Answer:
the radius of the protons path is r = 0.85 m.
Explanation:
the force due to magnetic fields lead to the cetripetal force, such that:
F = q×v×B = m×(v^2)/r
q×B = m×v/r
then:
r = m×v/q×B
r = p/q×B
then, the kinetic energy of the proton:
K = 1/2×m×v^2 = p^2/(2×m)
q×B = \sqrt{2×m×K}/r
r = \sqrt{2×m×K}/(q×B)
= \sqrt{2×(1.67×10^-27)×(5.3×1.60×10^-13)}/(1.60×10^-19×0.39)
= 0.85 m
Answer:
Thrust is the force acting normally to a surface. So, Thrust = 1200 N. And , Pressure = Thrust / Area. = 1200 / 0.001. = ...
Explanation:
this right like me
Answer:
0.63
Explanation:
We are given that
Radius of earth,
Radius of orbit A,
Radius of orbit B,
We have to find the ratio of the potential energy of satellite B to that of satellite A in orbit.
Potential energy of orbit A=
Potential energy of orbit B=
Hence,the ratio of the potential energy of satellite B to that of satellite A in orbit=0.63
Answer:
(1) 1×10⁻⁴
Explanation:
From the question,
α = (ΔL/L)/(ΔT)............. Equation 1
Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.
Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C
Substitute these values into equation 1
α = 1×10⁻⁴/10
α = 1×10⁻⁵ °C⁻¹ .
β = (ΔA/A)/ΔT................... Equation 2
Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.
make ΔA/A the subject of the equation
ΔA/A = β×ΔT.......................... Equation 3
But,
β = 2α.......................... Equation 4
Substitute equation 4 into equation 3
ΔA/A = 2α×ΔT................ Equation 5
Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹
Substitute into equation 5
ΔA/A = ( 2)×(1×10⁻⁵)×(5)
ΔA/A = 10×10⁻⁵
ΔA/A = 1×10⁻⁴
Hence the right option is (1) 1×10⁻⁴
