Answer:
c.
Explanation:
Initial velocity of cheetah,u=1 m/s
Time taken by cheetah =4.8 s
Final velocity of cheetah,v=28 m/s
We have to find the acceleration of this cheetah.
We know that
Acceleration,
Where v=Final velocity of object
u=Initial velocity of object
t=Time taken by object
Using the formula
Then, we get
Acceleration, a=
Acceleration=
Hence, the acceleration of cheetah=
Answer:
mass of box 1 = 2.20 kg
mass of box 2 = 5.93 kg
Explanation:
Let the mass of box 1 and box 2 is respectively
and 
so we will have
Force applied on box 1 then acceleration



Now we know that contact force between them in above case is given as



now we have

Los Angeles lies on the Pacific plate, San Francisco lies on the North American plate, and the meeting point of the two cities is mathematically given as
T = 120 x 105 years
<h3>What is the meeting point of the two plates?</h3>
Generally, the equation for Distance is mathematically given as
D = Rate x Time
Therefore
T = D/R
T = (600 x 105) / 5
T = 120 x 105 years
In conclusion, the meeting point of the two plates will be
T = 120 x 105 years
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The speed at which sound travels through the gas in the tube is 719.94m/s
<u>Explanation:</u>
Given:
Frequency, f = 11999Hz
Wavelength, λ = 0.03m
Velocity, v = ?
Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.
As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:
λ/2 = 0.03/2
λ = 0.06m
We know,
v = λf
v = 0.06 X 11999Hz
v = 719.94m/s
Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s