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sattari [20]
2 years ago
9

Motors are used to convert electrical energy into mechanical work and the output mechanical work of a motor is rated in horsepow

er. 1 hp equals _____.
Engineering
1 answer:
Vladimir [108]2 years ago
5 0

Answer:

Which discovery did Gregor Mendel make to the 9=086

Explanation:

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How might an operations manager alter operations to meet customer demand? Name at least two ways.
Citrus2011 [14]
One way is manager changes itself and the other one is the same thing i think.
4 0
3 years ago
A compound sliding miter saw can be used to make
Sunny_sXe [5.5K]

Answer:

D

Explanation:

3 0
2 years ago
How many hours should it take an articulated wheel loader equipped with a 4-yd^3 bucket to load 3000 yd^3 of gravel (average mat
densk [106]

Answer:

17 hours 15 minutes

Explanation:

See attached picture.

4 0
3 years ago
If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0
3 years ago
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