Answer:
900 cm/s or 9 m/s.
Explanation:
Data obtained from the question include the following:
Length (L) = 30 cm
frequency (f) = 60 Hz
Velocity (v) =.?
Next, we shall determine the wavelength (λ).
This is illustrated below:
Since the wave have 4 node, the wavelength of the wave will be:
λ = 2L/4
Length (L) = 30 cm
wavelength (λ) =.?
λ = 2L/4
λ = 2×30/4
λ = 60/4
λ = 15 cm
Therefore, the wavelength (λ) is 15 cm
Now, we can obtain the speed of the wave as follow:
wavelength (λ) = 15 cm
frequency (f) = 60 Hz
Velocity (v) =.?
v = λf
v = 15 × 60
v = 900 cm/s
Thus, converting 900 cm/s to m/s
We have:
100 cm/s = 1 m/s
900 cm/s = 900/100 = 9 m/s
Therefore, the speed of the wave is 900 cm/s or 9 m/s.
Answer:
Explanation:
initial velocity, u = 0
final velocity, v = 60 mph = 26.8 m/s
time t = 10 s
Let a be the acceleration and s be he distance traveled.
Use first equation of motion
v = u + a t
26.8 = 0 + a x 10
a = 2.68 m/s
Use second equation of motion
s = ut + 1/2 at²
s = 0 + 0.5 x 2.68 x 10 x 10
s = 134 m
As, 1 m = 3.28 ft
So, s = 134 x 3.28 ft
s = 439.6 ft
We are 8 light minutes from the sun. That means two things, we see the sun as it was 8 minutes ago, and we WOULD continue to see the sun for 8 minutes after it disappeared.
Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.
Answer:
The average acceleration of the bearings is 
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval 
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration

Put the value into the formula



Hence,The average acceleration of the bearings is 