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3 years ago

Any clue on this one

Physics

1 answer:

rodikova [14]3 years ago

3 0

Answer: First option

Explanation: The higher the frequency, the higher the energy.

λν=c where λ is the wavelength, ν is the frequency and c is the speed of light. So when wavelength decreases, v increases and so does energy.

An AM radio wave has a very long wavelength. It therefore has a very low frequency and low energy.

A light wave has a very short wavelength. It therefore has a high frequency and high energy.

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The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago

According to Hooke's law

SOVA2 [1]

Answer:

According to Hook's law, we know,

strain/stress =Constant

Explanation: So, the ratio between stress and strain is always constant.

So, if stress is increased, then strain changes in that way so that this ratio always remains constant.

7 0

3 years ago

What is the direction of the centripetal force when applied to an object?

mihalych1998 [28]

For an object moving in a path that's a circle or a part of one,
the centripetal force acts in the direction toward the center of
the circle. That direction is perpendicular to the way the object
is moving.

3 0

3 years ago

Read 2 more answers

A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How

Sati [7]

0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>

Mathematically, the work done to strech a string is given as 1/2 ×K×x².
K is the spring constant.

<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>

The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
Then K = Force / x.
Here x = 15 - 10 = 5 cm = 0.05 m
K = 40/0.05 = 800N/m.

<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>

Work done = 1/2×k×x²
Here x= 18-15=3cm or 0.03 m
So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

#SPJ4

6 0

1 year ago

HELPP WILL MARK BRAINLIEST!!!

guapka [62]

I guess it’s B cause that maybe is the output

3 0

3 years ago

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