A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep
1 answer:
Answer:
0.20
Explanation:
The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.
There are two forces acting in the horizontal direction:
- The pushing force: F = 99 N, forward
- The frictional force: , backward, with
coefficient of kinetic friction
m = 50 kg mass of the box
g = 9.8 m/s^2 gravitational acceleration
The net force must be zero, so we have
which we can solve to find the coefficient of kinetic friction:
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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:
The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°
Explanation:
According to Snell's Law,
n₁ sin θ₁ = n₂ sin θ₂
When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n
n sin θ₁ = n sin (90° - θ₁)
Note that from trigonometric relations,
Sin (90° - θ₁) = cos θ₁
n sin θ₁ = n cos θ₁
(sin θ₁)/(cos θ₁) = 1
tan θ₁ = 1
θ₁ = arctan 1 = 45°
Hope this Helps!!!
