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 .
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Answer:
a) 2.5m/s
b) 0.91m/s
c) 0m/s
Explanation:
Average velocity can be said to be the ratio of the displacement with respect to time.
Average speed on the other hand is the ratio of distance in relation to time
Thus, to get the average velocity for the first half of the swim
V(average) = displacement of first trip/time taken on the trip
V(average) = 50/20
V(average) = 2.5m/s
Average velocity for the second half of the swim will be calculated in like manner, thus,
V(average) = 50/55
V(average) = 0.91m/s
Average velocity for the round trip will then be
V(average) = 0/75, [50+25]
V(average) = 0m/s
Answer:
upwards
Explanation:
Torque is the vector cross product of the force and radial distance.


The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.
Answer:
The ratio is
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is 
Generally from Kepler's third law

Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that 
=> ![[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%5E2%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E3)
Here
is the period of Deimos
and
is the period of Phobos
So
![[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D)
=> ![\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%7D%7BT_2%7D%20%20%3D%20%20%5B%5Cfrac%7B23500%20%7D%7B9380%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D)
=>