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saul85 [17]
4 years ago
12

In a nuclear power plant, the nuclear reaction is kept from going critical by keeping the rate of reaction safe how to do the co

ntrol rod figure into this.
Physics
1 answer:
Svetlanka [38]4 years ago
5 0
The control rods absorb some of the neutrons that are
flying around, so those neutrons don't get to smash into
more Uranium nuclei and start more fissions.

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Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction
andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

3 0
3 years ago
The vibration produced days or even years before an earthquake
Margaret [11]
I don't know I guess its the plate tectonics 
7 0
3 years ago
What is a gravitational wave and why was it so hard to detect?
Alika [10]

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5 0
3 years ago
A truck accelerates from a stop to 27m/sec in 9 minutes. What was the trucks acceleration?
olchik [2.2K]
The acceleration is 3 m/s per minute, or 0.05 m/s per second.
5 0
4 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
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