Calculate the decrease in gravitational potential energy of the ball as it moves down through the 0.90m the mass of the ball is
1 answer:
Answer:
G.P.E = 5.292 Joules
Explanation:
Given the following data;
Mass = 0.6 kg
Height = 0.9 m
To find the gravitational potential energy;
Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.
Mathematically, potential energy is given by the formula;
Where,
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
But we know that acceleration due to gravity is equal to 9.8m/s²
G.P.E = 0.6*9.8*0.9
G.P.E = 5.292 Joules
You might be interested in
Answer:
a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft
Explanation:
For this exercise let's use the projectile launch relationships.
The initial height is I = 5 ft and the final height y = 4 ft
y = y₀ + t - ½ g t²
The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft
x = v₀ₓ t
t = x / v₀ₓ
We replace
y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²
v_{oy} = v₀ sin θ
v₀ₓ = vo cos θ
y –y₀ = x tan θ - ½ g x² / v₀² cos² θ
5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)
1 = 120 tan θ - 0.0213 sec² θ
Let's use the trigonometry relationship
Sec² θ = 1 - tan² θ
1 = 120 tan θ - 0.0213 (1 –tan²θ)
0.0213 tan²θ + 120 tanθ -1.0213 = 0
We change variables
u = tan θ
u² + 5633.8 u - 48.03 = 0
We solve the second degree equation
u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2
u = [- 5633.8 ± 5633.82] / 2
u₁ = 0.0085
u₂= -5633.81
u = tan θ
θ = tan⁻¹ u
For u₁
θ₁ = tan⁻¹ 0.0085
θ₁ = 0.487º
For u₂
θ₂ = -89.99º
The launch angle must be 0.487º
b) let's look for the time it takes for the arrow to arrive
x = v₀ₓ t
t = x / v₀ cos θ
t = 120 / (300 cos 0.487)
t = 0.400 s
The deer must be at a distance of
v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s
x = v t
x = 29.33 0.4
x = 11.73 ft
Answer:
Frequency,
Explanation:
Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,
f is the frequency of this light
So, the frequency of this light is .