Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:
where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as
where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
Answer:
d
Explanation:
d because a b c are energy equation not work
Explanation:
It is given that,
Wavelength, 
Slit width, 
Order, m = 2
If the diffracted light projects onto a screen at distance 1.50 m, L = 1.5 m
For the diffraction of light,
y = 0.0037 m
So, the distance from the center of the diffraction pattern to the dark band is 0.0037 meters. Hence, this is the required solution.
Answer:
y₁ = 37.2 m, y₂ = 22,6 m
Explanation:
For this exercise we can use the kinematic equations
For the globe, with index 1
y₁ = y₀ + v₁ t
For the shot with index 2
y₂ = 0 + v₂ t - ½ g t²
At the point where the position of the two bodies meet is the same
y₁ = y₂
y₀ + v₁ t = v₂ t - ½ g t²
14 + 8.40t = 27.0 t - ½ 9.8 t²
4.9 t² - 18.6 t + 14 = 0
t² - 3,796 t + 2,857 = 0
Let's look for time by solving the second degree equation
t = [3,796 ±√(3,796 2 - 4 2,857)] / 2
t = [3,796 ± 1,727] / 2
t₁ = 2.7615 s
t₂ = 1.03 s
Now we can calculate the distance for each time
y₁ = v₂ t₁ - ½ g t₁²
y₁ = 27 2.7615 - ½ 9.8 2.7615²
y₁ = 37.2 m
y₂ = v₂ t₂ - ½ g t₂²
y₂ = 27 1.03 - ½ 9.8 1.03²
y₂ = 22,612 m
