E
1 answer:
You might be interested in
a)
The gravitational field strength near the surface of the asteroid is given by:
where
G is the gravitational constant
M is the mass of the asteroid
R the radius of the asteroid
h is the distance from the surface
Substituting the data of the asteroid:
is the mass
is the radius of the asteroid
is the distance from the surface
We find
b) i)
The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):
So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:
where
s = 3 km = 3000 m is his displacement to reach the surface
u = 0 is his initial velocity
t is the time
Solving for t,
b) ii)
Again, we can use another suvat equation:
where
v is the final velocity
u is the initial velocity
g is the acceleration of gravity
t is the time
Since we have
u = 0
The velocity of the astronaut at the surface will be
b) iii) 175 years
The duration of one year here is
And the time it takes for the astronaut to reach the surface of the asteroid is
Therefore, to find the number of years, we just need to divide the total time by the duration of one year:
So, the astronaut will take 175 years to reach the surface.
Answer:
Force(F) = -80,955.01 N
Explanation:
We need to first determine the impulse that the truck driver received from the car during the collision
So; m₁v₁ - m₂v₂ = (m₁m₂)v₀
where;
m₁ = mass of the truck = 4280 kg
v₁ = v₂ = speed of the each vehicle = 7.69 m/s
m₂ = mass of the car = 810 kg
Substituting our data; we have:
(4280×7.69) - (810×7.69) = (4280+810)v₀
32913.2 - 6228.9 = (5090)v₀
26684.1 = (5090)v₀
v₀ =
v₀ = 5.25 m/s
NOW, Impulse on the truck = m (v₀ - v)
= 4280 × (5.25 - 7.69)
= 4280 × (-2.44)
= -10,443.2 kg. m/s
Force that the seat belt exert on the truck driver can be calculated as:
Impulse = Force × Time
-10,443.2 kg. m/s = F (0.129)
F =
Force(F) = -80,955.01 N
Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N
Answer:
The centripetal acceleration at highway speed is greater.
Explanation:
We assume the motion of the car is uniformly accelerated. Let the highway speed be v.
By the equation of motion,
u is the initial velocity, a is acceleration and t is time
Because the car starts from rest, u = 0.
This is the tangential acceleration of the thread of the tire.
The centripetal acceleration is given by
r is the radius of the tire.
Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about
The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.
The spring-mass system moves by simple harmonic motion, where there is a continuous conversion from elastic potential energy to kinetic energy and viceversa.
The total mechanical energy of the system at any moment of the motion is
where the first term U is the elastic potential energy, with k being the spring constant and x the displacement of the spring with respect to its rest position, and the second term K is the kinetic energy, with m being the mass of the object attached to the spring and v its speed.
The total energy E is constant during the oscillation of the spring, but the values of U and K change. In fact, when the displacement of the spring is maximum (x is maximum), then all the energy is potential energy U, because the speed of the object is zero (it's the moment when the mass is changing direction). On the contrary, when the mass crosses the equilibrium position (rest position) of the spring, then the potential energy is zero (U=0) because the displacement is zero (x=0), and so all the energy is kinetic energy of the motion, and so K is maximum.
The total mechanical energy of the system at any moment of the motion is
where the first term U is the elastic potential energy, with k being the spring constant and x the displacement of the spring with respect to its rest position, and the second term K is the kinetic energy, with m being the mass of the object attached to the spring and v its speed.
The total energy E is constant during the oscillation of the spring, but the values of U and K change. In fact, when the displacement of the spring is maximum (x is maximum), then all the energy is potential energy U, because the speed of the object is zero (it's the moment when the mass is changing direction). On the contrary, when the mass crosses the equilibrium position (rest position) of the spring, then the potential energy is zero (U=0) because the displacement is zero (x=0), and so all the energy is kinetic energy of the motion, and so K is maximum.