Given:
The angle of projection of the basketball, θ=35°
The height at which the ball leaves the hand, h=7 ft
The initial velocity of the basketball, v=20 ft/s
To find:
The parametric equations describing the shot.
Explanation:
The range, x of the basketball is given by,
On substituting the known values,
The change in the height, y of the basketball is given by,
Where g is the acceleration due to gravity.
On substituting the known values,
Final answer:
The parametric equations describing the shot are
85.689 kg is the astronaut's mass.
<u>Explanation:</u>
Given data: m = 22.5 kg and T = 1.3 sec
So, using the below formula,
Now, after putting the values of m and T in the above equation, we will find out the value of k which is as,
To remove the square root, take square on both sides, we get,
Now, we have the same string but this time we have different mass and different time. So, let the mass of the astronaut is and = 2.54 sec, k= 525.7 kg. Apply these values in the equation, we get,
Taking squares on both sides, we get,
Answer:
These are Diffraction Grating Questions.
Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:
Given as
y = nDλ/w Eqn 1
where
w = width of slit
D = distance to screen
λ = wavelength of light
n = order number
Making x the subject of the formula gives,
w = nDλ/y
Given
y = 0.0149 m
D = 0.555 m
λ = 588 x 10-9 m
and n = 3
w = 6.6x10⁻⁵m
Hence, the width of the slit w, in micrometers (μm) = 66μm
Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen
i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx
Recall Eqn 1, y = nDλ/w
given, D = 27cm = 0.27m
λ = 632 x 10-9 m
w = 0.1mm = 1.0x10⁻⁴m
For the 9th order, n = 9,
y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m
Similarly, for n = 5,
y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m
Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m
Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm
The mass on the left has a downslope weight of
W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N
The mass on the right has a downslope weight of
W2 = 8kg * 9.8m/s² * sin35º = 45.0 N
The net is 25.3 N pulling downslope to the right.
(a) Therefore we need 25.3 N of friction force.
Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º
25.3N = µ * 92.3 N
µ = 0.274
(b) total mass is 11.5 kg, and the net force is 25.3 N, so
acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s²
tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N
Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √
hope this helps. :)