An ice cube causes hot coffee to become cool because the amount of coldness contrasts the hot coffee to make it a little cooler
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V = 
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.