Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
I think it’s 15cm
Might be 7cm
Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
now, wavelength calculation
n = 1
λ₁ = 4 m
n = 2
λ₂ = 2 m
n =3
λ₃ = 1.333 m
now, frequency calculation
n = 1
f₁ = 12.5 Hz
n = 2
f₂= 25 Hz
n = 3
f₃ = 37.5 Hz
Answer:
5,000J
Explanation:
Work = Force x Distance
Distance back and forth is canceled out, so either the answer is + or -
5.0m + 5.0m = 10.0m
500N x 10.0m = 5,000J
This is a uniform rectilinear motion (MRU) exercise.
To start solving this exercise, we obtain the following data:
<h3><u>
Data:</u></h3>
- v = 4.6 m/s
- d = ¿?
- t = 10 sec
To calculate distance, speed is multiplied by time.
We apply the following formula: d = v * t.
We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:
Therefore, the speed at 10 seconds is 46 meters.
