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Olegator [25]
3 years ago
9

1. A 4000-kg truck traveling with a velocity of 20 m/s due south collides headon with a 1350-kg car traveling with a velocity of

10 m/s due north. The two vehicles stick together after the collision. a. What is the momentum of each vehicle prior to the collision? b. What are the size and direction of the total momentum of the two vehicles after they collide?
Physics
1 answer:
velikii [3]3 years ago
4 0

(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) The size of momentum is 93500 kgm/s and it will be directed towards South.

Explanation:

The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.

(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.

Momentum of truck = Mass * Speed = 4000 * 20 =80000  kgm/s

Similarly, the momentum of car will be

Momentum of car = 1350 * 10 = 13500 kgm/s

So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.

Momentum of vehicles after collision = total momentum before collision

Momentum after collision = 80000+13500 = 93500 kgm/s.

The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.

Thus, the size of momentum is 93500 kgm/s and it will be directed towards South.

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F_2 = \frac{G*100M*m}{(10R)^2}

F_2 = \frac{GMm}{R^2}

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\frac{F_1}{F_2} = 1

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The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the
ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

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A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when
Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

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  • <em>The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²</em>
  • <em>Air pressure at sea level = 101.325 kPa</em>

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896

From supersonic isentropic table, at \frac{A}{A^*} = 2.896, we can determine the following;

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The tank pressure is calculated as follows;

\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn  more about converging-diverging nozzle design here: brainly.com/question/13889483

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