Answer:
52 mL
Explanation:
Since the pressure of the gas remains constant, we can use the following gas law:
where:
is the initial volume
is the final volume
is the initial temperature
is the final temperature
Substituting numbers and re-arranging the equation, we find the final volume of the gas:
So the correct answer is
1. 52 mL
The distance of tiger's leap from the base of rock is 5.58 m
It is a question of two dimensional motion
The time of motion in two dimensional motion is given by:
t=
where y is the height and g is the acceleration due to gravity
y is given to be 7.5m and let us assume g to be 9.8 m/s^2
t =
= 1.24s
Using time and speed,
We know that distance is the product of speed and time,
Distance= speed x time
speed is given to be 4.5 m/s
distance from the base of rock = 4.5 x 1.24
= 5.58m
Hence the distance of tiger's leap from the base of rock is 5.58 m
Disclaimer:
The acceleration due to gravity is assumed to be 9.8 m/s^2
For further reference:
brainly.com/question/11213880?referrer=searchResults
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Explanation:
Both slinky waves and seismic waves are mechanical waves. Mechanical waves requires a medium to travel. These are of two types i.e. transverse wave and longitudinal waves.
Slinky wave is transverse in nature. The medium particle move to and fro during wave propagation.
Seismic waves are longitudinal in nature. These waves moves in the form of compression and rarefaction. Seismic waves are of two types i.e. P waves and S waves.
Hence, we have concluded that both waves are mechanical waves.
Answer:
N₂/N₁ = 45
Explanation:
The relation of the number of turns with the voltage is given by the EMF equation of transformer:
(1)
<em>where V₁: voltage of the primary winding, V₂: voltage of the secondary winding, N₁: number of turns in the primary winding, and N₂: number of turns in the secondary winding </em>
Assuming that the primary winding is connected to the input voltage supply, and using equation (1), we can calculate the ratio of N₂ to N₁:
So, the ratio N₂/N₁ is 45.
Have a nice day!
Answer:
Speed of a child on the rim is 4.25 m/s.
Explanation:
Given;
diameter of the merry-go-round, d = 5.0 m
period of the motion, t = 3.7 s
one complete rotation of the merry-go-round = πd
one complete rotation of the merry-go-round = π(5) = 15.71 m
Speed is given as distance / time
speed of a child on the rim = 15.71 / 3.7
speed of a child on the rim = 4.25 m/s
Therefore, speed of a child on the rim is 4.25 m/s.