Answer:
b. less than w.
Explanation:
In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).
This is supported by the length contraction equation below:
L = 
Here, L is the observed length
is the original length of the object
v is the relative speed between the object and the observer
and c is the speed of light
Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0
This effectively decreases the length that is observed.
Solution:
According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.
So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.
v_f=v_o + at ……..(a)
[where v_f and v_o are final velocity and initial velocity, respectively]
Now ,
Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.
Applying this value in equation (a)
v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction
For calculating the magnitude of the equation we have to square root the given value
(16.6i - 17.165y)
\\
\left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ =
\sqrt{275.56+294.637225}\\=
\sqrt{570.197225}\\=
23.87[/tex]
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
<u>Answer:</u>
The final velocity of the two railroad cars is 1.09 m/s
<u>Explanation:</u>
Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by
where
V= Final velocity
M1= mass of the first object in kgs = 12000
M2= mas of the second object in kgs = 10000
V1= initial velocity of the first object in m/s = 2m/s
V2= initial velocity of the second object in m/s = 0 (given at rest)
Substituting the given values in the formula we get
V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s
Which is the final velocity of the two railroad cars
The chemical energy in Jay's body, to kinetic energy in the car
