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2 years ago

A high frequency sound will have a ?

Physics

2 answers:

stepladder [879]2 years ago

7 0

Answer:

The frequency of a sound wave is what your ear understands as pitch. A higher frequency sound has a higher pitch and the lower the period

masya89 [10]2 years ago

6 0

Answer:

High-frequency sound waves produce high-pitched sounds, and low-frequency sound waves produce low-pitched sounds.

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Look at the potential energy diagram for a chemical reaction. Which statement correctly describes the energy changes that occur

lukranit [14]

The activation energy is 10 kJ and the reaction is exothermic.

8 0

2 years ago

How is voltmeter connected in the circuit to measure the potential difference between two points?

34kurt

Voltmeter is used to find the potential difference between two points.

We always connect it in parallel to the points where we need the potential difference.

Here in order to make the reading accurate we can increase the resistance of voltmeter so that it can not withdraw any current from the circuit.

7 0

3 years ago

What’s the difference between the resistance of 1,000 feet of aluminum No. 14 wire and 1,000 feet of No. 14 copper wire? A. The

Lapatulllka [165]

the answer is A. The aluminum has 0.84 ohms more resistance.

3 0

3 years ago

A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc

Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

$F=\dfrac{mv^2}{r}$

v is maximum tangential speed

$v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s$

So, the maximum tangential speed is 11.7 m/s.

8 0

2 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago