Answer:Rolling friction is friction that acts on objects when they are rolling over a surface. Rolling friction is much weaker than sliding friction or static friction. This explains why most forms of ground transportation use wheels, including bicycles, cars, 4-wheelers, roller skates, scooters, and skateboards.
Explanation:
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket, and the drive
Answer:
The minimum coefficient of static friction required, µ = 0.10
Note. The question is incomplete. The complete question is given below:
While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.
The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.
Explanation:
First, velocity in mph is converted to m/s
1 mph = 0.447 m/s
55 mph ≈ 24.6 m/s
The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²
Force that can be generated by the truck, F = ma
F = 8850kg * 1.07 m/s² = 9469.5 N
However, with the added mass of the log on it, the acceleration of the truck will become;
a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²
Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N
Normal reaction on the truck due to the weight of the log, R = mg
R = 929 kg * 9.8m/s² = 9104.2 N
Coefficient of static friction, µ = F/R
µ = 901.13/9104.2
µ = 0.098 ≈ 0.10
Therefore, the minimum static friction required is µ = 0.10
mester Exam 1 11 of 35
A car has an oil drip. As the car moves, it drips oil at a regular rate, leaving a trail of spots on the road. Which diagram shows the spots
of car that is continuously slowing down?
What is the volume of a brick that is 30 cm long, 8 cm wide, and 10 cm tall?
2. Using Graph 2, calculate the net force experienced by the particle between 4 and 6 seconds. The
particle has a mass of 0.25 kg.
A +5.0 N
B. +0.5 N
C. -0.5 N
D. -2.0 N
Using Newtons Second Law:
F = m×a
F = (0.25 kg)(-2 m/s²)
F = -0.5 N
The correct option is CThis 200-kg horse ran the track at a speed of 5 m/s. What was the average kinetic energy?
Answer:
2500 JExplanation:
The average kinetic energy can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]k = \frac{1}{2} \times 200 \times {5}^{2} \\ = 100 \times 25[/tex]
We have the final answer is
2500 JHope this helps you
Define a rotation of the earth answer fast
Answer:
here u go
Explanation:
Earth's rotation is the rotation of planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north pole star Polaris, Earth turns counterclockwise.
An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?
Answer:
The fraction of the object that is below the surface of the water is ¹⁷/₂₀
Explanation:
Given;
specific gravity of the object, γ = 0.850
Specific gravity is given as;
[tex]specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3[/tex]
Fraction of the object's weight below the surface of water is calculated as;
[tex]= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}[/tex]
Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀
When 26400j of energy is supplied to a 2.0kg bloom of aluminum it temperature rise from 20oc to 35oc.The block is well so there is no energy lost to sorround determine the specific heat capacity of aluminum
Answer:
880J/kelvin
Explanation:
Q =MC ×change in t
c =C/m
C=Q/change in t
c= Q/ m× change in t
c = 26400 / 2.0 × 15
c = 880 J/kelvin
what is the difference between alcoholic and Mercury thermometer based on their function?
What kind of electricity does turning wheel generates? Please help!
Answer: Kinetic Energy to Electrical.
Explanation: The magnet is rotated as a result of the spinning wheels, and this results in a powerful stream of electrons, therefore converting kinetic to electrical.
12 seconds after starting from rest a frewly falling cantaloupe has a speed of
Answer:
The cantaloupe has a speed of 117.6 m/s
Explanation:
Free Fall Motion
It occurs when an object falls under the sole influence of gravity. Any object that is being acted upon solely by the force of gravity is said to be in a state of free fall. Free-falling objects do not face air resistance.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2[/tex].
The final velocity of a free-falling object after a time t is given by:
vf=g.t
The cantaloupe has been dropped from rest. We are required to find the speed after t=12 seconds.
Calculate the final speed:
vf=9.8 * 12 = 117.6 m/s
The cantaloupe has a speed of 117.6 m/s
How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase in the emitted frequency?
Answer:
v0 = 24.42 m/s (Approx)
Explanation:
Given:
Increase in frequency = 7.1% =
Computation:
Assume n = 100%
n1 = [(v+v0)/(v+v1)]n
[100 + 7.1] = [(344+v0)/(344+0)]100
107.1 = [(344+v0)/(344)]100
v0 = 24.42 m/s (Approx)
A 300 g bird flying along at 6.0 m/s sees a 10 g insect heading straight toward it with a speed of 30 m/s. the bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?
Answer:
6.77m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses of the object
u1 and u2 are the velocities before collision
v is the final collision
Given
m1 = 300g = 0.3kg
u1 = 6.0m/s
m2 = 10g = 0.01kg
u2 = 30m/s
Required
The bird's speed immediately after swallowing v
Substitute the given values into the formula
m1u1 + m2u2 = (m1+m2)v
0.3(6) + 0.01(30) = (0.3+0.01)v
1.8+0.3 = 0.31v
2.1 = 0.31v
v = 2.1/0.31
v = 6.77m/s
Hence the bird's speed immediately after swallowing is 6.77m/s