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Naddik [55]
2 years ago
10

A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets lo

cated on the perimeter of a wheel. Determine the power generation potential of this water jet.
Physics
1 answer:
snow_tiger [21]2 years ago
5 0

Answer:

216000 W or 216 kW

Explanation:

Power: This can be defined as the rate at which energy is consumed or used, The S.I unit of power is Watt (W)

Generally,

Power = Energy/time

P = E/t........................ Equation 1.

But

E = 1/2mv²..................... Equation 2

Where m = mass, v = velocity.

Substitute equation 2 into  equation 1

P = 1/2mv²/t...................... Equation 3

Let flow rate (Q) = m/t

Q = m/t................ Equation 4

Substitute equation 4 into equation 3

P = Qv²/2........................ Equation 5

Where Q = flow rate, v = velocity, P = power.

Given: Q = 120 kg/s, v = 60 m/s

Substitute into equation 5

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus the power generation potential of the water jet = 216000 W or 216 kW

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

5 0
2 years ago
Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh
ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0
2 years ago
An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t
Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

\omega = positive

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

\alpha = negative

so here correct answer is

B. W is positive and a is negative

8 0
2 years ago
A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied​
never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman  = 300N/m²

Area = 0.02m²

Unknown:

Force applied  = ?

Solution:

Pressure is the force per unit area on a body

        Pressure  = \frac{force}{area}

         Force  = Pressure x area

        Force  = 300 x 0.02  = 6N

8 0
2 years ago
An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars
Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}

the car did not exceed the speed limit

Hope this helps!!

7 0
2 years ago
Read 2 more answers
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