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tigry1 [53]
3 years ago
11

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

n container with 22 holes, each 0.15 cm in diameter. If the water in the hose has a speed of 0.95 m/s, at what speed does it leave the sprinkler holes
Physics
1 answer:
trapecia [35]3 years ago
6 0

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden d_1=1.1cm

Speed of water in the hose is v_1=0.95m/sec

Number of holes n = 22

Diameter of each holes d_2=15cm

According to continuity equation A_1v_1=A_2v_2

d_1^2\times v_1=22\times d_2^2v_2

1.1^2\times 0.95=22\times 0.15^2\times v_2

v_2=2.322m/sec

So water leaves the sprinkler at a speed of 2.322 m/sec

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A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
pochemuha
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
7 0
4 years ago
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio
Harman [31]

Answer:

The  angular acceleration is  \alpha  = 3.235 \ rad/s ^2

Explanation:

From the question we are told that

    The moment of inertia is  I  =  0.034\ kg \cdot m^2

     The  net torque is  \tau  =  0.11\ N \cdot m

Generally the net torque is mathematically represented as

           \tau =  I  *  \alpha

Where \alpha is the angular acceleration so  

        \alpha  =  \frac{\tau }{I}

substituting values

         \alpha  =  \frac{0.1 1}{ 0.034}

        \alpha  = 3.235 \ rad/s ^2

6 0
3 years ago
he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?
balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

U = K

\frac{kQe}{R} = \frac{1}{2}mv^2

\frac{2kQe}{mR} = v^2

now take square root of both sides

v =\sqrt{\frac{2kQe}{mR}}

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

k = 9 * 10^9


5 0
3 years ago
Read 2 more answers
A car drives 100 km north, 50 km east, then 10 km south. What is the car's displavement?​
Over [174]

Answer:

\sqrt{ ({40}^{2} }  +  {50}^{2} ) =  \sqrt{(1600 + 2500)}  =  \sqrt{4100}  = 64.03

6 0
4 years ago
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