Answer: 345600 J
Explanation: solution attached
KE= 1/2 mv²
Answer:
Explanation:
First of all, let's convert from nanometres to metres, keeping in mind that
So we have:
Now we can convert from metres to centimetres, keeping in mind that
So, we find:
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
Emf = d (phi-B) / dt
<span>B dA/dt, where dA/dt is the area swept out by the wire per unit time. </span>
<span>0.88 V = (0.075 N/(A m)) (L)(4.20 m/s), so </span>
<span>L = (0.88 J/C) / [ (0.075 N s/C m)(4.2 m/s) ] = about 3 meters</span>
