Answer:
a=0.212 m/s²
Explanation:
Given that
q= 10⁻⁹ C
m = 5 x 10⁻⁹ kg
Magnetic filed ,B= 0.003 T
Speed ,V= 500 m/s
θ= 45°
Lets take acceleration of the mass is a m/s²
The force on the charge due to magnetic filed B
F= q V B sinθ
Also F= m a ( from Newton's law)
By balancing these above two forces
m a= q V B sinθ
a=0.212 m/s²
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
Answer:
Explanation:
Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree
From the formula of resultant vector
(1.703I)² = I² + I² + 2 I² cosФ
2.9 I² = 2I² + 2 I² cosФ
.9I² = 2 I² cosФ
cosФ = .9 / 2
= .45
Ф = 63.25 .
Answer:
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