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mart [117]
3 years ago
5

Explain what chain rule is . workout appropriate examples to support your explanation.

Physics
1 answer:
Luda [366]3 years ago
5 0
The chain rule is a formula for computing the derivative of the composition of two or more functions.

<span> (f o g(x)<span>)1</span>=<span>g′</span>(x)<span>f′</span>(g(x))</span>
or
<span> dy/dx=dy/du   du/d<span>x</span></span>
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To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If
arlik [135]

Answer:

Part a)

a = 1260.3 m/s^2

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

v_f^2 - v_i^2 = 2 a d

v_1^2 - 0^2 = 2(9.81)(4.0)

v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

v_f^2 - v_i^2 = 2 a d

0 - v_2^2 = 2(-9.81)(2.00)

v_2 = 6.26 m/s

Part a)

Average acceleration is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}

a = 1260.35 m/s^2

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
AHH please help-
umka2103 [35]

Answer:

Radiation

Explanation:

I hope that help's

5 0
3 years ago
What is the dependent variable in this hypothesis?
frosja888 [35]

Answer:

The height at wich the object floats

Explain

took on edge 2020

4 0
3 years ago
(kg)
stellarik [79]

Answer:

(C) an increase in tue distance between the ibject causes a greater change in the gravitational force than the same increase in mass

Hope this helps

5 0
2 years ago
Read 2 more answers
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