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3 years ago

I'm bored. I want to talk to you humans! weird. STUPID QUARANTINE!!! So. What are yall up to?

Engineering

1 answer:

kiruha [24]3 years ago

3 0

Answer:

nothing much what class r u in

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A particular motor rotates at 3000 revolutions per minute. What is its speed in rad/sec, and how many seconds does it takes to m

Leno4ka [110]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

$\omega=\dfrac{2\pi N}{60}\ rad/s$

Now by putting the values in the above equation

$\omega=\dfrac{2\pi \times 3000}{60}\ rad/s$

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

$\omega=\dfrac{ 3000}{60}\ rad/s$

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

$\dfrac{1}{50}=0.02\ s$

4 0

3 years ago

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

8 0

3 years ago

Who play 1v1 lol unblocked games 76

nikklg [1K]

Answer:

I did not what is it about?

8 0

3 years ago

Read 2 more answers

To check for ripple voltage from the alternator, connect a digital multimeter and select

Roman55 [17]

Answer:

isn't it summer? sjsushsiansudndd

8 0

3 years ago

While driving down the highway, your car must overcome the forces of friction and air resistance in order to keep the car moving

ElenaW [278]

Answer:

Efficiency of the engine equals 20%

Explanation:

We know that when the car moves it must do work against the resisting forces to keep moving and this work is spend as energy by the engine to keep the car moving.

we know that

$Work=Force\times Displacement$

Thus to keep the car moving for 100,000 meters the theoretical work that requires to be done equals

$Work=560N\times 100,000m=56\times 10^{6}Joules$

Now the actual energy spend by the car equals the energy spend by burning 2.8 gallons of gasoline.

Thus the energy produced by burning 2.8 gallons of gasoline equals

$2.8\times 100\times 10^{6}=280\times 10^{6}Joules$

Thus the efficiency is calculated as

$\eta =\frac{56\times 10^{6}}{280\times 10^{6}}\times 100\\\\\therefore \eta =20percent$

5 0

3 years ago