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Leya [2.2K]
3 years ago
10

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.

00 m/s?B) If she then breaks to a stop in 0.800 s, what is her deceleration?
Physics
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

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22.2 m/s

Explanation:

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The average speed can be found by using the equation \frac{distance}{time}. After substitution, this gives the fraction \frac{320 000}{14 400}, which reduces to 22 \frac{2}{9} m/s, or about 22.2 m/s.

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A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t
MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

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A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of
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Answer:

The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}

\dfrac{1}{f}=\dfrac{347}{5852}

f=16.86\ cm

We need to calculate the focal length

Using formula of magnification

m= \dfrac{-v}{u}

Put the value into the formula

2=\dfrac{v}{u}

v = -2u

Using formula of for focal length

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}

\dfrac{1}{16.86}=\dfrac{1}{2u}

2u=16.86

u=\dfrac{16.86}{2}

u=8.43\ cm

Hence, The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

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