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3 years ago

Why should we not use nuclear energy

Physics

1 answer:

Softa [21]3 years ago

5 0

Answer:

Barriers to and risks associated with an increasing use of nuclear energy include operational risks and the associated safety concerns, uranium mining risks, financial and regulatory risks, unresolved waste management issues, nuclear weapons proliferation concerns, and adverse public opinion.

Explanation:

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Is work being done if you push a child in a wagon

Leto [7]

yes. youre using energy to make the wagon with the child in the wagon

6 0

4 years ago

Read 2 more answers

As an electron approaches a proton, the electron's force of attraction...

Tems11 [23]

Answer: B. Increases

Explanation:

An electron is a negatively charged particle, while a proton is positively charged.

Opposites attract so as it approaches the proton the force of attraction will increase.

7 0

3 years ago

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

6 0

3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }