Answer:
The true weight of the aluminium is
4.5021 kg
Explanation:
Given data
= 4.5 kg
= 1.29 
= 2.7× 
The true mass of the aluminium is given by

Put all the values in above equation we get

4.5021 kg
Therefore the true weight of the aluminium is
4.5021 kg
Answer:
10.2 m
Explanation:
The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

where
y is the position of the m-th minimum
m is the order of the minimum
D is the distance of the screen from the slit
d is the width of the slit
is the wavelength of the light used
In this problem we have:
is the wavelength of the light
is the width of the slit
m = 13 is the order of the minimum
is the distance of the 13th dark fringe from the central maximum
Solving for D, we find the distance of the screen from the slit:

Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Become more positive instead of negative
Answer:
47.8 °C
Explanation:
Use the heat equation:
q = mCΔT
where q is the heat absorbed/lost,
m is the mass of water,
C is the specific heat capacity,
and ΔT is the change in temperature.
Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.
100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT
ΔT = 47.8 °C