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2 years ago

What is the volicity of a rocket?

Engineering

2 answers:

Marysya12 [62]2 years ago

8 0

Answer:

7.9 kilometers per second

Explanation:

Bas_tet [7]2 years ago

7 0

Answer:

I'm gonna assume you meant velocity.

Explanation:

If a rocket is launched from the surface of the Earth, it needs to reach a speed of at least 7.9 kilometers per second (4.9 miles per second) in order to reach space. This speed of 7.9 kilometers per second is known as the orbital velocity, it corresponds to more than 20 times the speed of sound.

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A 3-phase induction motor with 4 poles is being driven at 45 Hz and is running in its normal operating range. When connected to

Dennis_Churaev [7]

Answer:

a) The slip for the given conditions is 0.2074 or 20.74% and the developed torque is 7.14 Nm.

b) The new slip after reducing the torque to 4 Nm is 0.1162 or 11.62% and the new motor speed is 1193.13 rpm.

Explanation:

In order to find the slip we can use the definition as the relative difference between Synchronous speed and the rotor speed and that the developed torque is the ratio between output power and rotor angular velocity.

Slip.

We can find the synchronous speed u sing the following formula

$N_s =\cfrac{120f}p$

Where f stands for the frequency and p the number of poles, so we have

$N_s = \cfrac{120(45)}{4}\\N_s =1350 \,rpm$

Replacing on the slip formula

$S = \cfrac{N_s-N_r}{N_s}$

we get

$S = \cfrac{1350-1070}{1350}\\ S = 0.2074$

Thus the slip for the given conditions is 0.2074 or 20.74%.

Developed torque.

We can find the angular velocity in radians per second

$\omega_r =1070 \cfrac{rev}{min} \times \cfrac{1 \, min}{60 \, s}\times \cfrac{2\pi rad}{1 \, rev}\\\omega_r =112.05 \, \cfrac{rad}{s}$

Thus we can replace on the torque formula

$\tau = \cfrac{P}{\omega_r}\\\tau=\cfrac{800 W}{112.05 \, \cfrac{rad}{s}}$

We get

$\tau = 7.14\, N m$

The developed torque is 7.14 Nm.

Slip for the reduced torque.

The torque is proportional to the slip, so we can write

$\cfrac{\tau_1}{\tau_2}= \cfrac{S_1}{S_2}$

Thus solving for the new slip $S_2$ we have:

$S_2 = S_1 \cfrac{\tau_2}{\tau_1}$

Replacing the values obtained on the previous part we have

$S_2 = 0.2074 \cfrac{4 Nm}{7.14 Nm}\\ S_2=0.1162$

So the new slip after reducing the torque to 4 Nm is 0.1162 or 11.62%

Motor speed.

We can use the slip definition

$S_2 =\cfrac{N_s-N_{r_2}}{N_s}$

Solving for the motor speed we have

$S_2N_s =N_s-N_{r_2}$

$N_{r_2}=N_s-S_2N_s \\N_{r_2}=N_s(1-S_2)$

Replacing values we have

$N_{r_2}=1350 \, rpm(1-0.1162)\\N_{r_2}=1193.13 rpm$

The new motor speed is 1193.13 rpm.

5 0

2 years ago

The 5 ft wide gate ABC is hinged at C and contacts a smooth surface at A. If the specific weight of the water is 62.4 lb/ft3 , f

pochemuha

Answer:

The solution is given in the attachments.

5 0

3 years ago

Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter ???

Andrej [43]

We need to define the variables,

So,

$F_x (x) = 1-e^{-\lambda x}\\F_x (x) = 1-e^{-0.5x}$

Therefore, the probability that the repair time is more than 4 horus can be calculate as,

$P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018$

The probability that the repair time is more than 4 hours is 0.136

b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,

$P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}$

$P(x\geq 12|x>7)=0.6321$

The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63

3 0

3 years ago

What are the horizontal structures beneath a slab that help transfer the load from the slab to the columns?

Delicious77 [7]

Answer: D) Beams

Explanation:

3 0

2 years ago

Read 2 more answers

The ________________ attraction between the Earth and the moon is ______________ on the side of the Earth that happens to be ___

Karolina [17]

Answer:

Gravitational; strongest; facing; closer; near side; toward.

Explanation:

The gravitational attraction between the Earth and the moon is strongest on the side of the Earth that happens to be facing the moon, simply because it is closer. This attraction causes the water on this “near side” of Earth to be pulled toward the moon. These forces of attraction and inertia tends to keep the water in place and consequently, leads to a bulge of water on the near side with respect to the moon.

Also, you should note that what is responsible for the moon being in orbit around the Earth is the gravitational force of attraction between the two planetary bodies (Earth and Moon).

7 0

3 years ago