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kiruha [24]
3 years ago
7

Nuclear power doesn't directly emit greenhouse gases, but it does leave behind dangerous nuclear _______.

Physics
1 answer:
Morgarella [4.7K]3 years ago
6 0

Answer:

radioactive waste?

Explanation:

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A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
2 years ago
A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
GuDViN [60]

Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

4 0
2 years ago
Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia
Gala2k [10]

Answer:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

5 0
3 years ago
A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?
creativ13 [48]

Answer:

s=800 m

Explanation:

Given that,

Acceleration of a runner, a = 4 m/s²

Time, t = 20 seconds

We need to find the distance covered by her. Initially, she was at rest. It means its initial velocity is equal to 0. So, using second equation of motion as follows :

s=ut+\dfrac{1}{2}at^2

Herre, u = 0

s=\dfrac{1}{2}at^2\\\\s=\dfrac{1}{2}\times 4\times (20)^2\\\\s=800\ m

So, she will cover a distance of 800 m.

8 0
3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
2 years ago
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