Explanation:
The lines in the spectra of star is due to absorption of light emitted by star by different elements present in atmosphere of the star .
If some element is present in the outer atmosphere of he star , it will absorb a particular portion of light , for which there will be black line in the spectrum .
These are called Fraunhoffer's lines .
Now there is no such line at the place of Helium in the light spectrum . This means that Helium element is not present in the outer atmosphere of the star.
Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
Answer:
a) T = 1,467 s
, b) A = 0.495 m
, c) v = 4.97 10⁻² m / s
Explanation:
The simple harmonic movement is described by the expression
x = A cos (wt + Ф)
Where the angular velocity is
w = √ k / m
a) Ask the period
Angular velocity, frequency and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2pi √ m / k
T = 2π √ (1.2 / 22)
T = 1,467 s
f = 1 / T
f = 0.68 Hz
b) ask the amplitude
The mechanical energy of a harmonic oscillator
E = ½ k A²
A = √2 E / k
A = √ (2 2.7 / 22)
A = 0.495 m
c) the mass changes to 8.0 kg
As released from rest Ф = 0, the equation remains
x = A cos wt
w = √ (22/8)
w = 1,658
x = 3.0 cos (1,658 t)
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum when without wt = ±1
v = Aw
v = 0.03 1,658
v = 4.97 10⁻² m / s
Answer:
k = 6.72
Explanation:
K of paper = 3.7
k of air = 1
Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So
V = Q / C = 2.5
Capacity becomes C / 3.7 in air .
capacity becomes C/3.7 when paper is replaced by air .
V₁ = Q / (C/3.7)
= 3.7 Q/C
3.7 x 2.5
= 9.25 V
In the second case ,
capacitance due to new unknown dielectric k
= C/3.7 x k
= kC / 3.7 ( Capacitance in air is C/3.7 )
V ( new ) = Q / ( kC/3.7 )
= 3.7 Q/kC
.55 x 2.5 = 3.7 x( 2.5 / k )
k = 3.7 / .55
= 6.72
