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3 years ago

5

I dropped an apple (mass 0.1kg) from the window because i'm weird. (15m above the ground). How fast was it going when it hit the

ground? plz yall what's the speed

1 answer:

Olegator [25]3 years ago

5 0

Answer:

I think the answer is 1 m per second.

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Boyles law<br> squeezing a balloon is one way to burst it. Why? ...?

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Which of the following objects exerts the greatest gravitational force on the Earth?

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Explanation:

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Because CMB radiation was _____ throughout the universe, scientists can only relate the source to the big bang.

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3 years ago

Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi

max2010maxim [7]

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

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3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

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3 years ago