Answer:
A) Its density will decrease
Explanation:
When an object is heated, its volume increases. This is due to the fact that the particles in the medium vibrate more (if it is a solid) or they move more (if it is a liquid or a gas), therefore they tend to occupy a larger space.
At the same time, the mass of the object does not change, because the mass just represents the amount of matter contained in the object, so it does not increase/decrease at different temperatures.
The density of an object is defined as the ratio between the mass (m) and the volume (V):
We said that the mass remains unchanged while the volume increases: since the density is inversely proportional to the volume, this means that the density decreases.
Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width
Where, D = width of slit
= wavelength
Put the value into the formula
Here, 
should be maximum.
So. maximum value of is 1
Put the value into the formula
(b). If the minimum number is 50
Then, the width is
(c). If the minimum number is 1000
Then, the width is
Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Answer:
1. Luminosity
2.Apparent brightness
Explanation:
There are two factors on which brightness of star appear to be in the sky
The two factors are
1. Luminosity
2.Apparent brightness
1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.
2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.
As we know that spring force is given as
here we know that
F = 4 N
x = 2 cm = 0.02 m
now from the above equation we will have
so the elastic constant of the spring will be 200 N/m
Answer:
110 m
Explanation:
Draw a free body diagram of the car. The car has three forces acting on it: normal force up, weight down, and friction to the left.
Sum of the forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum of the forces in the x direction:
∑F = ma
-F = ma
-Nμ = ma
Substitute:
-mgμ = ma
-gμ = a
Given μ = 0.40:
a = -(9.8 m/s²) (0.40)
a = -3.92 m/s²
Given that v₀ = 30 m/s and v = 0 m/s:
v² = v₀² + 2aΔx
(0 m/s)² = (30 m/s)² + 2 (-3.9s m/s²) Δx
Δx ≈ 110 m
