An object on Earth weighs 600 N and on Mars 100 N. If its mass on Earth is approximately 60 kg what is the mass of the object on
1 answer:
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Answer:
a)
b)
c)
Explanation:
Our most valuable tool in solving this problem will be the conservation of mechanical energy:
That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this mechanical energy will remain constant. (as long as there is no dissipation)
For a point particle, we have that kinetic energy is:
Where m is the mass, and V is the particle's velocity,
Potential energy on the other hand is:
where g is the acceleration due to gravity () and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which
, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.
Let's calculate the mechanical energy just before the ball is dropped:
As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:
That's it, the actual value of m is not important now, as we will see.
Now, what's the potential energy at the bottom? Let's see:
At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set
So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:
And so we have found the velocity of the ball as it hits the floor.
Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:
where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:
Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.
As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:
what's the average speed when the ball is descending?
so the time it takes the ball to go down is:
Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:
and the time it takes to go up is:
When we add both times , we get:
Answer:
4.0 m/s
Explanation:
In the first part of the run, the athlete runs a distance of
at a speed of
So, the time he/she takes is
In the second part of the run, the athlete covers an additional distance of
with a speed
So, the time taken in this second part is
So, the total distance covered is
d = 300 m + 300 m = 600 m
And the total time taken
t = 100 s + 50 s = 150 s
Therefore, the average speed for the entire trip is
Answer:
1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s
2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions
3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s
4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²
Explanation:
The given parameters are;
The initial velocity of the fan, n = 0.220 rev/s
The magnitude of the angular acceleration = 0.920 rev/s²
The direction of the angular acceleration and the angular velocity = Clockwise
The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m
1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s
The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²
The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;
ω = ω₀ + α·t
From which we have;
ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s
The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s
2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;
θ = ω₀·t + 1/2·α·t²
θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians
2·π radians = 1 revolution
∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution
The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions
3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;
The tangential speed, = ω × r = ω × D/2
At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;
= ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413
The tangential speed, = 1.0338604413 m/s
The tangential speed ≈ 1.034 m/s
4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;
a = α × r = α × D/2
a = 5.78053048261 × 0.800/2 = 2.31221219304
The tangential acceleration, a ≈ 2.312 m/s²