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IRISSAK [1]
2 years ago
11

Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0

.05 seconds

Physics
1 answer:
GuDViN [60]2 years ago
8 0

Answer:

20 hertz of frequency produced.

Explanation:

\boxed{frequency = \frac{1}{period} }

Here we will find frequency and period should be in second, here given: 0.05 seconds

using the formula:

\boxed{frequency = \frac{1}{0.05} }

\boxed{frequency =20}

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When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of
Fittoniya [83]

Answer:

  75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

  KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

  KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

__

The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

4 0
2 years ago
The escape speed from the moon is much smaller than from earth. True or False
Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

v_{e} = \sqrt{\frac{2GM}{r}}  (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is 5.972x10^{24}kg and its radius is 6371000m

Then, replacing those values in equation 1 it is gotten.

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}  

v_{e} = 11.18m/s

For the case of the Moon:

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}  

v_{e} = 2.38m/s

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0
3 years ago
A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief
fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

x_1 = -120cos70^o = -41.04 km

y_1 = -120sin70^o = -112.763 km

After the 2nd flight its coordinate would be:

x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km

y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km

So in order to fly back to its HQ it must fly a distance and direction of

s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km

tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71

\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o north of east

3 0
2 years ago
Is the Earth bigger than the moon​
sesenic [268]
Yes the Earth is bigger than the Moon.
The moon is one-quarter the size of Earth.
4 0
1 year ago
Read 2 more answers
A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t
marishachu [46]

Answer:

t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

        y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

      864 = 240 t + 16 t²

Let's solve the second degree equation

    16 t² + 240 t - 864 = 0

Let's divide by 16

    t² + 15 t - 54 = 0

The solution of this equation is

     t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

     t = [-15 ±√(225 +216)] / 2

     t = [-15 + - 21] / 2

We have two solutions.

     t₁ = [-15 +21] / 2

     t₁ = 3 s

     t₂ = -18 s

Since time cannot have negative values, the correct t₁ = 3s

4 0
3 years ago
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