Answer:
162500000.
Explanation:
Given that
Diameter of the wire , d= 1.8 mm
The length of the wire ,L = 15 cm
Current ,I = 260 m A
The charge on the electron ,e= 1.6 x 10⁻¹⁹ C
We know that Current I is given as
I=Current
q=Charge
t=time
q= I t
q= 260 m t
The total number of electron = n
q= n e
n=162500000 t
The number of electron passe per second will be 162500000.
Answer:
a) 24.692 m/s
b) 19.4 m
Explanation:
To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:
We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:
a) The velocity (V2) is:
Substituting the known values we can get the velocity at the out:
Atmospheric pressure= 101000 Pa
Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3
b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:
We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:
Substituting the known values, the height (Z3) is:
Z3=Maximum Height=19.376=19.4 m
Answer:
Explanation:
Given:
Thickness of the paperweight cube,
apparent depth from one side of the inbuilt paper in the plastic cube,
apparent depth from the other side of the inbuilt paper in the plastic cube,
Now as we know that refractive index is given as:
- Let the real depth form first side of the slab be,
- Then the depth from the second side of the slab will be,
Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.
Now the refractive index:
Total energy before the toss:
Total energy at the roof:
Energy must be conserved so :
If v₁ is the initial velocity and v₂ is the final velocity, which is zero at the roof: