Answer: The correct option is A ( horizontally towards the east)
Explanation:
Magnetic field is a region around a magnet or a current- carrying conductor, where a magnetic force is experienced. The magnetic effect of electric current was first discovered in the early 1820 by Oersted. Using a wire that had current flowing through it and a pivoted magnetic needle, he discovered that the direction of deflection depended on the direction of the current and whether the wire was above or below the needle.
From the way the needle turns when current when current carrying wire is held parallel to it, he therefore concluded that:
--> a current has magnetic field all round it,
--> the magnetic field is in a direction perpendicular to the current.
The above discovery was now modified in Fleming's left hand rule which states that when conductor carrying current is placed in a magnetic field, the conductor will experience a force perpendicular to both the field and the flow of current.
Therefore from the question, a vertical wire carrying current in DOWNWARD direction is placed in a HORIZONTAL magnetic field directed to the NORTH. The direction of the force on the wire is to the EAST.
Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:
Potential difference is the work done in moving a positive test charge from infinity to the point in question.
Voltage is an expression of PD. (Joules / Coulomb)
Say that a capacitor has a PD of 5 Volts. The work in moving a positive test charge from the positive plate to the negative plate is -5 Joules/Coulomb or -5 volt. (At the positive plate the positive test charge (1 Coulomb) already has a PD of + 5 Volts.)
Answer:
<h2>b) 4230 J
</h2>
Explanation:
Step one:
given data
extension= 40cm
Spring constant K= 52.9N/cm
Step two:
Required
the Kinetic Energy KE
the expression to find the kinetic energy is
KE= 1/2ke^2
substituting our data we have
KE= 1/2*52.9*40^2
KE=0.5*52.9*1600
KE= 42320Joules
<u>The answer is b) 4230 J
</u>
Answer:
a. 
b. 
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c. 
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,
- speed of lagging car,
- distance between the cars,
- deceleration of the leading car after braking,
a.
Time taken by the car to stop:
where:
, final velocity after braking
time taken
b.
using the eq. of motion for the given condition:
where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:
is the time taken to stop after braking
