Answer:
14.7 m/s
Explanation:
a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²
v₀ = initial speed of the driver = 0 m/s
v = final speed of the driver after 30 ms
t = time interval for which the acceleration is experienced = 30 ms = 0.030 s
Using the equation
v = v₀ + a t
Inserting the values
v = 0 + (490) (0.030)
v = 14.7 m/s
Answer:
F = - 59.375 N
Explanation:
GIVEN DATA:
Initial velocity = 11 m/s
final velocity = 1.5 m/s
let force be F
work done = mass* F = 4*F
we know that
Change in kinetic energy = work done
kinetic energy = 
kinetic energy =
= -237.5 kg m/s2
-237.5 = 4*F
F = - 59.375 N
Answer:
The average magnitude of magnetic field B= 0.0433/ d Tesla
(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)
Explanation:
Induced emf
where B= magnetic field
d= breadth of rectangular piece
V= velocity with which the rectangular piece = o.o6m/s
n= no of turns = 10
EMF = 26mV
since d (breadth of the frame) is not given, I will use it as a variable
EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)
From eq 1, we get
B= (EMF)/(n d V)
B= (26 X 0.001) / (10 d 0.06)
B= 0.0433/ d Tesla
V = u + a*t = 1100ft/s + (1000*10) ft/s = 11100 ft/s
Answer is <span>11,100 ft/s </span>
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.