Help pweaseeee ill give u a brainlest
1 answer:
You might be interested in
1) Current: 4.5 A
2) Time taken: 4.7 s
Explanation:
1)
The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:
where
I is the current
q is the amount of charge passing a given point in a time t
For the wire in this problem, we have
q = 9.0 C is the amount of charge
t = 2.0 s is the time interval
Solving for I, we find the current:
2)
To solve this problem, we can use again the same formula
where
I is the current
q is the amount of charge passing a given point in a time t
In this problem, we have:
I = 3.0 A (current)
q = 14.0 C (charge)
Therefore, the time taken for the charge to move past a particular spot in the wire is
Learn more about electric current:
#LearnwithBrainly
1 kilometre=1000 metre
1 hour = 3600 second
The initial velocity of car A is 35.0 km/hr i.e
= 9.72 m/s
The initial velocity of car B is 45 km/hr =12.5 m/s
The initial velocity of car C is 32 km/hr = 8.89 m/s
The initial velocity of car D is 110 km/hr=30.56 m/s
The acceleration of car A is given as
The time taken by car A = 15 min.
From equation of kinematics we know that-
v= u+at [Here v is the final velocity and a is the acceleration and t is the time]
Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s
=11.456111106 m/s
The acceleration of B is given as
The time taken by car B =20 min
The final velocity of B is -
v= u+at
= u-at [Here a is negative due to deceleration]
=12.5 m/s +[0.0011574074074×20×60]
=13.8888888.....
=13.9
The acceleration of C is given as
The time taken by car C =30 min
The final velocity of C is-
v = u+at
=8.89 m/s+[0.003086419753×30×60] m/s
=14.4455555555..m/s
=14.45 m/s
The car C is decelerating.The deceleration is given as-
The time taken by car D= 45 min.
The final velocity of the car D is-
v =u+at
=30.56 -[0.00462962962962×45×60]m/s
=18.06 m/s
Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.