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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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A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0
3 years ago
Ask Your Teacher The height h in feet reached by a dolphin t seconds after breaking the surface of the water is given by h = −16
kirza4 [7]

Answer:

1 second

Explanation:

h = −16t² + 32t

When, h = 16

16 = −16t² + 32t

Divide each of the numbers by 16

1 = -1t² + 2t

Rearrange the equation

1t²-2t+1 = 0

Solving by the quadratic formula, we get

t=\frac{-(-2)\pm \sqrt{(-2)^2-4\times 1\times 1}}{2\times 1}\\\Rightarrow t=\frac{2\pm 0}{2}\\\Rightarrow t=1\ s

So, time taken by the dolphin to jump out of the water and touch the trainer's hand is 1 second.

3 0
4 years ago
Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin
maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

cos60 = \frac{F_x}{F}

now we have

F_x = F cos60

F_x = 0.50 F

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

cos45 = \frac{F_y}{F}

now we have

F_y = F cos45

F_y = 0.707 F

Now for the component along z axis we know that

F_x^2 + F_y^2 + F_z^2 = F^2

now plug in all components

(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2

0.5 F^2 + 0.25 F^2 + F_z^2 = F^2

F_z^2 = F^2(1 - 0.75)

F_z^2 = 0.25 F^2

F_z = 0.5 F

5 0
3 years ago
A 15 N force and a 45 N force act on in object in opposite directions. What is the net force on the object?
vovikov84 [41]

Answer:

30N in the direction the 45N acts.

Explanation:

Fnet = F1 + F2 (the vector sum of the forces)

Assigning a positive direction to the 45N force and a negative direction to the 15N force gives:

Fnet = 45 - 15

Fnet = 30N

Since the answer is positive, it is in the direction the 45N force acts.

7 0
3 years ago
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