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Dmitry [639]
4 years ago
13

Mike shoots a large marble (Marble A, mass: 0.05 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marble

A was initially moving at a velocity of 0.6 m/s, but after the collision it has a velocity of −0.2 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer
Physics
1 answer:
SpyIntel [72]4 years ago
7 0

<u>Answer</u>

1 1/3 m/s


<u>Explanation</u>

The momentum before collision and after collision is always conserved.

m₁v₁ = m₂v₂

(0.05×0.6) + (0.03×0) = (0.03 × v) + (0.05 × -0.2)

0.03 + 0 = 0.03v - 0.01

0.03v = 0.03+0.01

0.03v = 0.04

v = 0.04/0.03

  = 4/3

   = 1 1/3 m/s

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Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel
Anon25 [30]

Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

5 0
3 years ago
A small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is rele
Lunna [17]

Answer:

Part a)

W = 0.15 J

Part b)

W = 0.11 J

Part c)

U = 0.19 J

Part d)

U = 0.038 J

Part e)

U = 0.075 J

Part f)

It is independent of the speed of the object so all part answers will remain the same

Explanation:

Part a)

As we know that Point P is at height 5R while point Q is at height R

so the work done by gravity from P to Q is given as

W = mg(5R - R)

W = 0.032(9.8)(4)(0.12)

W = 0.15 J

Part b)

When it reaches to the top of the loop then its final height from ground is

h = 2R

so work done from P to Q is given as

W = mg(5R - 2R)

W = 3mgR

W = 0.11 J

Part c)

Potential energy at P point is given as

U = mgH

U = 0.032(9.8)(5)(0.12)

U = 0.19 J

Part d)

Potential energy at Q point is given as

U = mgH

U = 0.032(9.8)(0.12)

U = 0.038 J

Part e)

Potential energy at top point is given as

U = mgH

U = 0.032(9.8)(2)(0.12)

U = 0.075 J

Part f)

Since all the answer from part a) to part e) depends only upon the position of the object.

So here we can say that it is independent of the speed of the object so all part answers will remain the same

8 0
4 years ago
Enhancing cardiorespiratory and circulatory efficiency; using your heart and lungs when exercising is called what
Vinil7 [7]
Answers : Agility number 2
4 0
3 years ago
Read 2 more answers
How much does a 0.15 kg baseball weigh on earth?
Gnesinka [82]
1.472 N

to get weight you multiply an object's mass in kilograms with the acceleration of gravity(9.81m/s) :)
7 0
3 years ago
Which object would be affected more by friction. A bowling ball or a ping pong ball? EXPLAIN WHY
Nadusha1986 [10]

Answer:

a bowling ball because its heavier

Explanation:

3 0
3 years ago
Read 2 more answers
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