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3 years ago

Which is a nonrenewable resource for the production of nuclear energy?

Physics

2 answers:

Anuta_ua [19.1K]3 years ago

5 0

Answer:

It is Uranium.

Option (D)

Explanation:

Uranium is a radioactive substance.

When uranium is bombarded with neutrons it emits very large amount of energy along with some stable substances. The energy emitted s called nuclear energy.

The reaction is given by

IRINA_888 [86]3 years ago

3 0

D. Uranium

Hope this helps :)

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speed is the ratio of the distance an object moves to a. the direction the object moves. b. the amount of time needed to travel

Dennis_Churaev [7]

It is the ratio of distance an object moves to the amount of time needed to travel the distance. Option b.

5 0

2 years ago

Read 2 more answers

In a Young's double-slit experiment, a set of parallel slits with a separation of 0.142 mm is illuminated by light having a wave

Alexandra [31]

Answer:

$1.152\ \mu m$

$1.44\ \mu m$

Explanation:

d = Gap between slits = 0.142 mm

$\lambda$ = Wavelength of light = 576 nm

L = Distance between light and screen = 3.5 m

m = Order = 2

Difference in path length is given by

$\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m$

The difference in path lengths is $1.152\ \mu m$

For dark fringe the difference in path length is given by

$\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m$

The difference in path length is $1.44\ \mu m$

5 0

3 years ago

Why do plants need to obtain carbon atoms?

Firdavs [7]

Answer:

A) Plants use carbon to build glucose during photosynthesis.

Explanation:

Plants produce their food by the process of photosynthesis.

hope this helps :}

5 0

3 years ago

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

3 0

3 years ago

Throughout the problem, take the speed of sound in air to be 343 .

anyanavicka [17]

Answer:

A pipe has a length of 1.15 m.

a) Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz

b) What is the frequency of the first harmonic if the pipe is closed at one end?

Answer in units of Hz

Explanation:

8 0

3 years ago