Answer:
The correct option is;
Amplitude
Explanation:
When transmitting picture signals over the air by broadcasting stations, the signals are shifted into high frequency channels of Very High Frequency (VHF) or Ultra High Frequency (UHF) carrier currents and imposing the the television signal by changing the amplitude of the high frequency carrier current to match the transmitted television signal waveform shape
Answer:

Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A =
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
Answer:
A.The vertical velocity is constantly increasing as the ball falls.
B.The horizontal velocity does not noticeably change as the ball falls.
G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.
H.The velocity vector of the ball changes as it travels through the air.
Explanation:
As the ball is projected horizontally so here the vertical component of the velocity is zero
So the time to reach the ground is given as

so we will have

so this is the same time as the ball is dropped from H height
Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.
So overall the velocity vector will change due to net acceleration g
Answer:
1 x 10¹⁷
Explanation:
Given data:
Radius of the earth = 6000km
Radius of an atom = 60pm
Now, how many orders is the radius of the earth larger than an atom
Solution:
To solve this problem, let us express both quantity as the same unit;
1000m = 1km
6000km = 6000 x 10³m = 6 x 10⁶m
60pm;
1 x 10⁻¹²m = 1pm
60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m
Now;
The order:
= 1 x 10¹⁷
The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz.