The answer is (a) because movement is acceleration
1). The equation is: (speed) = (frequency) x (wavelength)
Speed = (256 Hz) x (1.3 m) = 332.8 meters per second
2). If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.
If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.
If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.
3). The equation is: Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)
Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.
That's ' 200 k Hz ' .
Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised. They broadcast on several different frequencies,
and one of them is 198 kHz !
Answer
A. the work done on the refrigerant in each cycle is 105kJ
B the coefficient of performance of the refrigerator is 4.8
Explanation
Given data
Work done at high temperature T2 Qh=610kJ
Work done at low temperature T1 Ql=505kJ
We know that the net work done by the refrigerator is expressed as
Wnet= Qh-Ql
=610-505
=105kJ
Also we know that the coefficient of performance is expressed as
COP= Ql/Wnet
COP= 505/105
= 4.8
The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.
Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y
Explanation:
According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
(1)
Where;
is the Gravitational Constant
is the mass of the Earth
is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
So for satellite X, the orbital period 
is:
(2)
Where 
(3)
(4)
For satellite Y, the orbital period 
is:
(5)
Where 
(6)
(7)
This means 
Now let's calculate the tangential speed for both satellites:
<u>For Satellite X:</u>
(8)
(9)
<u>For Satellite Y:</u>
(10)
(11)
This means 
Therefore:
Satellite X has a greater period and a slower tangential speed than Satellite Y
