Answer:
49.07 miles
Explanation:
Angle between two ships = 110° = θ
First ship speed = 22 mph
Second ship speed = 34 mph
Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b
Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c
Here the angle between the two sides of a triangle is 110° so from the law of cosines we get
a² = b²+c²-2bc cosθ
⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110
⇒a² = 2408.4
⇒a = 49.07 miles
Observe that the given vector field is a gradient field:
Let , so that
Integrating the first equation with respect to , we get
Differentiating this with respect to gives
Now differentiating with respect to
gives
Putting everything together, we find a scalar potential function whose gradient is ,
It follows that the curl of is 0 (i.e. the zero vector).
Answer:
∅=
Explanation:
z=impedance
=2R
R=R
The resultant of the resistances in the circuit is called impedance
is inductive reactance of the circuit
R is the resistance of the resistor
z=
z=
Z=
Z=Rohms
tan∅=2R/R
tan∅=2
∅=Tan^-1(2)
∅=
phase angle is ∅=
The time interval for which the proton remains in the field is -
Δt = .
We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.
We have to determine time interval during which the proton is in the field.
<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>The magnitude of force on the charged particle moving in a uniform magnetic field is given by -
F = qvB sinθ
According to the question, we have -
Entering Velocity (v) = 20 i m/s
Magnetic field intensity (B) = 0.3 T
Leaving velocity (u) = - 20 j m/s
Now -
The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -
d = =
Since, the radius of circular path is not given, we will assume it R.
Therefore, time for which proton remained in the field is -
t =
Hence, the time interval for which the proton remains in the field is -
Δt =
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