Answer:
6.25 m/s
Explanation:
mass of man (m1) = 80 kg
mass of boy (m2) = 20 kg
mass of man and boy after collision (m12)= 20 + 80 = 100 kg
velocity of man and boy after collision (v) = 2.5 m/s
angle θ = 60 °
How fast was the boy moving just before the collision
?
- From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
- M₁₂ = total momentum after collision = m12 x v = 100 x 2.5 = 250
- Mboy = momentum of the boy before collision = m2 x Velocity of boy
- Mman = momentum of the man before collision = m1 x velocity of man
- from the triangle, cos θ =
cos 60 = 
Mboy = 250 x cos 60 = 125
- recall that momentum of the boy (Mboy) also = m2 x Velocity of boy
therefore
125 = 20 x velocity of boy
velocity of boy = 125 / 20 = 6.25 m/s
Answer:
10mm
Explanation:
According to Hooke's law which states that "the extension of an elastic material is directly proportional to the applied force provided the elastic limit is not exceeded. Direct proportionality there means, increase/decrease in the force leads to increase/decrease in extension.
Mathematically, F = ke where;
F is the applied force
k is the elastic constant
e is the extension
from the formula k = F/e
k = F1/e1 = F2/e2
Given force of 1N indents the spring inwards by 2mm, this means force of 1N generates extension of 2mm
Let F1 = 1N e1 = 2mm
The extension that will be produced If force of 5N is applied to the string is what we are looking for. Therefore F2 = 5N; e2= ?
Substituting this values in the formula above we have
1/2=5/e2
Cross multiplying;
e2 = 10mm
This shows that we must have dent it by 10mm before it pushes outwards by a 5N force
Answer
given,
Stress for plastic deformation = 267 MPa
modulus of elasticity = 115 GPa
cross sectional area = 377 mm²
a) maximum load (in N) that may be applied to a specimen
= σ x A
= 267 x 10⁶ x 377 x 10⁻⁶
= 100659 N
b) modulus of elasticity = stress/strain
115 x 10⁹ =
L = 127 mm
115 x 10⁹ =
Δ l = 0.295 mm
maximum length after the stretched = 127 mm + 0.295 mm
= 127.295 mm
Answer:
Explanation:
Each piece is analyzed by using the Principle of Momentum Conservation and the Impulse Theorem:
Heavier object:
Lighter object:
After the some algebraic handling, the following relationship is found:
Given that both pieces have horizontal velocities only and both are modelled as projectiles, the horizontal component of velocity remains constant and directly proportional to travelled distance. Then:
Answer:
Assuming the emissivity is 1.0, the temperature of the wood stove will be about 417 C
Explanation:
You can use the Stefan-Boltzmann formula tying energy with temperature of a blackbody:
P - rate of radiation
is the blackbody's emissivity
is the Stefan-Boltzmann constant
A - area
T - temperature
So,
![P=\epsilon \sigma AT^4\implies T = \sqrt[4]{\frac{P}{\epsilon \sigma A}}\\T = \sqrt[4]{\frac{18000W}{1.0\cdot 5.67\cdot 10^{-8} \frac{W}{m^2\cdot K^4}\cdot 1.4 m^2}}\approx 690K \approx 417^\circ C](https://tex.z-dn.net/?f=P%3D%5Cepsilon%20%5Csigma%20AT%5E4%5Cimplies%20T%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7BP%7D%7B%5Cepsilon%20%5Csigma%20A%7D%7D%5C%5CT%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B18000W%7D%7B1.0%5Ccdot%205.67%5Ccdot%2010%5E%7B-8%7D%20%5Cfrac%7BW%7D%7Bm%5E2%5Ccdot%20K%5E4%7D%5Ccdot%201.4%20m%5E2%7D%7D%5Capprox%20690K%20%5Capprox%20417%5E%5Ccirc%20C)
Assuming the emissivity ( a number between 0 and 1) is 1.0, the temperature of the wood stove will be about 417 C. The emissivity number was not given in your question. If you determine what it is, you can divide this result by the fourth power of that value to get an updated result.
