Answer:
Approximately (assuming that the projectile was launched at angle of above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing is the same as the altitude at which this projectile was launched: .
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is (upwards,) the vertical velocity right before landing would be (downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be . In other words, .
Hence, the time it takes to achieve a (vertical) velocity change of would be:
.
Hence, this projectile would be in the air for approximately .
Answer:
When there is nearsightedness or myopia
Explanation:
As, in myopia the image is formed in front of the retina.
Which makes things looking at things near very easy, but looking at far away things very difficult.
So concave lens makes things look bigger, so therefore, it is used during myopia, to make things look bigger when they are far away
The answer for the following answer is answered below.
- <u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
- <u><em>Therefore the option for the answer is "B".</em></u>
Explanation:
Frequency (f):
The number of waves that pass a fixed place in a given amount of time.
The SI unit of frequency is Hertz (Hz)
Time period (T):
The time taken for one complete cycle of vibration to pass a given point.
The SI unit of time period is seconds (s)
Given:
frequency (f) = 100 Hz
wavelength (λ) = 2.0 m
To calculate:
Time period (T)
We know;
According to the formula;
<u>f =</u><u></u>
Where,
f represents the frequency
T represents the time period
from the formula;
T =
T =
T = 0.01 seconds
<u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
Beaker would be most appropriate for measuring the approximate volume of a liquid.