Answer:
Explanation:
At t = 1 velocity = 0
At t = 3 velocity = 3
slope of the line = 3-0 / 3-1 = 3/2
At t = 2
velocity = 3/2 x ( 2 - 1 )
= 1.5 m /s
velocity at t = 2 is 1.5 m /s
Position at t = 2 :
displacement at t = 2
area of graph upto t = 2
= 1 / 2 x 1 x 1.5 = .75
position at t = 2 :
= initial position + displacement
= 10 + .75 = 10.75 m
position at 6 s :
displacement at t = 6
area of curve upto t = 6
= 1 / 2 x 2 x 3 + 3 x 3 + 1/2 x 3 x ( 4.5 - 3 )
= 3 + 9 + 2.25
= 14.25 m
position at t = 6
= initial position + displacement
= 10 + 14.25 = 24.25 m
position at 9 s :
displacement at t = 9
area of curve upto t = 9
= 1 / 2 x 2 x 3 + 4 x 3 + 1/2 x 4 x ( 5 - 3 )- 1/2 x 2 x 1.5
= 3 + 12 + 4 - 1.5
= 17.5 m
position at t = 9
= initial position + displacement
= 10 + 17.5 = 27.5 m
By utilizing principles of projectile motion, it is found that the football clears the crossbar by approximately 10.75 meters.
To determine by how much the ball clears or falls short of clearing the crossbar, we need to use the physics principles of projectile motion. The maximum height 'h' of the football can be given by equation of motion: h = (v²sin²θ) / (2g), where 'v' is the initial velocity, 'g' is the acceleration due to gravity and 'θ' is the angle of projection.
Substituting the given values: h = [(20)²sin²53°] / (2*9.8) ≈ 13.8 m above the ground when it was kicked. However, the football was kicked from ground level, so we need to subtract the height of the crossbar from this value, which is 3.05 m. Thus, the ball clears the crossbar by approximately: 13.8 - 3.05 = 10.75 m.
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Answer: The frequency of the wave is 0.02 Hertz.
Explanation:
Frequency is a parameters which describe a wave. This is defined as the number of oscillations happening per unit second.
Mathematically,
In the question,
Number of oscillations happening per second = 50
Putting this value in above equation, we get:
Hence, the frequency of the wave is 0.02 Hertz.
The correct answer to the question will be that the electric force between them will be decreased.
EXPLANATION:
Let us consider the charge contained by A and B are denoted as Q and Q' respectively.
Let R is the separation distance between them.
As per Coulomb's law in electrostatics, the force of attraction or repulsion between the two charge bodies A and B will be -
Coulombic force F =
Here, is the permittivity of the medium in which the charges are present.
From above, we see that electric force is inversely proportional to the square of separation distance between them.
Mathematically it can be written:
As per the question, the distance between A and B is increased.
Hence, the electric force between A and B will be decreased.