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Vlad1618 [11]
2 years ago
15

calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​

Physics
1 answer:
zheka24 [161]2 years ago
5 0

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

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A man runs at an average speed of 5.0m/s how long will it take him to run 5.2km on a perfectly straight line
Anni [7]

The time taken is 1040 s.

<h3>What is speed?</h3>

The term speed refers to the rate at which the distance changes per unit time. This is why we define speed as the ratio of the distance to time for a body that is moving along a straight line.

Now;

We must first convert the distance to meters;

distance = 5.2km  or 5200m

Speed = distance/time

time = distance/speed

time =  5200m/5 m/s

time = 1040 s

Learn more about speed:brainly.com/question/28224010

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8 0
1 year ago
……………………………………………………………..?
kotegsom [21]

Answer:

C

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

3 0
2 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

5 0
3 years ago
Picture a ball traveling at a constant speed around the inside of a circular Structure. is the ball acceleration? Explain why?
baherus [9]
Yes. On a circular path, the direction of motion is constantly changing. Change of direction is acceleration, even at constant speed.
8 0
3 years ago
Read 2 more answers
Which substance cannot be seperated physically or chemically?
scoray [572]

Answer:

What is a Pure Substance?

Explanation:

It is something which cannot be divided into parts by physical means, as it's all made up of the same thing. Pure substances are either elements or compounds. Elements can NOT be separated into other types of matter (physically or chemically).

4 0
3 years ago
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