me no speak spanish
We shall apply law of conservation of mechanical energy for projectile being thrown .
Total energy on the surface = total energy at height h required
a ) At height h , velocity = .351 x ( 2 GM/R x h )
h = .14 R
h = .54 R
c ) least initial mechanical energy required at launch if the projectile is to escape Earth
= GMm / R + 1/2 m (2GM/R)
Initial speed = 11.2m/s
Time of running = 2s
Acceleration of the puppy = ?
Acceleration is the rate of change of velocity with time.
final speed = 0m/s
Acceleration = = -5.6m/s²
It takes the car 2.5 s to slow down to a final speed of 10.0 m/s.
The car has moved 31.25 m during the braking period.
Assuming that motion is a straight line and knowing that the accelaration is a constant (-2.0 m/s²), these equations apply:
v(t) = v0 +a*t
X(t) = X0 + v0*t + 1/2*a*t²
v(t) = velocity at time "t".
v0 = initial velocity.
a = acceleration.
t = time.
X(t) = position at time "t"
X0 = initial position
v0 = 15.0 m/s
a = 2.0 m/s²
v(t) = 10.0 m/s
We need to find "t" at which the speed is 10 m/s.
From the equation of velocity above:
v(t) = v0 + a*t
solving the equation for "t"
v(t) - v0 = a*t
(v(t) - v0)/a = t
(10 m/s - 15 m/s) / (-2m/s²) = t
2.5 s = t
Now that we know how long it takes to slow down to a speed of 10 m/s, we can calculate how far has the car moved in that period using the equation for position at a given time:
Since we only want to know how many meters has the car moved during 2.5 s with an acceleration of -2m/s², we can make X0(initial position) = 0.
X(2.5 s) = 0 m + 15 m/s * 2.5 s + 1/2 * (-2m/s²)* (2.5 s)²
X(2.5 s) = 37.5 m + (-6.25 m) = 31.25 m
The torque τ on an electric dipole with dipole moment p in a uniform electric field E is given by τ = p × E where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment.