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
The answer is A. <span>Some work input is used to overcome friction. </span>
Answer:
A)
B)
C)
Explanation:
Given that:
- no. of turns i the coil,

- area of the coil,

- time interval of rotation,

- intensity of magnetic field,

(A)
Initially the coil area is perpendicular to the magnetic field.
So, magnetic flux is given as:
..................................(1)
is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.


(B)
In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.
∴ 
From eq. (1)


(C)
According to the Faraday's Law we have:



Answer:A- mass charge.
This can also be called current.
Explanation:
This is Kirchhoff’s 2nd law.
Kirchhoff’s junction law states that the sum of current(mass charge) flowing in and out of the junction must be equal to zero. This law emphasizes conservation of charge and energy. Charge is also a form of energy and it can neither be created nor destroyed.