If you increase the mass of an object and want to move an object a specific distance, then you need to do extra work than the earlier
<h3>What is work done?</h3>
The total amount of energy transferred when a force is applied to move an object through some distance
Work Done = Force * Displacement
For example, let us suppose a force of 10 N is used to displace an object by a displacement of 5 m then the work done on the object can be calculated by the above-mentioned formula
work done = 10 N ×5 m
=50 N m
Thus, when an object's mass is increased and it is desired to move it a certain distance, more work must be done than previously.
Learn more about work done from here
brainly.com/question/13662169
#SPJ1
<span>A light-year measures the distance that light travels in 1 year.
Answer : B ) Distance
-Hope this helps.</span>
Answer:
100
Explanation:
take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time) = 100
Answer:
0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.
Explanation:
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
