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AleksAgata [21]

3 years ago

7

A dragster accelerated to a speed of 112m/s over a distance of 398m. Determine the acceleration (assume uniform) of the dragster

.

1 answer:

iragen [17]3 years ago

7 0

. Given- vf-112m/s, d-398m, Vi-0m/s (112 m/s)2 =

(0 m/s)2 + 2*(a)*(398 m)12544 m2/s2 = 0 m2/s2 + (796 m)*a12544 m2/s2 =

(796 m)*a(12544 m2/s2)/(796 m) = aa = 15.8 m/s26.

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If the moon moved closer to the earth, how would the gravitational force between them change?.

hichkok12 [17]

If the moon moved closer to the earth, The gravitational force between them will increase.

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2 years ago

URGENT! Please define these, and give an example.

adelina 88 [10]

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6 0

3 years ago

A 6.7 kg object moves with a velocity of 8 m/s. what is its kinetic energy

blagie [28]

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3 years ago

Read 2 more answers

You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the

Jet001 [13]

Answer:

Explanation:

Given

initial velocity $u=11.2\ m/s$

At maximum height velocity of ball is zero

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(0)^2 -(11.2)^2=2\times (-9.8)\times (s)$

$s=\frac{11.2^2}{2\times 9.8}$

$s=6.4$

time taken by the ball to reach the maximum height

$v=u+at$

$0=11.2-9.8\times t$

$t=\frac{11.2}{9.8}$

$t=1.142\ s$

At $t=2\ s$ height of ball is

$h=ut+\\frac{1}{2}at^2$

$h=11.2\times 2-\frac{1}{2}9.8\times (2)^2$

$h=22.4-19.6$

$h=2.8\ m$

i.e. ball is moving downward

height at $v=5\ m/s$

$v^2-u^2=2as$

$s=\frac{25-125.4}{2\times (-9.8)}$

$s=5.12\ m$

8 0

3 years ago

How fast is a car going in m/5 if speedometer reads 45 miles?

fredd [130]

Answer:

i think you would have to divide to 2 numbers to get the sum

Explanation:

4 0

3 years ago