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:
v = -10⁵ m/s
Explanation:
given,
speed of asteroid,v' = 100 m/s
mass of superman = m
mass of asteroid,M = 1000 m
recoil velocity of superman,v= ?
using conservation of momentum.
m u + M u' = m v + M v'
initial velocity of asteroid and superman is equal to zero
0 + 0 = m v + 1000 m x 100
m v = -100000 m
v = -10⁵ m/s
superman's velocity after throwing the asteroid is equal to v = -10⁵ m/s
Well, density is mass/volume. So what's 115 g / 16 cm3? That'll get you your density. Remember that density will be in g/cm3!
Answer:
11.) g = 3.695 m/s^2
12.) g = 8.879 m/s^2
13.) E = 8127 N/C
Explanation:
11.) Given that the
Mercury mass M = 3.3 × 10^23kg
Radius r = 2.44 ×10^6 m
Gravitational constant G = 6.67408 × 10^-11 m3kg-1 s^-2
Gravitational field strength g can be calculated by using the formula below
g = GM/r^2
Substitutes all the parameters into the formula
g = (6.67408 × 10^-11 × 3.3 × 10^23)/(2.44×10^6)^2
g = 2.2×10^13/5.954×10^12
g = 3.695 m/s^2
12.) Given that the
Venus mass M = 4.87×10^24kg
Radius r = 6.05 × 10^6 m
Using the same formula for gravitational field strength g
g = GM/R2
Substitute all the parameters into the formula
g = (6.67408 × 10^-11 × 4.87×10^24)/(6.05×10^6)^2
g = 3.25×10^14/3.66×10^13
g = 8.879 m/s^2
13.) Given that the
Charge = 2.26 nC = 2.26×10^-9
Distance d = 0.05m
Electric field strength E can be calculated by using the formula below
E = Kq/d^2
Where
K = electrostatic constant 8.99 × 10^9 Nm2/C2
Substitutes all the parameters into the formula
E = (8.99 × 10^9 × 2.26×10^-9)/0.05^2
E = 20.3174/2.5×10^-3
E = 8126.96 N/C
