Answer:
Perfectly inelastic collision
Explanation:
There are two types of collision.
1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.
2. Inelastic collision: When the momentum the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.
For a perfectly elastic collision, the two bodies stick together after collision.
Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.
Pressure is measured as force per unit area, which is the third option.
<h3>What is pressure?</h3>
Pressure is amount of force that is applied over a given area divided by the size of this area.
Pressure is calculated by multiplying the force applied on the object by the area covered.
Since force is measured in Newtons (N) and area is measured in m², the unit of pressure is Nm².
Therefore, pressure is measured as force per unit area.
Learn more about pressure at:
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Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct
Answer:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
Answer:
Explanation:
Q₁ = 20 X 10⁻⁶ C
Q₂ = -45 X 10⁻⁶ C
d is required distance.
F = Force between the = 7 N
F = k x Q₁ X Q₂ / d²
d² = k x Q₁ X Q₂ / F = 20 X 45 X 10⁻¹² X 9 X 10⁹ /7
=1.157
d = 1.075 m
