
Ivan

Answer:

The distance s of how far the ball will go at the highest setting = 2.25m

Explanation:

Let consider x to be the representative of the compression and the distance to be s

Recall that:

\dfrac{1}{2}\times K \times  x^2 = mgs +c

By cross multiplying

K \times  x^2 = 2(mgs +c)

K \times  x^2 = 2\times 9.81(ms) +2c

K \times  x^2 = 19.62(ms) +2c

x^2 = A \times  s+B

Thus, for the first setting

x = 1 , s = 0.25

for the second setting

x = 2,   s = 1

1 = 0.25A + B ---  (1)

4 = A + B    ----- (2)

From (1); let B =  1 - 0.25A  and substitute it into (2)

4 = A + 1 - 0.25 A

4 - 1 = A - 0.25 A

3 = 0.75 A

A = 3/0.75

A = 4

From (2)

4 = A + B

4 = 4 + B

B = 4 - 4

B = 0

Therefore, for the highest setting, where x = 3

Then :

x^2 = A \times  s+B will be:

3² =   4s + 0

9 = 4s

s = 9/4

s = 2.25 m

∴

The distance s of how far the ball will go at the highest setting = 2.25m

6 0
3 years ago
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