Answer:
Velocity=1.1m/s
Amplitude=0.35m
Explanation:
Given:
time 't' = 2.9s
wavelength 'λ'= 5.5m
distance 'd'=0.7m
The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel from its highest point to its lowest is a half period.
So, T = 2 x 2.9 => 5.8 s
As we know that frequency is the reciprocal of time period, we have
f= 1/T = 1/5.8 =>0.2 Hz
In order to find how fast are the waves traveling, the velocity is given by
Velocity = f λ
V= 0.2 x 5.5 =>1.1m/s
The distance between the boat's highest point to its lowest point is double the amplitude.
Therefore , we can write
Amplitude 'A'= d/2 =>0.7/2 =>0.35m
Answer: 2.4×10^-3 v/m
Explanation: distance between plates of capacitor (d) =5.0×10^-3m
Potential difference between plates (v) = 12v
Force on electronic charge (f) = 3.8×10^-16 N
Strength of electric field (E) =?
The formulae that relates potential difference, eoectiic field strength and distance between plates is given as
v = Ed
By substituting the parameters, we have that
12 = E × 5.0×10^-3
E = 12/ 5.0 × 10^-3
E = 2.4×10^-3 v/m
5.972 × 10^24 kg
it is the weight of earth
hope it is helpful to you
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
