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Anna35 [415]
3 years ago
10

Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.

Physics
1 answer:
LuckyWell [14K]3 years ago
7 0

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

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Yes very very fast needs to be on time
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3 years ago
Please help on answer
Ludmilka [50]

Answer:

D. 1

Explanation:

I believe this is the answer

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2 years ago
The oscillating electric field in a plane electromagnetic wave is given by <img src="https://tex.z-dn.net/?f=%7B50%5Csin%20%28%5
Scorpion4ik [409]

Here, we are given with:

{:\implies \quad \longrightarrow \begin{cases}\sf E_{0}= 50\: V/m\\ \sf \nu = 2\times 10^{7}Hz\end{cases}}

(a) So now, we can thus obtain

{:\implies \quad \sf \omega =2\pi \nu}

{:\implies \quad \sf \omega =2\pi (2\times 10^{7})}

{:\implies \quad \boxed{\bf{\omega = 4\pi \times 10^{7}\: s^{-1}}}}

Now, finding \lambda

{:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}

{:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}

Now, finding k

{:\implies \quad \sf k=\dfrac{2\pi}{\lambda}=\dfrac{200\pi}{239}}

{:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}

Thus, expression for the electric field is:

{:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}

(b) Now, here

{:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}

{:\implies \quad \boxed{\bf{B_{0}\approx 16.67×10^{-7}\:T}}}

Thus, expression for the magnetic field:

{:\implies \quad \boxed{\bf{B_{y}=16.67\times 10^{-7}\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)\:T}}}

(c) The electromagnetic wave propagates along Z-axis

4 0
2 years ago
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.09 mm apart and position
kow [346]

Answer:

pretty dark

Explanation:

i dont know

4 0
3 years ago
What is the drawback to using superconductors?
Fittoniya [83]

Answer:

Option A

The cost of keeping the semiconductor below the critical temperature is unreasonable

Explanation:

First of all, we need to understand what superconductors are. Superconductors are special materials that conduct electrical current with almost zero resistance. This means that there is little or no need for a voltage source to be connected to them. As a matter of fact, once a superconductor is connected to a power supply, one can remove the power supply and the current will still flow.

However, most superconducts can only conduct at very low temperatures up to -200 degrees Celcius. This is because, at that temperature, their atoms and molecules are relatively settled, hence they pose little or no resistance to the flow of current.

This as you can guess is extremely difficult to do, as you will need a lot of effort to cool it to that temperature and maintain it.

This makes option a the answer:

The cost of keeping the semiconductor below the critical temperature is unreasonable.

7 0
3 years ago
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