Answer:
2 1/2 m/s^2 or 2.5m/s^2
Explanation:
From the question
final velocity v =25m/s
Initial velocity u =10m/s
time = 6 seconds
acceleration a = ?
Using the equation for linear motion
v = u + at
25 = 10 + a x 6
25 = 10 + a6
Subtract 10 from both sides
25-10 = 10 + a6 -10
25 - 10 = 10 -10 + a6
15 = a6
Divide both sides by 6
15/6 = a6/6
5/2 = a
a = 2 1/2 m/s^2 or 2.5m/s^2
Explanation:
s = ut + 1/2 a t^2
375 = 0 * 5 + 1/2 * a * (5)^2
375 = 1/2 * a * 25
a = 375*2/25
a = 15* 2
a = 30m/sec^2
v = u + at
v = 0 + 30 * 5
v = 150 m/sec
hope it helps you
Answer: Sound Energy
Sound Energy
Explanation:The vibrations produced by the ringing bell causes waves of pressure that travel or propagate through the medium that is air. Sound energy is a form of mechanical energy that is generally associated with the motion and position of the ringing bell.
Answer:
The inductance of solenoid A is twice that of solenoid B
Explanation:
The inductance of a solenoid L is given by
L = μ₀n²Al where n = turns density, A = cross-sectional area of solenoid and l = length of solenoid.
Given that d₁ = 2d₂ and l₂ = 2l₁ and d₁ and d₂ are diameters of solenoids A and B respectively. Also, l₁ and l₂ are lengths of solenoids A and B respectively.
Since we have a cylindrical solenoid, the cross-section is a circle. So, A = πd²/4.
Let L₁ and L₂ be the inductances of solenoids A and B respectively.
So L₁ = μ₀n²A₁l₁ = μ₀n²πd₁²l₁/4
L₂ = μ₀n²A₂l₂ = μ₀n²πd₂²l₂/4
Since d₁ = 2d₂ and l₂ = 2l₁, sub
L₁/L₂ = μ₀n²πd₁²l₁/4 ÷ μ₀n²πd₂²l₂/4 = d₁²/d₂² × l₁/l₂ = (2d₂)²/d₂² × l₁/2l₁ = 4d₂²/d₂² × l₁/2l₁ = 4 × 1/2 = 2
L₁/L₂ = 2
L₁ = 2L₂
So, the inductance of solenoid A is twice that of solenoid B
Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g 
let's calculate
a = 0.65 9.8 
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.
