Answer:
A. 4,9 m/s2
B. 2,0 m/s2
C. 120 N
Explanation:
In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:
If the package is barely lifted, that means that T=m_2*g; then:
Solving the equation for a_mín, we have:
Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:
For the monkey: 
For the package: 
The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:
For the package: 
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:
Solving a, we have
We can then replace this value of a in one for the sums of force and find the tension T:
The correct option that can be deduced for both Object P and Q is Option b) I and II only
To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.
<h3>What is Density?</h3>
Density is a physical property of an object and can be expressed by using the relation:
From the given parameters, we are being told that:
This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.
However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.
Therefore, we can conclude that:
- 1. Q is heavier than P
- II. 1cm³ of Q has a greater mass than 1cm³ of P
Learn more about density here:
brainly.com/question/6838128
From what I can see it's D, I did this by simply examining the other answers and seeing that they are beneficial, so, from that information, this one must not be.
Recall that average velocity is equal to change in position over a given time interval,
so that the <em>x</em>-component of 
is
and its <em>y</em>-component is
Solve for 
and 
, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.
Note that I'm reading the given details as
so if any of these are incorrect, you should make the appropriate adjustments to the work above.
The speed of light is: c
= 3x10^8 m/s <span>
or
c = 186,000,000 miles/sec = 1.86x10^8 mi/s
1 furlong = 0.125 mile
1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h)
= 1209600s = 1.2096x10^6 s
Therefore,
c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort)
<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>
