Answer:
a = 10.07m/s^2
Their acceleration in meters per second squared is 10.07m/s^2
Explanation:
Acceleration is the change in velocity per unit time
a = ∆v/t
Given;
∆v = 50.0miles/hour - 0
∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour
∆v = 22.352m/s
t = 2.22 s
So,
Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s
a = 10.07m/s^2
Their acceleration in meters per second squared is 10.07m/s^2
Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.
Let a = the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
0.9 = 4.5a
a = 0.2 m/s²
Answer: 0.2 m/s²
Answer:
Explanation:
given that
Radius =0.75m
Cnet=0.13nC
a. Electric field inside the sphere located 0.5m from the center of the sphere.
The electric field located inside the sphere is zero.
b. The electric field located 0.25m beneath the sphere.
Since the radius is 0.75m
Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m
Then
E=kq/r2
K=9e9Nm2/C2
q=0.13e-9C
r=1m
Then,
E= 9e9×0.13e-9/1^2
E=1.17N/C. Q.E.D
Answer:
at y=6.29 cm the charge of the two distribution will be equal.
Explanation:
Given:
linear charge density on the x-axis, 
linear charge density of the other charge distribution, 
Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.
Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.
<u>we know, the electric field due to linear charge is given as:</u>

where:
linear charge density
r = radial distance from the center of wire
permittivity of free space
Therefore,





∴at y=6.29 cm the charge of the two distribution will be equal.
Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm
