ddawgg:
A football player kicks a 0.94 kg football with a force of 2.4 N. Calculate the acceleration of the football as the player kicks the ball. Round your answer to the nearest tenth of a m/s2.
jimthompson5910:
We use Newton's second law F = m*a In this case, F = 2.4 newtons = force applied m = 0.94 kg = mass a = unknown acceleration
jimthompson5910:
So you need to solve for 'a' F = m*a 2.4 = 0.94*a
ddawgg:
whats the answer\
jimthompson5910:
Divide both sides by 0.94 to solve for 'a'
ddawgg:
thx merry christmas
jimthompson5910:
No problem
ddawgg:
so divide 2.4 by 0.94
jimthompson5910:
correct
ddawgg:
2.5531914
ddawgg:
rounded is 2.6
jimthompson5910:
You have the correct answer
ddawgg:
thank you very much
jimthompson5910:
You're welcome
ddawgg:
here is another one A model rocket accelerates at 15.3 m/s2 with a force of 44 N. Calculate the mass of the rocket. Round your answer to the nearest tenth of a kg. The mass of the rocket is kg.
ddawgg:
the massof the rocket is _______kg.
jimthompson5910:
So we have a = 15.3 and F = 44
jimthompson5910:
Solve for m F = m*a 44 = m*15.3 To solve for m, we need to divide both sides by 15.3 to undo the multiplication.
ddawgg:
2.9
jimthompson5910:
Correct answer
ddawgg:
thx bud
jimthompson5910:
Sure thing
ddawgg:
edginuity sucks
ddawgg:
What is the mass of a crate if a force of 200 N causes it to accelerate at 8 m/s2? (Formula: F=ma)
jimthompson5910:
So we have F = 200 and a = 8 this time
ddawgg:
i divid 200 by 8
jimthompson5910:
Yes those are the correct steps. What does that lead to?
ddawgg:
25 kg
jimthompson5910:
correct
ddawgg:
you are really good
jimthompson5910:
Thanks I appreciate it
ddawgg:
no thanks you
jimthompson5910:
sure thing
ddawgg:
Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table. A 3-column table with 4 rows. The first column labeled Object has entries 1, 2, 3, 4. The second column labeled Mass has entries 10 kilograms, 100 grams, 10 grams, 1 kilogram. The third column labeled Force (Newtons) has entries 4 Newtons, 20 Newtons, 4 Newtons, 20 Newtons. Based on the data, which object has the greatest acceleration?
ddawgg:
i described the table
ddawgg:
well it described the table
jimthompson5910:
it would be better to post a screenshot
ddawgg:
how o you do that on a chroam book
jimthompson5910:
I'm not sure I never used them before. This site might help https://www.digitaltrends.com/computing/how-to-take-a-screenshot-on-a-chromebook/
jimthompson5910:
Thanks
ddawgg:
np
jimthompson5910:
So what you'll do is divide each force over its corresponding mass. Then you'll pick the largest result.
jimthompson5910:
For row 3, you need to convert 10 grams to kilograms
jimthompson5910:
1 kg = 1000 grams 1 gram = 0.001 kg 10 grams = 0.01 kg
jimthompson5910:
Row 2 is a similar story, 100 grams = 0.1 kg
ddawgg:
A ball has a mass of 140 g. What is the force needed to accelerate the ball at 25 m/s2? (Formula: F=ma)
jimthompson5910:
F = ma F = 0.140*25 F = ?? Note how I converted 140 grams to 0.140 kg
ddawgg:
so divide 0.140 by 25
jimthompson5910:
No you compute 0.140*25 the * means multiply
ddawgg:
3.500000000000...
jimthompson5910:
3.5 yes
jimthompson5910:
Your thoughts?
jimthompson5910:
hint: think of a tug of war
ddawgg:
wrong one
ddawgg:
mb
jimthompson5910:
Not sure what you mean
jimthompson5910:
Choice C is correct
ddawgg:
Laura sees a horse pulling a buggy. She wonders how it can accelerate if the action of the horse pulling the cart would cause an equal and opposite reaction of the cart pulling on the horse. Which explanation best answers her question?
ddawgg:
a.The net forces exerted on the horse and cart are the same, but the action force is larger. b.The net forces exerted on the horse and cart are the same, but they are in the same direction. c.The net forces exerted on the horse and cart are not the same, so the two forces cancel out. d.The net forces exerted on the horse and cart are not the same, so they are not balanced forces.
jimthompson5910:
I'm not sure about this one
ddawgg:
ok how about
jimthompson5910:
The smooth ice reduces friction, which in this case is the reactionary force. This is because it reacts to the force of the puck being pushed. For example, if the puck is pushed to the right, then the reactionary force friction pushes to the left to slow down the puck. |dw:1608315485846:dw|
jimthompson5910:
|dw:1608315567263:dw|
ddawgg:
so it slowes down the puck
jimthompson5910:
yes
jimthompson5910:
Smooth ice doesn't eliminate friction entirely
ddawgg:
ok then
jimthompson5910:
Friction is a good thing or else the players wouldn't be able to stop, slow down, or change direction.
jimthompson5910:
Your thoughts?
jimthompson5910:
Hint: think of something that isn't moving and what it takes to get it moving
ddawgg:
i got it
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