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Romashka [77]
3 years ago
14

In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg m/s. After the

two balls collide, the final momentum of the cue ball is 0.01 kg m/s
Physics
1 answer:
ella [17]3 years ago
5 0

Answer: 0.22

Explanation: Subtract 0.01kg from 0.23kg

You might be interested in
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's
Dmitrij [34]

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0
3 years ago
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what
Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

E=\dfrac{k}{d^2}

We can write it as follows :

\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2

Put all the values,

\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}

So, the correct option is (a).

5 0
3 years ago
Early humans used rocks as tools to make other things as well as to construct buildings. Why was it better for them to use rocks
babymother [125]
Wood isn't as tough as rock. Wood also breaks down in weather and cracks under pressure. Plus rocks were more accessible.
6 0
3 years ago
Read 2 more answers
What quantity measures the number of complete cycles an oscillation makes per second? A. period B. amplitude C. frequency D. for
Free_Kalibri [48]

Hello, there Jcparris

Your answer is going to be C. Frequency

If my answer helped please leave a thank rate 5 stars and the most important rank me braiinliest thank you and have a great day!

8 0
3 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
3 years ago
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