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myrzilka [38]
2 years ago
10

The stone falls a further 25 m in the next 1.0 s of its fall. Calculate the stone's average speed during the 3 s of its fall.​

Physics
1 answer:
Angelina_Jolie [31]2 years ago
6 0

Answer:

5 minutes sweetheart

Explanation:

thank me later-also give me brainless if you want too when you get the answer correct!

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A substance that will conduct electricity only under certain conditions is called a(n)
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A photon has a momentum of 5.55 x 10-27 kg-m/s. (a) What is the photon's wavelength? nm (b) To what part of the electromagnetic
Brrunno [24]

Explanation:

It is given that,

Momentum of the photon, p=5.55\times 10^{-27}\ kg-m/s

(a) We need to find the wavelength of this photon. It can be calculated using the concept of De-broglie wavelength.

\lambda=\dfrac{h}{p}

h is the Planck's constant

\lambda=\dfrac{6.67\times 10^{-34}\ Js}{5.55\times 10^{-27}\ kg-m/s}

\lambda=1.2\times 10^{-7}\ m

or

\lambda=120\ nm

(b) The wavelength lies in the group of ultraviolet rays. The wavelength of UV rays lies in between 400 nm to 10 nm.

3 0
3 years ago
Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
fomenos

Answer:

(a) \vec{R}= 4.83\ m\ \hat{i}+1.47\ m\ \hat{j}

(b) (5.05 m, 16.93 degrees wrt x-axis)

Explanation:

Given:

  • \vec{D} = (3.00 m, 315 degrees wrt x-axis)
  • \vec{E} = (4.50 m, 53.0 degrees wrt x-axis)

Let us first fond out vector D and E in their rectangular form.

\vec{D} = (3\cos 315^\circ\ \hat{i}+3\sin 315^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.12\ \hat{i}-2.12\ \hat{j})\ m\\

Similarly,

\vec{E} = (4.5\cos 53^\circ\ \hat{i}+4.5\sin 53^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.71\ \hat{i}+3.59\ \hat{j})\ m\\\because \vec{R}=\vec{D}+\vec{E}\\\therefore \vec{R} = (2.12\ \hat{i}-2.12\ \hat{j})\ m+(2.71\ \hat{i}+3.59\ \hat{j})\ m\\\Rightarrow \vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m

Part (a):

We can write the resultant vector R as below:

\vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m

Part (b):

Magnitude\ of\ resultant = \sqrt{4.83^2+1.47^2}\ m = 5.05\ m\\\textrm{Direction in angle with the x-axis} = \theta = \tan^{-1}(\dfrac{1.47}{4.83})= 16.93^\circ

Since both the components of the resultant lie on the positive x and y axes. So, the resultant makes an acute angle with the positive x-axis.

So, R = (5.05 m, 16.93 degrees wrt x-axis)

3 0
3 years ago
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