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svetoff [14.1K]

3 years ago

14

Suppose 3 neutrons are released by an atom of U-235 in the first stage of a fission chain reaction. Each of these neutrons has e

nough energy to cause another fission reaction in another nucleus of U-235, releasing more neutrons. And if all of those 2nd stage neutrons cause other nuclei to fission in a 3rd stage, how many total U-235 atoms will have undergone fission after these 3 stages of the chain reaction?
A. 27
B. 9
C. 11
D. 12

2 answers:

8090 [49]3 years ago

5 0

Is not A.) 27 don't know the answer but it is not (A)

DIA [1.3K]3 years ago

4 0

The answer is A.) 27<span />

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A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free

Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R from equation (1)

T = I x a / R² .... (2)

mg - T = ma .... (3)

Substitute the value of T from equation (2) in equation (3)

$mg - \frac{Ia}{R^{2}}=ma$

$a=\frac{mg}{m+\frac{I}{R^{2}}}$

T is the acceleration in the system

Substitute the value of a in equation (2)

$T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}$

$T=\frac{I\times mg}{I+mR^{2}}$

This is the tension in the string.

4 0

3 years ago

What is the gravitational potential energy of a 1.3 kg book that is 473

Dmitrij [34]

Answer:

3047710272

Explanation:

556ijnhvcdse4456679908

7 0

3 years ago

A rock falls from rest off a cliff and hits the ground in 2 seconds. If there is no air resistance, determine the rock's velocit

mihalych1998 [28]

Answer:

19.6m/s

Explanation:

A Rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for certain period of time, for such motion following equation of motion can be used.

$v(t) = v_{0} +at$

here in our case $v_{0} =0$ because object starts off from rest and $a = g =9.8m/s^2$ is acceleration because of gravity ( Motion under gravity).

and of course t = 2 second.

Now by substituting all this information in equation of motion we get.

$v(2s) = 0+9.8m/s^2 *2s = 19.6m/s$

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

A rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for a certain period of time, for such motion following equation of motion can be used.

here in our case because object starts off from rest and is acceleration because of gravity ( Motion under gravity).

and of course t = 2 seconds.

Now by substituting all this information in equation of motion we get.

V = 19.6m/s

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

5 0

3 years ago

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the

aivan3 [116]

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^{\circ}$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$

$W=50\times 9\times \ cos(135)$

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

7 0

3 years ago

A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif

Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

$R=\frac{v^{2}sin(2\theta)}{g}$

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

$gR=v^{2}sin(2\theta)$

$\frac{gR}{v^{2}}=sin(2\theta)$

$sin^{-1}(\frac{gR}{v^{2}})=2\theta$

$\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}$

so now we can substitute the given data:

$\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}$

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

$tan \theta = \frac{h}{45.5m}$

so we can solve this for h, so we get:

$h=45.5m*tan(0.05614^{o})$

which yields:

$h=0.0445m$

or

$h=4.45cm$

5 0

3 years ago