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Serhud [2]
3 years ago
6

Which reaction takes place in a nuclear fission reactor?

Physics
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

Last option in the list of possible answers, with U235 and n (neutron) in the left (originators) of the reaction diagram.

Explanation:

Uranium 235 (which is a fissile isotope of uranium) plus slow neutrons is what produce the chain reaction that feeds nuclear reactors.

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A shell is shot with an initial velocity v with arrow0 of 18 m/s, at an angle of θ0 = 60° with the horizontal. At the top of the
BARSIC [14]

Answer:

D = 43 m

Explanation:

given,

initial velocity = 18 m/s

angle θ = 60°

total horizontal distance covered by the shell is

R = \dfrac{v_0^2sin 2\theta}{g}

applying conservation of momentum in horizontal direction

m v₀ cos θ = m₁v₁ + m₂ v₂

m v₀ cos θ = 0.5 m v₂

v₂ = 2 v₀ cos θ.

distance covered by the shell from point of explosion

R' = v t

  = (2 v_0 cos \theta) (\dfrac{v_0^2sin \theta}{g})

  =(2 \dfrac{v_0^2cos \theta sin \theta}{g})

  = \dfrac{v_0^2sin 2\theta}{g}

  = R

total distance traveled by the shell is

D = \dfrac{R}{2}+R'

   = 1.5 R

   = 1.5\dfrac{v_0^2sin 2\theta}{g}

D = 1.5\dfrac{18^2sin 2\times 60}{9.81}

   = 42.9 ≅ 43 m

D = 43 m

3 0
3 years ago
Give an example of a situation in which you would describe an object's position in
notka56 [123]
Incomplete question.  Full text is:

"<span>Give an example of a situation in which you would describe an object's position in (a) one-dimension coordinates (b) two-dimension coordinates (c) three-dimension coordinates"

Solution
(a) One dimension example: a man walking along a metal plank. We just need to specify one coordinate, the distance from the beginning of the plank.

(b) Two-dimension example: a ball moving on a circle. In this case, we need two coordinates: (x,y) to specify the position of the ball at every instant, since it is moving on a 2-D plane.

(c) The position of an airplane in the air: in this case we need 3 coordinates, the height, the latitude and the longitude of the airplane.</span>
8 0
4 years ago
A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the proce
Grace [21]
  • The angular acceleration of the wheel approximately <u>-0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s</u>.
  • It need approximately <u>14.474 s</u> to come to rest.

<h2>Introduction</h2>

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

  • 1 rotation = 2π rad
  • 1 rps = 2π rad/s
  • 1 rpm = \sf{\frac{1}{60} \: rps} = \sf{\frac{1}{30}\pi \: rad/s}

<h2>Formula Used</h2>

The following equations apply to proportionally changes circular motion:

<h3>Relationship between Angular Acceleration and Change of Angular Velocity </h3>

\boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}

With the following conditions:

  • \sf{\omega_t} = final angular velocity  (rad/s)
  • \sf{\omega_0} =  initial angular velocity (rad/s)
  • \sf{\alpha} = angular acceleration (rad/s²)
  • t = interval of the time (s)
<h3>Relationship between Angular Acceleration and Change of \sf{\theta} (Angle of Rotation) </h3>

\boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}

Or

\boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}

With the following condition :

  • \sf{\theta} = change of the sudut (rad)
  • \sf{\alpha} = angular acceleration (rad/s²)
  • t = interval of the time (s)
  • \sf{\omega_t} = final angular velocity  (rad/s)
  • \sf{\omega_0} =  initial angular velocity (rad/s)

<h2>Problem Solving</h2>

We know that :

  • \sf{\omega_t} = final angular velocity  = 0 rad/s >> see in the sentence "in the process of coming to rest."
  • \sf{\omega_0} =  initial angular velocity = 11 rad/s
  • \sf{\theta} = change of the sudut = 80.0 rad

What was asked :

  • \sf{\alpha} = angular acceleration = ... rad/s²
  • t = interval of the time = ... s

Step by step :

  • \sf{\alpha} = ... rad/s²

\sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}

\sf{0^2= (11)^2 + 2 \times \alpha \times 80}

\sf{0 = 121 + 160 \alpha}

\sf{-160 \alpha = 121}

\sf{\alpha = \frac{121}{-160}}

\sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}

  • t = ... s

\sf{\alpha = \frac{\omega_0 - \omega_t}{t}}

\sf{-0.76 = \frac{0 - 11}{t}}

\sf{-0.76t = -11}

\sf{t = \frac{- 11}{-0.76}}

\boxed{\sf{t \approx 14.474 \: s}}

<h3>Conclusion</h3>

So :

  • The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
  • It need approximately 14.474 s to come to rest.
5 0
2 years ago
Why doesn't the motor work?
exis [7]

Answer:

c

Explanation: its weird

8 0
3 years ago
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