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Vedmedyk [2.9K]
2 years ago
5

A kid on a playground swing makes a complete to-and-fro swing each 2 seconds.

Physics
1 answer:
sashaice [31]2 years ago
4 0

Answer:

The frequency of the swing: 1/2 Hertz

The period is: 2 Seconds

Explanation:

The time the kid takes to make a complete to-and-fro swing = 2 seconds

The period, T, is the time it takes to make one complete cycle of an oscillatory motion, therefore, we have;

The frequency, f, is the number of cycles completed each second, therefore, we have;

The time for 1 cycle = 2 seconds

2 seconds = 1 cycle

Dividing both sides by 2 gives;

2/2 seconds = 1/2 cycles

2/2 = 1

In 2/2 = 1 seconds The number of cycles completed = 1/2 cycles

Therefore, the number of cycles completed per (one) second = 1/2 cycles

Therefore the frequency of the swing, f = 1/2 cycle/seconds = 1/2 Hertz

The period, T, is the time it takes to complete one to-and-fro swing which is one cycle which is 2 seconds

Therefore, the period is 2 Seconds.

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A baseball rolls off of a .7 m high desk and strikes the floor .25 m always how fast was the ball rolling
labwork [276]

Answer:

the ball's velocity was approximately 0.66 m/s

Explanation:

Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.

Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.

Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:

0.7 = (1/2) g t^2

solve for t:

t^2 = 1.4 / g

t = 0.3779  sec

which we can round to about 0.38 seconds

No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:

horizontal distance covered = vi * t

0.25 = vi * (0.38)

solve for vi:

vi = 0.25/0.38  m/s

vi = 0.65798  m/s

Then the ball's velocity was approximately 0.66 m/s

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2 years ago
The period of a 261-Hertz sound wave is
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The period of a 261 hertz sound wave is
.000383 seconds
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3 years ago
You purchase a ten-year 1,000 bond with semiannual coupons for 982. The bond has a 1,100 redemption payment at maturity, a nomin
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3 years ago
Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

D=1.5(10)^{9}  m This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

1 AU=149,597,870,700 m

So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

8 0
3 years ago
a car's engine is only 12 % efficient at converting chemical energy in gasoline to mechanical energy. If it takes 18000 N of for
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