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olchik [2.2K]
3 years ago
5

Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air

resistance, which of the following statements are true when the two balls hit the ground?
(a) The kinetic energy of A is the same as the kinetic energy of B
(b) The kinetic energy of A is half the kinetic energy of B.
(c) The kinetic energy of A is twice the kinetic energy of B.
(d) The kinetic energy of A is four times the kinetic energy of B Explain your answer why.
What is the definition of General Plan Motion? What would be the effective methodology or approach to solve a rigid body kinematics problem?
Engineering
1 answer:
statuscvo [17]3 years ago
5 0

Answer:

The kinetic energy of A is twice the kinetic energy of B

Explanation:

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A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle
Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

X_{s} =1.23(\frac{V_{line}^{2}  }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms

the network voltage phase is

V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV

the power transmitted is equal to:

|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV

the line induced voltage is

|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV

7 0
3 years ago
Free brainlist because im new and i just want to but you have t friend me first
Amiraneli [1.4K]
Okay sure.









I’ll 1)chords
2)pulse
3)aerophone
4) the answer is C
5)rhythm

Pretty sure those are the answers
4 0
3 years ago
Xyxyydfufggivivihogcufuf​
Genrish500 [490]

Answer:

ummm why is you doing this

Explanation:

It doesnt make sense.

7 0
3 years ago
What are the three most common types of relearn procedures?
WITCHER [35]

Answer:

The three types of relearn procedures are auto relearn, stationary and OBD.

Explanation:

In TPMS system, after the direct service like adjustment of air pressure, tire rotation or replacement of sensors etc, is performed then maximum vehicle  often needs TPMS system relearn that needs to be performed.

For performing these relearn procedure, there are mainly three types:

  1. auto relearn
  2. stationary relearn
  3. OBD

After applying the relearn process, the TPMS system will again be in proper function.

8 0
3 years ago
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