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Aloiza [94]
3 years ago
13

how is momentum conserved when a cue ball moving with a velocity of 1.5 m/s strikes another billiard ball white playing pool? Sh

ow the work.
Physics
1 answer:
mote1985 [20]3 years ago
8 0

according to conservation of momentum , total sum of momentum of the objects taking part in a collision is same before and after the collision.

Total momentum before collision = Total momentum after the collision

when a cue ball moving at velocity 1.5 m/s hits a billiard ball , transfer of momentum takes place between the balls such that total sum of momentum of cue ball and billiard ball remain same before and after the collision.

hence we say that the momentum is conserved.

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Gauss's law is usualy written as :
snow_tiger [21]

Answer:

(a) the net charge inside the closed surface.

Explanation:

In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.

4 0
3 years ago
The question states: two large, parallel conducting plates are 12cm
AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, E=\frac{F}{q}

E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0
2 years ago
Solve each of these problems, remembering to include formula, calculations, and units!
KATRIN_1 [288]

Answer:

I nEeD heLp

Explanation:

HelP

5 0
2 years ago
A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf
kirza4 [7]

Answer:v=\sqrt{\frac{FL}{m}}

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

F=T=\frac{mv^2}{L}

v=\sqrt{\frac{FL}{m}}

8 0
3 years ago
A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c
tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

B=\dfrac{\mu_{0}NI}{l}

Where, N = Number of turns

I = current

l = length

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}

B=0.005026=5.026\times10^{-3}\ T

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is 5.026\times10^{-3}\ T.

4 0
3 years ago
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