To solve this problem we will apply the concepts related to wave velocity as a function of the tension and linear mass density. This is
Here
v = Wave speed
T = Tension
= Linear mass density
From this proportion we can realize that the speed of the wave is directly proportional to the square of the tension
Therefore, if there is an increase in tension of 4, the velocity will increase the square root of that proportion
The factor that the wave speed change is 2.
Explanation:
7.7 m/s
1st determine you subject slade to you have a mass of 15 kg being subjected to a force of 180N
so.
180N/15kg=180(kg m)/s^2/15kg=12 ms^2
now determine how long u pushed it
d=0.5 A T^2
sutable the known values getting
2.5 m = 0.5 12 ms^2 T^2
2.5m= 0.6 ms^2 T^2
0.41667 s^2=T^2
0.645497224s=T
the velocity multiply the time by the acceleration giving
0.645497224 s 12 ms^2 =7.745966692 ms.
after rounding the 2 significant u get 7.7 ms
Answer:
a). The truck's distance covered for the trip is
(60) + (60 - 15) = 105 kilometers .
b). Its displacement for the whole trip is the distance
and direction from the start-point to the end-point.
15 kilometers east .
Explanation:
Answer:4.05 s
Explanation:
Given
First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s
Both hit the ground at the same time
Let h be the height of cliff and it reaches after time t
For second stone
---2
Equating 1 &2 we get
Explanation:
Acceleration is change in velocity over change in time.
a = Δv / Δt
a = (10 m/s − 5 m/s) / 4 s
a = 1.25 m/s²
Work = force × distance
First, find the distance traveled.
v² = v₀² + 2aΔx
(10 m/s)² = (5 m/s)² + 2 (1.25 m/s²) Δx
Δx = 30 m
Now find the work:
W = (600 kg × 1.25 m/s²) (30 m)
W = 22,500 J
Alternative, work = change in energy.
W = ΔKE
W = ½ mv² − ½ mv₀²
W = ½ m (v² − v₀²)
W = ½ (600 kg) ((10 m/s)² − (5 m/s)²)
W = 22,500 J
