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Georgia [21]
3 years ago
7

water is pumped from a stream at the rate of 90kg every 30s and sprayed into a farm at a velocity of 15m/s. Calculate the power

of the pump.​
Physics
1 answer:
nikitadnepr [17]3 years ago
7 0

Answer:

340 W

Explanation:

Power = change in energy / change in time

P = ΔKE / Δt

P = ½ mv² / Δt

P = ½ (90 kg) (15 m/s)² / (30 s)

P = 337.5 W

Rounded to 2 significant figures, the power is 340 W.

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Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force const
IRISSAK [1]

First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:

\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}

Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:

\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}

Therefore the speed is 1.026 m/s.

5 0
1 year ago
For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou
Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

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here activation energy decrease whereas change in enthalpy remains same.

5 0
3 years ago
A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40
Vinvika [58]

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

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3 years ago
The acceleration of an object would increase if there was an increase in:
kolezko [41]

Answer:

Its either B or A

Explanation:

6 0
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