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ioda
2 years ago
12

Each vertical line on the graph is 1 millisecond (0.001 s) of time. What is the period and

Physics
1 answer:
Mekhanik [1.2K]2 years ago
6 0

Explanation:

Given that,

Each vertical line on the graph is 1 millisecond (0.001 s) of time.

We need to find the period and the frequency of the sound wave. The period of a wave is equal to the each vertical line on graph i.e. 0.001 s.

Let f be the frequency of the sound wave. So,

f = 1/T

i.e.

f=\dfrac{1}{0.001 }\\\\f=1000\ Hz

So, the period and the frequency of the sound waves is 1 milliseond and 1000 Hz respectively.

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A i think so

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To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh
sp2606 [1]

Answer:

v = 15.45 m/s

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

W_f = (U_i + K_i) - (U_f + K_f)

here we know that

W_f = F_f . d

W_f = 40 \times 4

W_f = 160 J

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3 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

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v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
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