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3 years ago

A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *

Physics

1 answer:

crimeas [40]3 years ago

8 0

Answer:

Given

mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

p=mv

2kgx3m/s

p=6kg.m/s

the momentum of ball's =6kg.m/s

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A basketball referee tosses the ball straight up for the starting tipoff. at what velocity must a basketball player leave the gr

sladkih [1.3K]

Mgh=mv²/2

v=√2gh=√2·9.8·1.25=4.95m/s

5 0

3 years ago

If opposing forces acting on an object are equal, the net force is

makkiz [27]

Answer:

0 N

Explanation:

suppose, you push a box with 5 N, and another person pushes the box on the opposite side of the box with 5 N, the net force (resultant ) is 0 N, the box will not move if it wasn't moving

hope this helps

8 0

2 years ago

A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to

Alborosie

Explanation:

If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :

$\epsilon=Blv$

The electric field in the conductor is given by :

$E=\dfrac{\epsilon}{l}\\\\E=\dfrac{Blv}{l}\\\\E=Bv$

It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.

6 0

2 years ago

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

mihalych1998 [28]

The radius of a nucleus of hydrogen is approximately $r_{n1}=1\cdot 10^{-15}m$ , while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: $r_{e1}=5.3 \cdot 10^{-11}m$

The radius of a dime is approximately $r_{n2} = 9\cdot 10^{-3}m$ : if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion
$r_{n1}:r_{e1}=r_{n2}:r_{e2}$
from which we find
$r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m$

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.

6 0

3 years ago

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb, stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

$W =k\int_{6}^{10.1} x$

$W = \dfrac{k}{2}[x^2]_6^{10.1}$

$W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)$

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

3 0

3 years ago