Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by
So from data
Now sub parts
(a)
(b)
(c)
Answer:
The Sun has a north and south pole, just as the Earth does, and rotates on its axis. However, unlike Earth which rotates at all latitudes every 24 hours, the Sun rotates every 25 days at the equator and takes progressively longer to rotate at higher latitudes, up to 35 days at the poles. This is known as differential rotation.
Explanation:
Answer:
B
Explanation:
if you sit up straight you will have a proper posture
The answer is B. Nutrients.
Answer:
The horizontally applied force = 2360 N
Explanation:
<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)
Fh = Fr + ma......... Equation 1
Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.
<em>Given: m = 400 kg, a = 1 m/s²</em>
Fr = 1/2 W, W = mg ⇒W = 400×9.8 = 3920 N
∴Fr = 1/2(3920), Fr = 1960 N
Substituting these values into equation 1
Fh = 1960 + 400×1
Fh = 1960 + 400
Fh = 2360 N
Therefore the horizontally applied force = 2360 N
