We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).
Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters
Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg
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and now we're ready to go !
Gravitational force =
G M₁ M₂ / R²
= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²² kg)/</span>(8.098 x 10⁶ m)²
= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons
= (I get ...) 4.463 x 10²³ Newtons
That's almost exactly 10²³ pounds
= 50,153,000,000,000,000,000 tons.
Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".
Answer: meter per second
Explanation: meter per second
Speed has the dimensions of distance divided by time. The SI unit of speed is the meter per second, but the most common unit of speed in everyday usage is the kilometer per hour or, in the US and the UK, miles per hour. For air and marine travel the knot is commonly used.
A wave is a result of the disturbance in the equilibrium state. There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.
Chapter name please then I can answer
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
![y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%20%3D%20%281.22%20cm%29sin%5B14.4%20m%5E%7B-%201%7Dx%5Dcos%5B%28166%5C%20rad%2Fs%29t%5D)
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K = 
where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a) 
where
K = propagation vector
(b) The length of the string is given by:
(c) Now, we first find the frequency of the wave:
Now,
Speed of the wave is given by:
