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2 years ago

A sound wave has a frequency of 425 Hz. What is the period of this wave

Physics

1 answer:

Oduvanchick [21]2 years ago

5 0

Answer: The time period of the given wave with frequency of 425 Hertz is 0.0023 seconds.

Explanation:

Frequency of the wave = 425 Hertz = $425 sec^{-1}$

$\text{Time period}=\frac{1}{Frequency}$

$\text{Time period}==\frac{1}{425 sec^{-1}}=0.0023 sec$

The time period of the given wave with frequency of 425 Hertz is 0.0023 seconds.

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A mass M tied to a light string is moving in a vertical circle. The tension in the string at the top is TT and TB at the bottom

natita [175]

Answer:

Explanation:

Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle

torque is zero as , centripetal force passes through axis of rotation that is center.

So work done by centripetal force = 0

So work done by tension on M = 0

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2 years ago

The amount of pressure exerted by a solid is based on

Veronika [31]

Answer:

pressure= force/area

A solid resting on a horizontal surfaceexerts a normal contactforce equals to its weight. The pressure of the solid on the surface depends on the area of contact. (b) the area of contact between the two surfaces. The greater the force or the smaller the area the greater the pressure.

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2 years ago

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A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

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2 years ago

An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of

stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

$a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2$

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

$a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2$

Option D is the correct answer.

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3 years ago

Tomorrow i have a science test, could someone please fill me in on the water cycle?

Lorico [155]

the cycle of processes by which water circulates between the earth's oceans, atmosphere and land involving precipitation as rain and snow drainage in streams and rivers and return to the atmosphere by evaporation and transpiration.

Hope this gives you a little bit more information!

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2 years ago