An iron cannon ball and a bowling ball are dropped at the same time from the top of a building. At the instant before the balls
1 answer:
Answer:
<em>C. kinetic energy</em>
Explanation:
<u>Free Fall </u>
When objects are dropped in free air (no friction), they are attracted to the Earth's surface by the force of gravity. Objects start from zero speed and gradually increase it because of the effect of the acceleration of gravity whose value can be considered as constant
The final speed of the object at time t is
Note that the final speed does not depend on the mass of the object. The kinetic energy is
The kinetic energy depends on the speed and the mass, so heavier objects dropped from the same height will have more kinetic energy. Let's analyze the options given in the question .
A. As shown above, the speed (or magnitude of the velocity) is the same regardless of the object's mass, so the heavier cannonball and the iron cannon will reach the ground at the same speed. Incorrect option
B. For every falling object in free air, the only acceleration is the acceleration of gravity. This option is not correct either
C. The kinetic energy is greater for the heavier cannonball because it has more mass, as discussed above. Correct option
D. Only the C. option is correct. This is not
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<h2>Answer:</h2>
(a) 10N
<h2>Explanation:</h2>The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
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Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
= -15 m/s
Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
Answer:
270 mi/h
Explanation:
Given that,
To the south,
v₁ = 300 mi/h, t₁ = 2 h
We can find distance, d₁
To the north,
v₂ = 250 mi/h, d₂ = 750 miles
We can find time, t₂
Now,
Average speed = total distance/total time
Hence, the average speed for the trip is 270 mi/h.