Answer:
7.93 lbs is equal to 3596.987 grams.
Explanation:
The weight in grams is equal to the pounds multiplied by 453.59237.
So... you would multiply 7.93 by 453.59237.
7.93 x 453.59237 = 3596.987
Hope that helped!
Power=Work/Time
The work done is the energy required to lift the box, fighting the force of gravity. So, Work=Potential energy of the box at 10 meters.
W=PE=mgh=(60)(9.8)(10)=5880J
Finally,
P=W/T=(5880)/(5)=1176Watt
So the answer is 1176 Watts
Answer:
it is sooo easy u need to use magnetic panels on the rail and put the magnet on the train it works under the principal of magnetic
Answer:
Induced current, I = 18.88 A
Explanation:
It is given that,
Number of turns, N = 78
Radius of the circular coil, r = 34 cm = 0.34 m
Magnetic field changes from 2.4 T to 0.4 T in 2 s.
Resistance of the coil, R = 1.5 ohms
We need to find the magnitude of the induced current in the coil. The induced emf is given by :
Where
is the rate of change of magnetic flux,
And 
Using Ohm's law, 
Induced current, 
I = 18.88 A
So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.
Answer:
Explanation:
Let the magnetic field be B = B₁i + B₂j + B₃k
Force = I ( L x B ) , I is current , L is length and B is magnetic field .
In the first case
force = - 2.3 j N
L = 2.5 i
puting the values in the equation above
- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]
= - 20 B₃ j + 20 B₂ k
comparing LHS and RHS ,
20B₃ = 2.3
B₃ = .115
B₂ = 0
In the second case
L = 2.5 j
Force = I ( L x B )
2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )
= - 20 B₁ k + 20B₃ i
2.3i−5.6k = - 20 B₁ k + 20B₃ i
B₃ = .115
B₁ = .28
So magnetic field B = .28 i + .115 B₃
Part A
x component of B = .28 T
Part B
y component of B = 0
Part C
z component of B = .115 T .
