A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.

Answers

Answer 1
Answer:

Answer:

 Recoil velocity of cannon = 2.92 m/s

Explanation:

By law of conservation of momentum, we have momentum of cannon = momentum of cannonball.

 Mass of cannon = 1200 kg

Mass of cannon ball = 100 kg

Velocity of cannon ball = 35 m/s

 We have, Momentum of cannon = momentum of cannon ball

                  1200 x v = 100 x 35

                            v =3500/1200 = 2.92 m/s

 Recoil velocity of cannon = 2.92 m/s

Answer 2
Answer:

Final answer:

The recoil speed of the cannon is 2.92 m/s.

Explanation:

To find the recoil speed of the cannon, we can use the conservation of momentum. The initial momentum of the cannon and cannonball system is zero since the cannon is at rest before firing. The final momentum is the sum of the momenta of the cannon and cannonball after firing. Using the equation:

Initial momentum = Final momentum

(mass of cannon) x (recoil speed of cannon) = (mass of cannonball) x (velocity of cannonball)

Plugging in the given values:

(1200 kg) x (recoil speed of cannon) = (100 kg) x (35 m/s)

Solving for the recoil speed of the cannon:

recoil speed of cannon = (100 kg x 35 m/s) / 1200 kg = 2.92 m/s

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QuestIuI(2 PUMILS)
How much power is needed to lift a 750 kg elephant 14.3 m in 30.0 seconds?

Answers

Given Information:

Mass of elephant = m = 750 kg

Height = h = 14.3 m  

time = t = 30 seconds

Required Information:

Power needed to lift elephant = P = ?

Answer:

Power needed to lift elephant ≈ 3507 watts

Explanation:

As we know power is given by

P = PE/t

Where PE is the potential energy and t is the time

Potential energy is given by

PE = mgh

Where m is the mass of elephant, g is the gravitational acceleration and h is the height to lift the elephant.

PE = 750*9.81*14.3

PE = 105212.25 Joules

Therefore, the required power to lift the elephant is

P = PE/t

P = 105212.25/30

P ≈ 3507 watts

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

Answers

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

Final answer:

The Cartesian components of the resultant force and the couple moment are calculated by summing up all the forces and moments acting on the object. The resultant force is 1724 N and the couple moment is 29.764 N*m.

Explanation:

The resultant force and couple moment in the Cartesian coordinate system can be obtained by summing up all the forces and moments acting on the object. In this case, we have the forces F1, F2, F3, F4 and the couple moment MA acting on the object. The resultant force (FR) can be calculated as the sum of all the forces, i.e., FR = F1 + F2 + F3 + F4. Using the values given, FR = 510 N + 306 N + 501 N + 407 N = 1724 N. The resultant moment (MR) can be calculated as the sum of all the moments, i.e., MR = d1*F1 + d2*F2 + d3*F3 + d4*F4 - MA. Using the values given, MR = 0.880 m * 510 N + 1.11 m * 306 N + 0.560 m * 501 N + 2.08 m * 407 N - 1504 N*m = 29.764 N*m. Therefore, the Cartesian components of the resultant force and the couple moment are 1724 N and 29.764 N*m respectively.

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A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam experience a phase change, and if so, by how much is the phase of the beam changed?

Answers

The reflected beam experienced a phase change of about 180°.

What is reflection in the glass surface?

According to Snell's law, the light that incident on the glass surface will be reflected and transmitted at an angle equals to the angle of incidence.  

By the observation of refractive index of the glass for the normal incidence only 4% of the light is transmitted or reflected.

The light passing through glass is not only reflected on the front surface, but also on the back. For several times the light will gets reflected back and forth. So, the total reflectance through a glass window can be calculated as

                                    2·R / (1+R).

Thus, A light wave travelling in air is reflected by a glass barrier will undergo a  phase change of 180°, while light travelling in glass will not undergo a phase change if it is reflected by a boundary with air.

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Answer:

180 degree phase change

Explanation:

Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

Energy can be transferred from a closed system to the surroundings by: (A) Internal chemical reactions (B) Heat (C) Shaft work (D) Change in pressure without changing volume (E) Mass transfer

Answers

Answer:

option the correct is B

Explanation:

Let's analyze the different options, for a closed system

- an internal reaction changes the system, but does not affect the surrounding environment

- Heat, is a means of transfer that occurs when two bodies are in contact, one of the body can be a closed system since the only thing that happens is thermal transfer, without movement of the system itself. This is the correct result.

- Work implies a movement whereby the system must be mobile, it is not an option

- Pressure change. change in the system, but does not affect the environment

- Mass transfer is not possible in a closed system

After analyzing each option the correct one in B

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

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