Answer:
Explanation:
First of all we shall find the velocity at equilibrium point of mass 1.2 kg .
It will be ω A , where ω is angular frequency and A is amplitude .
ω = √ ( k / m )
= √ (170 / 1.2 )
= 11.90 rad /s
amplitude A = .045 m
velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s
= .5355 m /s
At middle point , no force acts so we can apply law of conservation of momentum
m₁ v₁ = ( m₁ + m₂ ) v
1.2 x .5355 = ( 1.2 + .48 ) x v
v = .3825 m /s
= 38.25 cm /s
Let new amplitude be A₁ .
1/2 m v² = 1/2 k A₁²
( 1.2 + .48 ) x v² = 170 x A₁²
( 1.2 + .48 ) x .3825² = 170 x A₁²
A₁ = .0379 m
New amplitude is .0379 m
Answer:
F= 0.009 N
Explanation:
Given that
Charge ,q= 5.13 μC
Velocity ,V= 8.64 x 10⁶ m/s
Magnetic field , B = 1.99 x 10⁻⁴ T
The force on a charge q moving with velocity v is given as follows
F= q V B
Now by putting the values in the above equation we get
[tex]F= 5.13\times 10^{-6}\times 8.64\times 10^{6}\times 1.99\times 10^{-4}\ N [\tex]
F=0.00882 N
F= 0.009 N
Therefore the force on the particle will be 0.009 N.
25 miles at 42.5 degrees east of south. (20^2+15^2 =C^2...You get 25 miles...East of South because she traveled a longer distance to the East. Then you find the angle between the original and ending point.) Hope it helps. You may ask me any other questions.
