Answer:
a)= 98kJ
b)=108kJ
c) = 10kJ
Explanation:
a. The work that is done by gravity on the elevator is:
Work = force * distance
= mass * gravity * distance
= 1000 * 9.81 * 10
= 98,000 J
= 98kJ
b)The net force equation in the cable
T - mg = ma
T = m(g+a)
T = 1000(9.8 + 10)
T = 10800N
The work done by the cable is
W = T × d
= 10800N × 10
= 108000
=108kJ
c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J
Work done by cable = PE +KE
108,100 J = KE + 98,100 J
KE = 10,000 J
= 10kJ
=
Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn
let the height of the person with marshmallow on her head be "h"
consider the motion of the marshmallow after it is dropped from bridge.
Y₀ = initial position of the marshmallow above the ground = 5.71 m
Y = final position of marshmallow on head of person = h
v₀ = initial velocity of the marshmallow = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
t = time of travel for marshmallow = 0.921 sec
Using the kinematics equation
Y = Y₀ + v₀ t + (0.5) a t²
inserting the values
h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²
h = 5.71 - 4.16
h = 1.55 m
Answer:
Acceleration of the bullet will be 1778835.6 
Explanation:
We have given length of the barrel refile s= 0.855 m
When the bullet leaves the muzzle its velocity is 553 m/sec
So final velocity v = 553 m/sec
Initial velocity will be 0 that is u = 0 m/sec
According to third equation of motion 
