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2 years ago

At a race car driving event, a staff member notices that the skid marks left by the race car are

1 answer:

5 0

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

$v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s$

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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A brick is dropped from a height of 31.9 meters

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Explanation:

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2 years ago

You drive on Interstate 10 from San Antonio to Houston, half the time at 72 km/h and the other half at 98 km/h. On the way back

zimovet [89]

Answer: a. 85km/hr b.82.3km/hr

c. 84km/hr

Explanation: first let take the total time from San Antonio to Houston to be 2hr.

Half time 1hr was covered with speed of 72km/hr

Distance = speed*time=72km/hr *1hr

=72km

So too with the second half of 1hr covered with speed of 98km/hr

Distance = 98km

Total distance from Houston to San Antonio is 98+72 =170km

a. Average speed from San Antonio to Houston is

S1 =170/2

=85km/hr

b.half distance from Houston to San Antonio which is 170km/2

= 85km was covered with speed of 72km/hr first half, so time

t = dist/speed

t = 85/72 = 1hr 12 mins

Remaining 85 km covered with a speed of 98km/hr

Time = 85/98 = 0.88*60min

= 52 mins

Total time = 1hr +12mins +52mins

=2hr4mins= 124/60 hr

So average speed = distance/time

=170/124/60

Using reciprocal law

Average speed S2= 170*60/124

= 82.3km/hr

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2 years ago

A ball is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 5.5 s. With what veloc

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V = u + at
0 = u -9.8 x 5.5
u = 9.8 x 5.5 = 53.9 m/s

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3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

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Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

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3 years ago

List at least 3 ways for you to represent your data graphically

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1. circle graph
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1 year ago

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