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Olin [163]
2 years ago
6

Identify the energy transformations in the image below:

Physics
1 answer:
zvonat [6]2 years ago
4 0

Answer:

B

Explanation:

Electrical to Radiant.

I think

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Switches, flashers, and similar devices controlling transformers and electronic power supplies shall be rated for controlling in
Reika [66]

Answer:

Option A - twice

Explanation:

5 0
2 years ago
If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square
antoniya [11.8K]

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

T =\frac{2\pi }{\omega }

Where \omega is the angular velocity and it is given by,

\omega = \sqrt{\frac{k}{m} }

Now if the spring constant is doubled then,

k_{2} = 2k

Thus,

T_{2} =\frac{2\pi }{\sqrt{\frac{2k}{m} } }

\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}

\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }

T_{2} = \frac{T}{\sqrt{2} }

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.

3 0
2 years ago
Read 2 more answers
Which compound is most likely a powerful and dangerous acid? H2S HBr Li2O LiBR
sattari [20]

HBr is the most powerful and dangerous acid .

Explanation:

A strong acid is one that instantly disunites or grants its protons in suspension. HBr is a strong acid. HBr is powerful than HCl or HF because the overlapping of a 1s-orbital and a 4p-orbital is surprisingly small, thus the binding is weak so splitting is very easy..

6 0
3 years ago
Read 2 more answers
What kind of proof is required when disproving old ideas?
GrogVix [38]

Answer:

D

Explanation:

In order to disprove an old theory you need to prove a new one that contradicts it using scientific research and experimentation

8 0
3 years ago
A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.
Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

v =  \sqrt{\frac{2GM}{R}}

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

v = 14.9km/h = 14900m/s

Through this equation we can find the mass of the Planet in function of the distance, therefore

M = \frac{v^2R}{2G}

M = \frac{14900^2R}{2(6.67*10^{-11})}

M = 16.64*10^{17}R

The orbital velocity is

v_o = \sqrt{\frac{GM}{R+h}}

9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}

11.1*10^7R = (R+15000*10^3)(9200)^2

2.64*10^7R = 12.69*10^{13}

R = 4.81*10^6m

The time period of revolution is,

T = \frac{2\pi(R+h)}{v_o}

T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}

T = 4307s

T = 72min = 1hour12min

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0
3 years ago
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