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Darina [25.2K]
3 years ago
10

A swimmer standing near the edge of a lake notices a cork bobbing in the water. While watching for one minute, she notices the c

ork bob (from up to down to back up) 240 times. What is the frequency in Hz of the water wave going by?
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

4Hz (240 cycles/60 seconds = 4 cycles/second)

Explanation:

hope this helped!

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the International System of Units, the fundamental units are: The meter (symbol: m), used to measure length. The kilogram (symbol: kg), used to measure mass. The second (symbol: s), used to measure time. The ampere (symbol: A), used to measure electric current.

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Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius.
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18 degrees celcius = 64.4

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Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun
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Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction = u\cos\theta

and the speed in the vertical direction is = u\sin\theta upward.

For A:

The speed in the horizontal direction = u\cos75^{\circ}

and the speed in the vertical direction is = u\sin75^{\circ} upward.

For B:

The speed in the horizontal direction = u\cos15^{\circ}

and the speed in the vertical direction is = u\sin15^{\circ} upward.

Let t_A and t_B are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A = u\cos75^{\circ}\times t_A\cdots(i)

The range for the projectile B = u\cos15^{\circ}\times t_B\cdots(ii)

At the highest point, the vertical velocity is 0.

Bu using the equation of motion v^2=u^2 +2a s.

Here, the final velocity v=0, the initial velocity u = u \sin \theta , h= vertical distance up to the highest point, and a= -g (as per sign convention).

So, s= \frac{u^2\sin^2 \theta}{2g}

For projectile A: The maximum height attained.

s_A= \frac{u^2\sin^2 75^{\circ}}{2g}

For projectile B: The maximum height attained.

s_B= \frac{u^2\sin^2 15^{\circ}}{2g}

As \sin^2 75^{\circ} > \sin^2 15^{\circ}, the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point =\frac {u\sin\theta}{g}

where g is the acceleration due to gravity.

So, the total time of flight

= 2 \times \frac {u\sin\theta}{g}

The total time of flight for A

=2 \times \frac {u\sin75^{\circ}}{g}

The total time of flight for A

=2 \times \frac {u\sin15^{\circ}}{g}

Now, from equations (i) and (ii),

The range for the projectile A =

u\cos75^{\circ}\times  \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}

The range for the projectile B =

u\cos15^{\circ}\times  \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.

Both the projectile have the same range.

Hence, option (A) is correct.

7 0
3 years ago
PLEASE ANSWER QUICK :) -
vekshin1

Answer:

At a convergent boundary where both plates are continental, mountain ranges grow and earthquakes are common. At a transform boundary, there is a transform fault and massive earthquakes occur but there are no volcanoes. Processes acting over long periods of time create Earth's geographic features.

Explanation:

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