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shepuryov [24]
1 year ago
8

One major disadvantage of breeder reactors is that they can do which of the following? Select one:a.explodeb.leak radioactivityc

.melt downd.go dormant
Physics
1 answer:
zimovet [89]1 year ago
8 0

definition of breeder reactors.

These are a type of nuclear reactors which produce more fissile material than they consume

Advantages: Breeder reactors produce Pu-239 which can be extensively used as a nuclear fuel. Also, Pu-239 can absorb neutron to form Pu-240, which is another fertile material.[1]

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The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0
3 years ago
Ну car has 4 wheels is a<br> observation. *<br> 1 point<br> O Quantitative<br> Qualitative
sergiy2304 [10]
Quantitative because it tells you how many wheels
7 0
3 years ago
For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2
Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

n = \frac{c}{v}

The speed of violet light in the object is 1.9 * 10^8 m/s.

The speed of red light in the object is 2 * 10^8 m/s

Hence, the refractive index for violet light is:

n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56

and for red light, it is:

n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

n = \frac{sin(i)}{sin(r)}

The angle of incidence is 30°.

The angle of refraction for violet light will be:

1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56}  = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o

And the angle of refraction for red light will be:

1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5}  = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56}  = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o

And the angle of refraction for red light will be:

1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5}  = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

6 0
3 years ago
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.70
Aleksandr-060686 [28]
Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s

Let ω₂ =  the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
      = (0.7/3.5)*(30.159)
      = 6.032 rad/s

ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s

Answer: 0.96 rev/s


3 0
3 years ago
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
DerKrebs [107]

Answer:

A) 26.5 m/s

B) 33.0 m/s

Explanation:

A)

  • Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.
  • So, we can apply the definition of average velocity to find this speed as follows:

       v_{x} = \frac{\Delta x}{\Delta t}  (1)

  • We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.
  • This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.
  • Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).
  • Since the acceleration is constant, we can use the following kinematic equation, as follows:

       \Delta y  = y_{f} - y_{o} = v_{o} * t + \frac{1}{2}  * g *t^{2}  (2)

  • if we take the river level as our x-axis, this means that yf = 1.3 m and

       y₀ = 20.8 m.

  • At the same time, due to in the vertical direction the car has no initial velocity, this means that  v₀ = 0.
  • Replacing by the values in (2) , and solving for t:

       t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} }  = 2 s  (3)

  • If we choose t₀ =0 ⇒ Δt = t = 2 s
  • Replacing Δx and Δt in (1):

       v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s  (4)

B)

  • When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.
  • Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:

      v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s  (5)

  • Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:

       v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}}  = 33.0 m/s  (6)

5 0
3 years ago
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