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3 years ago

5

Explain the difference between impulse and impulsive force?

1 answer:

ExtremeBDS [4]3 years ago

7 0

An impulsive force is a force that is acting only during a short time, I mean, for an instant. Impulse is a physics magnitude define by the product of the impulsive force and the time that it was acting.

Is there any mistake in my English? Please, let me know.

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Which statement is true? *

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The last paragraph is the correct statement. Please don't make me type it all out.

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3 years ago

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A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

mario62 [17]

Answer:

$0.2$ mm

Explanation:

As we know that

$Y = \frac{m\lambda * D}{d}$

where

m represents the order of minimum

y represents the distance on the screen of the minimum from central axis

λ is the wavelength of the light

D is the distance between screen-to-slit

d represents the width of the slit

For first minima

$y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}$

For second minima

$y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}$

$Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm$

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3 years ago

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

geniusboy [140]

Answer:

$E=\frac{K}{ed}$

Explanation:

We are given that

Initial kinetic energy of an electron=K

Distance=d

Final velocity=v=0

Charge,q=-1e

We have to find the magnitude of electric field.

Work done= $Force\times displacement$

Using the formula

Work done= $qE\times d=-eEd$

Using work energy theorem

Work done=Final K.E-Initial K.E=0-K

Work done=-K

Substitute the values

-K=-eEd

K=eEd

$E=\frac{K}{ed}$

Hence, the magnitude of the electric field= $E=\frac{K}{ed}$

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3 years ago

5) What type of substance can seep<br> groundwater and make it unusable?

sergeinik [125]

Answer:

I believe <u>chemicals</u>, though I might be wrong.

Explanation:

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2 years ago

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

-BARSIC- [3]

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62 distance traveled after collision .. crumpling of truck

Part a

$V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s$

Part b

$vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s$

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3 years ago

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