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lapo4ka [179]
2 years ago
4

In a game of pool, a 0.4 kg cue ball is traveling at +0.80 m/s when it hits a slower striped ball moving at +0.38 m/s. After the

collision, the striped ball moves off at +0.62 m/s. What is the magnitude of the final velocity of the cue ball? Assume all pool balls have the same mass.
A. 0.20 m/s
B. 0.56 m/s
C. 1.0 m/s
D. 1.8 m/s
Physics
2 answers:
ziro4ka [17]2 years ago
4 0

Answer:b

Explanation I took the quizz and got it rightttttttttttt yessirrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

-BARSIC- [3]2 years ago
3 0

Substitute the known values:

(0.4 kg)(0.80 m/s) + (0.40 kg)(0.38 m/s) = (0.40 kg)(0.62 m/s) + (0.40 kg)(x m/s)

The value of x from the equation is: x = 0.56 m/s

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<h2><u>Answer:</u></h2>

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:

1. (There would be variables that would be hard to control, leading to less reliable data.)

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2 years ago
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For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

7 0
2 years ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
2 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i
Debora [2.8K]

Answer:

(A) Total energy will be equal to 0.044\times 10^{-5}J

(b) Energy density will be equal to 0.0175J/m^3

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate A=\pi r^2=3.14\times 0.02^2=0.001256m^2

Distance between the plates d = 0.50 mm = 0.50\times 10^{-3}m

Permitivity of free space \epsilon _0=8.85\times 10^{-12}F/m

Potential difference V =200 volt

Capacitance between the plate is equal to C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F

(a) Total energy stored in the capacitor is equal to

E=\frac{1}{2}CV^2

E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J

(b) Volume will be equal to V=Ad, here A is area and d is distance between plates

V=0.001256\times 0.02=2.512\times 10^{-5}m^3

So energy density =\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3

7 0
2 years ago
A lamp uses a 230 V mains supply and transfers 96 J of energy every second. Work out the current through the lamp. Give your ans
sertanlavr [38]

Answer:

0.4 A

Explanation:

From the question,

Electric power = Voltage×current

P = VI.......................... Equation 1

Make I the subject of the equation

I = P/V..................... Equation 2

Given: P = 96 J/s, V = 230 V.

Substitute into equation 2

I = 96/230

I = 0.4 A.

Hence the current is 0.4 A

8 0
2 years ago
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