Answer:
μ = 0.0315
Explanation:
Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)
N = m*g (1)
The friction force is equal to the product of the normal force by the coefficient of friction.
F = μ*N (2)
This way replacing 1 in 2, we have:
F = μ*m*g (2)
Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.
where:
vf = final velocity = 0
vi = initial velocity = 85 [km/h] = 23.61 [m/s]
d = displacement = 900 [m]
F = friction force [N]
The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.
Now replacing:
(1/2)*m*(23.61)^2 = μ*m*g*d
0.5*(23.61)^2 = μ*9,81*900
μ = 0.0315
Answer:
4000 N/m²
Explanation:
Area = πr²
Area = π x 0.12² (diameter/2)
Area = 0.05 m²
Pressure = force/area
Pressure = 200/0.05
Pressure = 4000 N/m²
Hope this helps!
No it would not be static friction.
Answer: 2.7 x 10^5 joules
Explanation:
Given that:
Time taken = 3 minutes
convert time in minutes to seconds
(Since 1 minute = 60 seconds
3 minutes = 3 x 60 = 180 seconds)
Power of toaster = 1500 watt
Work done by the toaster = ?
Recall that power is the rate of work done per unit time
i.e Power = work/time
work = Power x Time
Work = 1500 watt x 180 seconds
Work = 270000J
Place the result in standard form
270000J = 2.7 x 10^5J
Thus, 2.7 x 10^5 joules of work is done by the toaster.
Answer:
t = 3 [s]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 15 [m/s]
g = gravity acceleration = 10 [m/s²]
t = time [s]
Now replacing we have:
![0 = 15 -10*t\\10*t=15\\t= 1.5[s]](https://tex.z-dn.net/?f=0%20%3D%2015%20-10%2At%5C%5C10%2At%3D15%5C%5Ct%3D%201.5%5Bs%5D)
Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.
The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:
t = 1.5 + 1.5 = 3 [s]
