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zhuklara [117]
3 years ago
9

How can you determine the diameter of water

Physics
1 answer:
katrin [286]3 years ago
8 0

Answer:

To test your hydraulic skills, your Boss has requested you calculate the difference in water surface elevation between two reservoirs that are connected.

Explanation:

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What is the force in Newtons when the mass of a truck is 1000Kg and the acceleration of the truck is 3m/s2?
olga nikolaevna [1]

Answer:

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3 years ago
Why are electromagnets used in metal scrap yards?
Andrej [43]
Because the electromagnets can pick up magnetic material and move it around, hope this helps
8 0
3 years ago
Read 2 more answers
A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n
uranmaximum [27]

The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

f = (1-\frac{v_0}{v})f_0

Here,

f_0 = Frequency of the sound from the Whistle

f = Frequency of sound heard

v = Speed of the sound in the Air

Replacing we have that

1- \frac{v_0}{343} = \frac{20kHz}{21kHz}

\frac{v_0}{343} = 1-\frac{20}{21}

\frac{v_0}{343} = \frac{1}{21}

v_0 = \frac{1}{21}(343)

v_0 = 16.33m/s

Therefore the minimum speed to know if the whistle is working is 16.33m/s

3 0
3 years ago
Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—th
krek1111 [17]

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

7 0
3 years ago
During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is tr
sergeinik [125]

Answer:

The total work on the ball is 36.25 Joules

Explanation:

There is an important principle on classical mechanics that is the work-energy principle it states that the total work on an object is equal the change on its kinetic energy, mathematically expressed as:

W_{net}=\Delta K = K_f -K_i (1)

With W net the total work, Kf the final kinetic energy and Ki the initial kinetic energy. We're going to use this principle to calculate the total work on the baseball by the force exerted by the bat.

Kinetic energy is the energy related with the movement of an object and every classical object with velocity has some kinetic energy, it is defined as:

K=\frac{mv^2}{2}

With m the mass of the object and v its velocity, knowing this we can use on:

W_{net}= \frac{mv_f^2 -mv_i^2}{2}=\frac{m(v_f^2 -v_i^2)}{2}

In our case vf is the velocity just after the hit and vi the velocity just before the hit. For an average baseball its mass is 145g that is 0.145 kg, then

W_{net}=\frac{0.145*(30.0^2 -20.0^2)}{2}

W_{net}=36.25 J

8 0
3 years ago
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