The value was determined to be 0.122 m/s. The velocity of a body or object determines its direction of motion. Speed is a scalar quantity in its most fundamental form.
Velocity is essentially a vector quantity. It is the rate of change in distance. The initial speed of the first train, which has a mass of 150,000 kg, is 0.3 m/s. The second train has an initial speed of -0.120 m/s and a mass of 110,000 kg.
Let v represent the post-collision speed of the connected mass.
Utilize the idea of momentum.
The speed of the trains is constant both before and after a collision.
150.000 + 110.000v 45.000 - 13200 = 260.000 v 31800 = 260.000 v v = 0.122 m/s 150000 x 0.3 - 110000 x 0.120
After colliding, they move at a speed of 0.122 m/s towards the direction of the right.
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Answer:
Explanation:
Given,
initial angular speed, ω = 3,700 rev/min
= 
final angular speed = 0 rad/s
Number of time it rotates= 46 times
angular displacement, θ = 2π x 46 = 92 π
Angular acceleration
<h2>
Option 3, 216 m is the correct answer.</h2>
Explanation:
We have initial velocity, u = 15 m/s
Time, t = 12 seconds
Final velocity, v = 21 m/s
We have equation of motion v = u + at
Substituting
21 = 15 + a x 12
a = 0.5 m/s²
Now we have equation of motion v² = u² + 2as
21² = 15² + 2 x 0.5 x s
s = 216 m
Displacement = 216 m
Option 3, 216 m is the correct answer.
<span>First of all, the maximum speed occurs when the object passes through the
equilibrium position
The kinetic energy when the object has this max speed is
K= 1/2 * mass * (1.25 m/s)^2
The potential energy in the spring when the speed is equal to zero
U= 1/2 * k * xmax^2
The maximun force of the spring is
mass*acceleration = k*xmax
m * 6.89 m/s2 = k * xmax
xmax = 6.89* m / k
0.5 * m * 1.56 = 0.5 * k * xmax^2
</span>m * 1.56 = k * (<span>6.89* m / k )^2 </span>
<span>
1.56 m = 47.47 m^2 / k
m/k = 0.032862
period = 2 *pi*sqrt[m/k]
= 2 pi </span><span>sqrt [ </span><span>0.032862]
= 1.139 s
A fourth of the period elapses between the instants of max acceleration and maximum speed
= 1/4* period
= 1/4 * </span><span><span>1.139 s </span>
= 0.284s </span>
Answer:
1.52m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute the given values into the formula
0.013(270)+2(130) = (270+130)v
3.51+260 = 400v
263.51 = 400v
v = 400/263.51
v = 1.52m/s
Hence the velocity after the bullet emerges is 1.52m/s
