Answer:
2.55 m/s
Explanation:
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.
Solution:
The work done by friction is given as:
The work done by gravity is:
Answer:
Explanation:
The force on a charge q made by an electric field
its
The electric charge of the electron its
.
Taking the unit vector pointing towards the east, the electric field will be:
.
So, the force will be:
So, the force its N to the west.
The potential energy of a 2-μc charge at that point in space is joules.
Given,
V=400v, q=2-μc=2*,
U(potential energy)=V*q=400*2*=
joules.
The energy that an item retains due to its position in relation to other objects, internal tensions, electric charge, or other reasons is known as potential energy in physics. The gravitational potential energy of an object is based on its mass and the distance from the centre of mass of another object. Other common types of potential energy include the elastic potential energy of an extended spring and the electric potential energy of an electric charge in an electric field. The joule, denoted by the sign J, is the SI's definition of an energy unit.
The vectors that are described as gradients of a particular scalar function known as potential can be used to represent these forces, also known as conservative forces, at any location in space.
At a certain point in space there is a potential of 400 v. what is the potential energy of a 2-μc charge at that point in space? group of answer choices'
Learn more about potential energy here:
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Answer:
The net work produced is 30.37 KJ
The efficiency of cycle is 72.3%
Explanation:
For the net work produced, we have the formula:
Work = (Th - Tc)(Sh - Sc)(M)
Where,
Th = higher temperature = 800° C + 273 = 1073 k
Tc = lower temperature = 25° C + 273 = 298 k
Sh = specific entropy at higher temperature
Sc = specific entropy at lower temperature
M = molar mass of air
Using, ideal gas table to find entropy. The table is attached.
therefore,
Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)
Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)
<u>Work = 30.37 KJ</u>
Now, for the efficiency (n), we have a formula:
n = 1 - Tc/Th
n = 1 - (298 k)/(1073 k)
<u>n = 0.723 = 72.3 %</u>
Answer:
0.356 times the mass pass through equilibrium per second.
Explanation:
Given that,
Spring constant = 10 N/m
Mass = 2 kg
Stretched spring = 2m
We need to calculate the frequency
Using formula of frequency
Where, m = mass
k = spring constant
Put the value into the formula
We know that,
Hertz = cycle per second
Hence, 0.356 times the mass pass through equilibrium per second.