Answer:
Vf = - 20 m/s ( -ve sign shows that the particle is moving opposite to positive x- direction).
Explanation:
Given:
Vi = 20 m/s, m= 10 mg =1 × 10⁻⁵ kg, q= -4.0 × 10⁻⁶ C , E= 20 N/C. t= 5.0 s
first to find Electric Force
F= Eq = 20 × -4.0 10⁻⁶ C = - 8 × 10⁻⁵ N (-ve sign shows that the field will push the particle opposite to positive x- direction)
We also have F=ma
⇒ a = F/m = - 8 × 10⁻⁵ N / 10 × 10⁻⁵ kg = -8 m/s² ( -ve sign shows that the particle is accelerated opposite to positive x- direction)
Now according the first equation of Motion.
Vf = Vi + at
Vf = 20 m/s + -8 m/s² × 5 s
Vf= -20 m/s ( -ve sign shows that the particle is moving opposite to positive x- direction)
Impulse is change in momentum. The initial momentum was
0.1kg*11m/s=1.1kg m/s.
The second momentum is 0.1*(-8.8m/s)=-0.88 (since it's moving backward now)
The difference is P1-P2 = 1.1-(-0.88)=1.98kg m/s
Answer:
<h2>hydrophobia </h2>
Explanation:
<h3>it is phobia of water</h3>
and please add me on sñap
<h2>arshad2623f</h2>
<h3>I hope u will add me </h3>
Kinetic energy since if its rolling is already using the stored potential energy that it had before it was given an amount of energy to release it for a given time
Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i = 
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>
