Answer:
[tex]l_o=550.055\ cm[/tex]
Explanation:
Given that,
Length of a street bridge, l = 5.5 m
The coefficient of bridge, [tex]\alpha =0.00001 ^0 C[/tex]
We need to find how much will it expand when temperatures is by 10°C.
The change in length per unit original length is given by :
[tex]\dfrac{\Delta l}{l}=\alpha \Delta T\\\\\Delta l = l\alpha \Delta T\\\\=5.5\times 0.00001 \times 10\\\\\Delta l=0.00055\\\\(l_o-l)=0.00055\\\\l_o=0.00055+5.5\\\\=5.50055\ m\\\\l_o=550.055\ cm[/tex]
Hence, the length will expanded 550.055 cm.
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
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
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.
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.
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
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?
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
what is the difference between alcoholic and Mercury thermometer based on their function?