A uniform magnetic field vector B is directed out of the page. A loop of wire of area 0.8 m2 is in the plane of the page. At a c
1 answer:
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Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α =
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹
Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as
Now for the vertical distance
vy_o=0
than the equation of motion becomes
Now using this acceleration the value of electric field is calculated as
Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation
So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
