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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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Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on

Andreas93 [3]

Answer:

0.82 m

Explanation:

We are given that

Weight, $w_1$ =225 N

Length,l=5 m

d=1.1 m

We have to find the distance x for which person weighs 385 N walk on the overhanging part of the plank before it just begins to tip.

Half length= $\frac{l}{2}=\frac{5}{2}=2.5 m$

$r=\frac{l}{2}-d=2.5-1.1=1.4 m$

$w_2=385 N$

When system in equilibrium then

$w_1r=w_2 x$

$225\times 1.4=385\times x$

$x=\frac{225\times 1.4}{385}$

$x=0.82 m$

6 0

4 years ago

A kid at the junior high cafeteria wants to propel an empty milk carton along a lunch table by hitting it with a 3.0g spit ball.

Natasha_Volkova [10]

Answer:

4.3 m/s

Explanation:

Using conservation of momentum, assume that the collision is elastic and that the 3.0 g ball stuck with the carton

3.0 g = 3 / 1000 = 0.003 kg = m₁

40 g = 40 / 1000 = 0.04 kg = m₂, u₂ = 0 since the body is stationary and v = 0.3m/s

m₁u₁ + m₂u₂ = v ( m₁ + m₂)

0.003 u₁ = 0.30 ( 0.003 + 0.04) = 0.0129

u₁ = 0.0129 / 0.003 = 4.3 m/s

5 0

3 years ago

What is the square root of 25x10^12

kirill115 [55]

$\sqrt{25 \times 10^{12} }$

In order to find it's square root, we could make it into two square roots.

$\sqrt{25 \times 10^{12} } = \sqrt{25} \times \sqrt{10^{12}}$

Let us find the square roots of both radicals seprately.

$\sqrt{25} =\sqrt{5*5}=5$

Each pair of a number inside square root gives a number out .

$\sqrt{10^{12}} = \sqrt{10*10*10*10*10*10*10*10*10*10*10*10} \ \ ( \ makes \ 6 \ pairs \ of \ 10 )$

$\sqrt{10*10*10*10*10*10*10*10*10*10*10*10} = 10*10*10*10*10*10$

$= 10^6.$

Therefore,

$\sqrt{25 \times 10^{12} }=\sqrt{25} \times \sqrt{10^{12}}={5 \times 10^6}$

$\sqrt{25 \times 10^{12} } = {5 \times 10^6}$

7 0

3 years ago

Chemical reactions at varying depths on other planets, such as Jupiter, result in what?(1point)

omeli [17]

I believe the answer would be "Different Colors," hope this helps :)

7 0

3 years ago

Two point charges are on the y axis. A 3.90-µC charge is located at y = 1.25 cm, and a -2.4-µC charge is located at y = −1.80 cm

ladessa [460]

Answer:

a) 1.6*10^6 V

b) 13.35*10^6 V

Explanation:

The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

q1 = 3.90µC = 3.90*10^-6 C

q2 = -2.4µC = -2.4*10^-6 C

r1 = 1.25 cm = 0.0125 m

r2 = -1.80 cm = -0.018 m

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

You replace all the parameters in the equation (1):

$V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V$

hence, the total electric potential is approximately 1.6*10^6 V

b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:

r1 = 0.0150m - 0.0125m = 0.0025m

r2= 0.015m + 0.018m = 0.033m

Then, you replace in the equation (1):

$V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V$

hence, for y = 1.50cm you obtain V = 13.35*10^6 V

4 0

3 years ago