Answer:
width of fringes are increased
Explanation:
The width of central maxima is given by the following expression
Width = 2 x Dλ / d
- D is distance of screen from source , d is slit width and λ is wavelength of light source.
- Here we see , on d getting decreased , width will increase because d is in denominator .
Due to increased width , position of a fringe moves away from the centre.
Answer:
Choices A, B, and C are correct.
Explanation:
Let us look at each of the choices one by one:
A. It is a vector
Yes. Velocity is a vector, or it's a speed with direction.
B. It is the change in displacement divided by the change in time.
Yes. The velocity can be written as
where 
is the displacement—a vector quantity.
C. It can be measured in meters per second.
Yes. The units of velocity are m/s, but also with a unit vector indicating the direction.
D. It is the slope of the acceleration vs. time graph.
Nope. The velocity is the slope of displacement vs. time graph.
Hence, only choices A, B, and C are correct.
Answer: 3,383.5 kg
Explanation:
from the question we were given the following
tension (T) = 4.5 x 10^4 N
maximum acceleration (a) = 3.5 m/s^2
acceleration due to gravity (g) = 9.8 m/s^2 ( it's a constant value )
mass of the car and its contents (m) = ?
we can get the mass of the car and it's contents from the formula for tension which is T = ma + mg
T = m (a + g)
therefore m = T / (a+g)
m = (4.5 x 10^4 / ( 3.5 + 9.8 )
m = 3,383.5 kg
Answer:
7557120 kg/hour
Explanation:
Given data;
Volume of air in one second = 1640 L
Density of air = 1.28 kg/L
Mass of air in 1 hour =?
Since mass = density × volume
==> Mass of air in one second = 1.28 ×1640 = 2099.2 kg
==> Mass of air in one minute = 2099.2×60=125952 kg
==> Mass of air in one hour = 125952× 60 = 7557120 kg
So rate of flow of air is 7557120 kg/hour
Sound at 70 dB is 70 dB louder than the human reference level. That's 10⁷ times as much as the reference sound power.
Sound at 73 dB is 73 dB louder than the human reference level. That's 10⁷.³ or 2 x 10⁷ times as much as the reference sound power.
Sound at 80 dB is 80 dB louder than the human reference level. That's 10⁸ or 10 x 10⁷ times as much as the reference sound power.
Now we can adumup:
Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)
Intensity = (13 x 10⁷) times the sound power reference intensity.
Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)
Intensity = 70 + 10 log(13)
Intensity = 70 + 10 (1.114)
Intensity = 70 + 11.14
Intensity = <em>81.14 dB</em>
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Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct. For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.
