All categories

3 years ago

Explain the law of conservation of momentum.

Physics

1 answer:

WITCHER [35]3 years ago

3 0

Answer:

It says that momentum cannot be created or destroyed. Momentum just passes through systems as opposed to just disappearing in thin air. Momentum can be changed to heat or energy.

Explanation:

You might be interested in

NaCl solid is an example of a/an A. Insulator B. Conductor OC. Nonmetal D. Metalloid

julia-pushkina [17]

It is an Insulator I’m pretty sure

6 0

1 year ago

In t-ball, young players use a bat to hit a stationary ball off a stand. The 140 g ball has about the same mass as a baseball, b

fomenos

Question is missing. Found on google:

"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units.

Part B

What is the net force on the ball during the hit?

Express your answer to two significant figures and include the appropriate units."

Solution:

A) $6000 m/s^2$

The acceleration of the ball is given by

$a=\frac{v-u}{t}$

where

v = 12 m/s is the final velocity

u = 0 is the initial velocity (the ball is stationary)

t = 2.0 ms = 0.002 s is the time of contact

Substituting,

$a=\frac{12-0}{0.002}=6.0 \cdot 10^3 m/s^2$

B) $8.4\cdot 10^2 N$

The force on the ball can be found by using Newton's second law:

$F=ma$

where

m = 140 g = 0.14 kg is the mass of the ball

$a=6.0\cdot 10^3 m/s^2$ is the acceleration

Substituting,

$F=(0.14)(6.0\cdot 10^3)=8.4\cdot 10^2 N$

3 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Ilia_Sergeevich [38]

Below is the attachment of the solutions.

5 0

4 years ago

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

gizmo_the_mogwai [7]

Let's see

Momentum be P

$\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c$

$\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c$

$\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c$

On comaparing

b-3c=1
b-3=1
b=1+3
b=4

and

-a-b=-1
-a-4=-1
-a=-1+4=3
a=-3

So the unit is

DV⁴/F³

5 0

3 years ago

Read 2 more answers

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil

Travka [436]

Complete question:

A small circular coil of 5 turns of wire lies in a uniform magnetic field of 0.8 T, so that the normal to the plane of the coil makes an angle of 100◦ with the direction of B~ . The radius of the coil is 4 cm, and it carries a current of 1 A.

What is magnitude of the magnetic moment of the coil? Answer in units of A · m2.

Answer:

The magnetic moment of the coil is 0.0252 A.m²

Explanation:

Given;

radius of the coil, r = 4 cm = 0.04 m

number of turns of the coil, N = 5 turns

magnetic field strength B = 0.8 T

current in the coil, I = 1 A

Area of the coil, A = πr² = π(0.04)² = 0.00503 m²

magnetic moment of the coil, μ = NIA

where;

N is the number of turns

I is the current in the coil

A is the area of the coil

magnetic moment of the coil, μ = 5 x 1 x 0.00503 = 0.0252 A.m²

Therefore, the magnetic moment of the coil is 0.0252 A.m²

8 0

3 years ago