A wave that moves up and down
Answer:
Explanation:
Given
mass of Jupiter is 
Density of Jupiter is same as Earth
considering Jupiter to be sphere of radius r
acceleration due to gravity is given by
Answer:
a) Total mass form, density and axis of rotation location are True
b) I = m r²
Explanation:
a) The moment of inertia is the inertia of the rotational movement is defined as
I = ∫ r² dm
Where r is the distance from the pivot point and m the difference in body mass
In general, mass is expressed through density
ρ = m / V
dm = ρ dV
From these two equations we can see that the moment of inertia depends on mass, density and distance
Let's examine the statements, the moment of inertia depends on
- Linear speed False
- Acceleration angular False
- Total mass form True
- density True
- axis of rotation location True
b) we calculate the moment of inertia of a particle
For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is
I = m r²
Answer:
2. + 2.1
Explanation:
Formula for resistance is as follows
R_t = R₀ ( 1 + α t )
R_t is resistance at changed temperature , Ro is resistance at initial temperature , α is temperature coefficient of resistivity . t is temperature change .
For iron , R₀ = 10 , α = 5 x 10⁻³ , t = - 100 degree
R_t = 10( 1 + 5 x 10⁻³ x - 100 )
= 10( 1 - 5 x 10⁻³ x 100 )
10 x ( 1 - .5 )
10 x .5
For carbon
, R₀ = 10 , α = - 0.5 x 10⁻³ , t = - 100 degree
R_t = 10( 1 - 0. 5 x 10⁻³ x - 100 )
= 10( 1 + .5 x 10⁻³ x 100 )
10 x ( 1 + .05 )
10 x 1.05
Required ratio
= 10 x 1.05 / 10 x .5
= 105 / 50
= 21 / 10
= 2.1
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm
