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ddd [48]
2 years ago
7

Two billiard balls of equal mass are traveling straight toward each other with the same speed. They meet head-on in an elastic c

ollision. What is the total momentum of the system containing the two balls before the collision? What about after the collision?
Physics
1 answer:
Rus_ich [418]2 years ago
5 0

Answer:

0 kg m/s before and after collision

Explanation:

Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:

P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

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Timmy drove 2/5 of a journey at an average speed of 20 mph.
mixer [17]

Answer:

4hr

Explanation:

5 0
2 years ago
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Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve
givi [52]

If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward.  So since the current is going at an angle then it has a x and y component.  So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

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2 years ago
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B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find
lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

4 0
2 years ago
(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
castortr0y [4]

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0
2 years ago
Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10
DochEvi [55]

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

             F =G \frac{m_1m_2}{r^2}  

let's calculate

             F = 6.67 10⁻¹¹  1  1 / 1²

             F = 6.67 10⁻¹¹ N

Electric force

             F = k \frac{q_1q_2}{r^2}  

indicates that the charge is q = 10 C

            F = 9 10⁹ 10 10 / 1²

            F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

3 0
2 years ago
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