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4 years ago

Which is not an issue with waste from nuclear energy?

Physics

1 answer:

Sergeeva-Olga [200]4 years ago

3 0

Answer:

A) atmospheric pollutants. Nuclear power plants do not produce air pollution as carbon dioxide, sulfur dioxide.

Hope this helps!

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Juggles the clown stands on one end of a teeter-totter at rest on the ground. Bangles the clown jumps off a platform 2.4 m above

tester [92]

Answer:

mj = 53.3kg

Explanation:

Since both gravitational energies are the same:

Eb = Ej => mb*g*hb = mj*g*hj

Solving for mj:

mj = mb*hb/hj = 80*2.4/3.6

mj = 53.3kg

5 0

3 years ago

A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm

joja [24]

Answer:17.44A

Explanation: Current density=I/Area

Area is given by 2.79mm^2=2.79×10^-6m^2

Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A

3 0

3 years ago

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

5 0

3 years ago

What provided the force that made the cart speed up

fgiga [73]

Answer:

Potential energy turn to kinetic energy

Explanation:

5 0

4 years ago

Read 2 more answers

Se deja resbalar por un plano inclinado de 30° un cuerpo de 15kg. Calcular la aceleración con la que desciende suponiendo que no

kupik [55]

Answer:

$4.9 m/s^2$

Explanation:

In order to find the acceleration of the block, we have to find the net force acting on it along the direction parallel to the incline.

However, there is only one force acting on the block along this direction: it is the component of the weight parallel to the plane, given by

$mg sin \theta$

where

m = 15 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30^{\circ}$ is the angle of the incline

According to Newton's second law, the net force is proportional to the acceleration, a:

$F=ma$

So we can write:

$ma = mg sin \theta$

And so, the acceleration is:

$a=g sin \theta = (9.8)(sin 30^{\circ})=4.9 m/s^2$

7 0

3 years ago