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3 years ago

6.) What force would evetually cause the ball to come to a stop after leaving the ramp

1 answer:

7 0

$\colorbox{aqua} A\colorbox{aqua} N\colorbox{aqua} S\colorbox{aqua} W\colorbox{aqua} E\colorbox{aqua} R\colorbox{aqua} S$

Forces of inertia

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A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu

serg [7]

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

5 0

3 years ago

1. Stops: Using the information learned in this course, explain three things you will not do when driving.

vazorg [7]

Don't text while driving
don't get your eyes off the road
don't get distracted

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3 years ago

A two-phase, liquidâ€“vapor mixture of h2o, initially at x 5 30% and a pressure of 100 kpa, is contained in a pistonâ€“ cylinder

Andrew [12]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

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2 years ago

The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the

Andru [333]

<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"

So, option B is your answer

Hope this helps!
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5 0

3 years ago

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

zavuch27 [327]

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.

3 0

3 years ago

6.) What force would evetually cause the ball to come to a stop after leaving the ramp​

6.) What force would evetually cause the ball to come to a stop after leaving the ramp