Answer:
Yes
Explanation:
Forces are pushing such as gravity, and forces from your own body and the couch are pushing against each other.
Answer: B. Concrete
Explanation:
Let N = reacting force pressing the bodies in context together (units in Newtons),
The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.
We need to start by finding the weight of the piece of wood.
friction is given by
f = μN
The value of f is 22.5,
and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.
22.5 = 75. 0.30
Putting the values into the equation: 22.5 = 0.30N.
Divide both sides by 0.30 to find the value of N:
N= 22.5/0.3 = 75
Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.
To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:
46.5 = µ(75).
Divide both sides by 75 to isolate μ.
The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).
Answer with Explanation:
We are given that
Length of wire 1=
Length of wire 2=
Resistivity of copper wire=
Resistivity of aluminum wire=
Wire 1=Copper wire
Wire 2=Aluminum wire
Diameter of both wires are same and resistance of both wires are also same.
We know that
Resistance =
When diameter of wires are same then their cross section area are also same .
When resistance and area are same then the length of wire depend upon the resistivity of wire .
The length of wire is inversely proportional to resistivity.
When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.
Therefore, 
Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).
Answer:
1.8x10⁻³m
Explanation:
From the question above, the following information was used to solve the problem.
wavelength λ = 4.5x10⁻⁷m
Length L = 2.0 meters
distance d = 5 x 10₋⁴m
ΔY = λL/d
= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m
= 0.00000045 / 0.0005
= 0.0000009/0.0005
= 0.0018
= 1.8x10⁻³m
from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m
thank you!
