B) The object's velocity doubled.
Explanation:
The graph is missing: find it in attachment.
The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as
where
m is the mass of the object
v is the velocity of the object
We notice that:
- The kinetic energy is directly proportional to the mass
- The kinetic energy is proportional to the square of the velocity
In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.
First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.
Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore
B) The object's velocity doubled.
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Answer:
v = 7934.2 m/s
Explanation:
Here the total energy of the Asteroid and the Earth system will remains conserved
So we will have
now we know that
now from above formula
now we have
now plug in all data
From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.
<h3>What is the gravitational force?</h3>
The gravitational force is an attractive force that acts between any two masses.
It is given by;
F = Gm1m2/r^2
F = 6.67 * × 10−11 * 2.5 * 5/(250)^2
F = 83.4 × 10−11 /62500
F= 1.33 * 10^-14 N
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Answer:
The law of inertia
Explanation:
A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
