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2 years ago

a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car

Physics

1 answer:

uysha [10]2 years ago

3 0

Answer:

= 201.53 meters

Explanation:

A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?

Use the formula d = $\frac{at^{2} }{2} ,$

where d is the distance, t is the time and "a" is the acceleration.
$d=\frac{9*54*6*5^{2} }{2} = 201.53 m$

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’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

How long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?

alexandr1967 [171]

Answer:

25 seconds

Explanation:

500/20

4 0

3 years ago

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

ale4655 [162]

Answer:

The last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

$cos X=\frac{x^2-y^2-z^2}{-2yz}$

$cos X=\frac{y^2 + z^2-x^2}{2yz}$

substituting the values in the equation we get,

$cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}$

$cos X=0.8951$

X = 26.47°

similarly,

$cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}$

$cos Y=0.649$

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0

3 years ago

A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

atroni [7]

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

bullet mass, Mb = 4.10 g

initial bullet velocity, Vbi = 837 m/s

wooden block mass, Mw = 820 g

initial wooden block velocity, Vwi = 0 m/s

final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

6 0

3 years ago

Calculate the efficiency of given pulley System if 40N

jonny [76]

Answer:

20n

Explanation:

60n - 40n = efficiency of pulley system

20n = efficiency of pulley system

4 0

3 years ago

a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car​

a car starts from the rest and accelerates at 9.54m/s for 6.5 seconds. what is the distance covered by the car