As the boy kicks the football with an angle, due to the effect of the gravitational force, the ball would follow a projectile path which is parabolic in nature. From this idea, we can derive equations pertaining to the maximum height that the ball would reach. At the maximum height of the ball, the velocity of the ball would be equal to zero. From the equations for projectile motion, we would obtain the equation as follows:
Maximum height = v0^2 sin^2 (theta) / 2g
Maximum height = (28.0 m / s )^2 sin^2 (30.0) / 2(9.8 m / s^2)
Maximum height = 10 m
The maximum height that the ball would reach would be 10 m.
No, I heavier object will fall much faster than something lighter than it. This is because it’s more dense and hard so it can cut through the air particles quicker than a lighter object which takes longer to cut through the air and fall
Example:
A rock vs a feather
The rock will fall quicker because it’s more dense and falls straight down and the feather will be slower because it flows slowly down through the air particles
Answer:
The amount of energy wasted by light bulb 2 = 70 J
Explanation:
Sankey diagrams are used to graphically indicate the proportional flow rate of a quantity
The labelled Sankey diagrams, here, indicate the energy transfers in two light bulbs, light bulb 1 and light bulb 2
From the labelled Sankey diagrams, we have;
For light bulb 1, the input energy = 80 J
The useful energy = 40 J
The wasted energy = 80 - 40 = 40 J
For light bulb 2, the input energy = 80 J
The useful energy = 10 J
Therefore, the wasted energy = 80 - 10 = 70 J
The amount of energy wasted by light bulb 2 = 70 J.
Answer:
38.7 units
Explanation:
a = 3.00 t² – 4.20 t
Integrate to find velocity as a function of time.
v = ∫ a dt
v = ∫ (3.00 t² – 4.20 t) dt
v = 1.00 t³ – 2.10 t² + C
The object starts at rest, so at t = 0, v = 0.
0 = 1.00 (0)³ – 2.10 (0)² + C
0 = C
v = 1.00 t³ – 2.10 t²
Integrate to get position as a function of time.
x = ∫ v dt
x = ∫ (1.00 t³ – 2.10 t²) dt
x = 0.250 t⁴ – 0.700 t³ + C
Find the difference in positions between t = 4.50 and t = 0.
Δx = [0.250 (4.50)⁴ – 0.700 (4.50)³ + C] – [0.250 (0)⁴ – 0.700 (0)³ + C]
Δx = 0.250 (4.50)⁴ – 0.700 (4.50)³
Δx = 38.7
The object moves 38.7 units.
Answer:
15.8m/s
Explanation:
This problem can be solved by taking into account the conservation of the momentum. In this case the momentum of the astronaut and the bag of tools must equal the momentum of the astronaut and the bag of tool after the astronaut throws the bag.
Hence, we have
where ma and va are the mass and velocity of the astronaut, mb and vb are the mass and velocity of the bag, after the astronaut throw the bag. The velocity v is the velocity where the astronaut has the bag of tool
By taking into account that the velocity of the astronaut must be zero to keep him near of the space station, we have that vb = 0.
Thus
