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astraxan [27]
2 years ago
9

A 0.2 kg baseball is struck with a force of 100 n from a baseball bat. according to newton's 2nd law, what can we determine abou

t the ball's motion while it is in contact with the bat? question 2 options: it has a speed of 500 m/s it has a speed of 0.002 m/s it has an acceleration of 20 m/s/s it has an acceleration of 500 m/s/s it has a speed of 20 m/s it has an acceleration of 0.002 m/s/s
Physics
1 answer:
Black_prince [1.1K]2 years ago
7 0

1. According to Newton's second law of motion, when ball contact with bat we can find the acceleration.

acceleration = 500 m/s^{2}

According to Newton's second law,

Force = mass * acceleration

acceleration = Force/Mass

acceleration = 100n/0.2kg

acceleration = 500 m/s^{2}

Learn more about acceleration here:

brainly.com/question/15295474

#SPJ4

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HELP DUE TONIGHT<br><br> What is friction? How does friction affect the motion of an object?
hram777 [196]

Answer:

Friction always acts in the direction opposing motion. This means if friction is present, it counteracts and cancels some of the force causing the motion (if the object is being accelerated).

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.

3 0
3 years ago
The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree
frosja888 [35]

Hi there!

We must begin by converting km/h to m/s using dimensional analysis:

\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s

Now, we can use the kinematic equation below to find the required acceleration:

vf² = vi² + 2ad

We can assume the object starts from rest, so:

vf² = 2ad

(17.22)²/(2 · 75) = a

a = 1.978 m/s²

Now, we can begin looking at forces.

For an object moving down a ramp experiencing friction and an applied force, we have the forces:

Fκ  = μMgcosθ  = Force due to kinetic friction

Mgsinθ = Force due to gravity

A = Applied Force

We can write out the summation. Let down the incline be positive.

ΣF = A + Mgsinθ - μMgcosθ

Or:

ma = A + Mgsinθ - μMgcosθ

We can plug in the given values:

22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))

A = 46.203 N

6 0
3 years ago
The tonga trench in the pacific ocean is 36,000 feet deep. assuming that sea water has an average density of 1.04 g/cm3, calcula
MakcuM [25]
1110 atm    

Let's start by calculating how many cm deep is 36,000 feet. 

 36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm   

 Now calculate how much a column of water 1 cm square and that tall would mass. 

 1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2   

 We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).  

 It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.   

 1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2 

 11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals   

 Now to convert to atm 

 111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm   

 Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result). 

 1107.274 atm + 1 atm = 1108.274 atm   

 And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.
8 0
3 years ago
A 14 kg block lifted 5 m has a gravitational potential energy of
harina [27]

Answer:

700 joules

Explanation:

potential energy = m × h × g

( m = mass, h = height, g = acceleration due to gravity )

  • P.E = 14 × 5 × 10
  • P.E = 700 J

hence, gravitational potential energy of the object is 700J

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

8 0
3 years ago
What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element
Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0
3 years ago
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