Answer:
t = 2 seconds
Explanation:
In 2nd question, the question is given the attached figure.
Initial speed of the bus, u = 0
Acceleration of the bus, a = 8 m/s²
Final speed, v = 16 m/s
We need to find the time taken by the car to reach the stop. Acceleration of an object is given by :
![a=\dfrac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv-u%7D%7Bt%7D)
t is time taken
![t=\dfrac{v-u}{a}\\\\t=\dfrac{16-0}{8}\\\\t=2\ s](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bv-u%7D%7Ba%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B16-0%7D%7B8%7D%5C%5C%5C%5Ct%3D2%5C%20s)
The bus will take 2 seconds to reach the stop.
Answer:
44.3 m/s
Explanation:
a) Draw a free body diagram of the mass M. There are three forces:
Weight force mg pulling down,
Normal force N pushing perpendicular to the ramp,
and tension force T pulling parallel up the ramp.
Sum of forces in the parallel direction:
∑F = ma
T − Mg sin 30° = 0
T = Mg sin 30°
T = Mg / 2
Draw a free body diagram of the hanging mass m. There are two forces:
Weight force mg pulling down,
and tension force T pulling up.
Sum of forces in the vertical direction:
∑F = ma
T − mg = 0
T = mg
Substitute:
mg = Mg / 2
m = M / 2
M = 2m
b) Velocity of a standing wave in a string is:
v = √(T / μ)
T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N. Therefore:
v = √(49 N / 0.025 kg/m)
v = 44.3 m/s
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Given that,
Initial velocity , Vi = 0
Final velocity , Vf = 40 m/s
Acceleration due to gravity , a = 9.81 m/s²
Distance can be calculated as,
2as = Vf² - Vi²
2 * 9.81 *s = 40² - 0²
s = 81.55 m
For half height, that is, s = 40.77m
Vf= ??
2as = Vf² - Vi²
2 * 9.81 * 40.77 = Vf² - 0²
Vf² = 800
Vf = 28.28 m/s
Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.
a . true hardness and density are physical properties