Answer:
10 m/s
Explanation:
The problem can be solved by using the law of conservation of momentum: the initial momentum has to be equal to the final momentum, so we can write the following
where
is the mass of the first car
is the initial velocity of the first car
is the mass of the second car
is the initial velocity of the second car
is the final velocity of the two combined cars after the collision
Re-arranging the equation and substituting the numbers, we find
(b) Determine the turning point of the mass. (Select all that apply.)
Point A
Point B
Point C
Point D
Point E
The speed at different points and the turning point of the mass can be determined using the principle of conservation of energy. However, concrete figures cannot be calculated without specified potential energy values or initial kinetic energy.
To compute the speed at points B, C, and D, we will use the principle of conservation of energy, which states that the total mechanical energy in a closed system—kinetic and potential energy—is conserved. In other words, energy cannot be created or destroyed, only transformed. Here, total energy = kinetic energy + potential energy. If the total mechanical energy decreases then that decrease in energy must go into another form of energy, such as heat from friction.
As for the turning point of the mass, it will occur when the kinetic energy is at a minimum, and the potential energy is at a maximum. This will happen when the velocity of the object is zero.
Without additional data points or numerical figures for instance the actual potential energy or initial kinetic energy, we cannot exactly compute the speed at points B, C, and D or determine the turning point of the mass.
#SPJ12
This question pertains to an exercise in physics, particularly related to the conservation of mechanical energy. The speed of the mass at different points can be calculated by considering changes in potential energy and applying the formula for kinetic energy. The turning point is when the mechanical energy equals the potential energy and the kinetic energy is zero.
The question is asking for the speed of a 2.5 kg mass at different points as it moves along the x-axis, as well as for the turning point of the mass. Without an illustration or explicit potential energy values, it's impossible to provide exact values. However, I can explain how to approach such a problem theoretically.
Firstly, the concept you need to apply here is the conservation of mechanical energy. This principle states that if there are no non-conservative forces doing work on the system, the total mechanical energy of the system (which is the sum of the kinetic and potential energy) remains constant.
To find the speed at different points, you'd need to know the potential energy at those points. The difference in potential energy between point A and any other point on the x-axis represents the change in kinetic energy (since the sum of potential and kinetic energy must remain constant if only conservative forces are acting). The speed at each point can be found using the formula for kinetic energy: KE = 1/2 * m * v^2.
Furthermore, the turning point of the mass will occur where the mechanical energy of the mass equals the potential energy of the system. This is because at the turning point, the mass stops momentarily before turning around, meaning its speed, and therefore its kinetic energy, will be zero. Therefore, the potential energy equals the total mechanical energy at the turning points.
#SPJ2
Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16
vB = 15.4 m/s : speed of the cart at B
Answer:
the guy above me is correct
Explanation:
You first subtract the speed at which the man is moving (11 m/s) from the rate the boat is moving (12.4 m/s). Which equals 1.4, then divide it by 6 meters, as the man is moving relative to the boat.
It therefore equals 4.29 s
Answer:
4.29 s
Explanation:
Apex
2100 m, how much time does it take for Tony to walk to
school?
min
Answer:
The answer is 30 min
Explanation:
t = s/v
v = 70 m/min, s = 2100 m
t = 2100/70 = 30 min.
Hope it helps you! \(^ᴥ^)/
A. It will be 2.54 g/cm².
B. It will be different from the other chemicals.
C. It will be 1.67 g/cm3.
D. It will be the same as the other reactant.
Answer:
The earth has a larger mass and therefore more inertia.
Explanation: