The gravitational force between the Earth and the Moon is
Explanation:
The gravitational force between two objects is given by:
where
:
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them objects
In this problem, we are given:
(mass of the Moon)
(mass of the Earth)
(distance between Earth and Moon)
Substituting into the formula, we find the force:
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Answer:
f Homework: Creating and Solving Linear Equations to Model Real World Problems Part I
he document will
1) Use the story below about Sanjeet and his friends' end-of-season basketball statistics to answer the
questions that follow.
Story
Points Scored
Terrence's points:
Sanjeet and his teanit members were looking
Sanjeet's points:
Cole's points:
at the total points scored by each player
during the season. Sanjeet scored twice ay
many pointy ay Terrence. Cole scored 12
more points than Sanjeet. Together the boys
scored 992 points during the season. How
many points did each boy score?
Equation:
Explanation:
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The correct answer is
<span>
C) 1200 g/m3. Let's see why. The relationship between liters and cube decimeters is
</span>
Therefore,
However, we also know that
Therefore,
and
Therefore, the density of the problem
becomes
Answer:
(a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). , and x-t graphs
Explanation:
Given that,
Time t = 1.0 s
Velocity = 5.0
The Acceleration equation is
We need to calculate the velocity
Using formula of acceleration
On integrating
Put the value into the formula
Put the value into the formula
We need to calculate the velocity at 2.0 sec
Put the value of initial velocity in the equation
(b). If the bus’s position at time t = 1.0 s is 6.0 m,
We need to calculate the position
Using formula of velocity
On integrating
The position at t = 2.0 s
Hence, (a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). , and x-t graphs
Answer:
2.42 seconds
Explanation:
Assume that air resistance is negligible, use trigonometry to find the vertical component of the velocity by using trigonometry:
<span>31⋅<span>sin50</span>=23.7</span>
Where 31 <span>m<span>s<span>−1</span></span></span> is the hypotenuse and by using sin to get the opposite component (vertical velocity) of the trajectory.
Now comes the use of the formula:
v = u + at
where v is the final velocity (0 <span>m<span>s<span>−1</span></span></span>), u is the initial velocity (31 <span>m<span>s<span>−1</span></span></span> ), a is the acceleration of gravity (9.81 <span>m<span>s<span>−2</span></span></span>) and t is the time it takes to arrive at the top of the trajectory.
By making t as the subject:
<span>t=<span><span>v−u</span>a</span></span>
You can calculate the value of t:
<span><span><span>0−23.7</span><span>−9.81</span></span>=2.42</span> (to 3 significant figures)
Better way to see it:
<span><span><span>0−<span>(31⋅<span>sin50</span>)</span></span><span>−9.81</span></span>=2.421</span> (to 4 significant figures)
Note: You must remember that you are dealing with velocity, not speed . Since velocity is a vector quantity, you must select the direction at which values will be positive. In my example, I set my upward direction as the positive value while my downward vectors as negative value (a, acceleration, 9.81 <span>m<span>s<span>−1</span></span></span>).