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3 years ago

How far will a rubber ball fall in 10 seconds?

Physics

1 answer:

Inessa [10]3 years ago

4 0

If it's not moving at all at the beginning of the 10 seconds, then it falls 490 meters straight down in 10 seconds.

(Note: This is true of all objects on Earth . . . rubber balls, feathers, grains of sand, school buses, battle ships . . . everything. As long as air doesn't hold them back. Anything falling from rest falls 490 meters in the first 10 seconds.)

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Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v

butalik [34]

Answer:

$u=34cm$

Explanation:

From the question we are told that:

Far point is $V=34 cm$

Near point is $u=17 cm$

Therefore

Focal Length

$f=-34cm$

Generally the equation for the Lens is mathematically given by

$\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$

$\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}$

$u=34cm$

5 0

2 years ago

Which of the following terms corresponds to #2 on the image?

Mila [183]

Answer:

Trough

Explanation:

cuz physics you see

7 0

3 years ago

A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m

mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

5 0

2 years ago

Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert

damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

$F = k \frac{q\cdot q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

$q+\frac{q}{2}=\frac{3}{2}q$

while the other charge will be

$q-\frac{q}{2}=\frac{q}{2}$

So, the new force will be

$F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F$

So, the force will decrease to 3/4 of its original value.

6 0

3 years ago

An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.

zlopas [31]

<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then the masses of the two pieces are also equal. Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90Â° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities. This is only possible, when the angle of their velocity with the initial direction is 60Â°. Because, cos (60Â°) = 1/2 = v/(2v).</span>

6 0

3 years ago