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mezya [45]
3 years ago
14

What is the relationship between the amount of force this player applies to the soccer ball and the motion of the ball? How migh

t a force be applied without causing any motion on the ball?
Chemistry
1 answer:
Airida [17]3 years ago
3 0

Answer:

The player then exerts a force that is equal in magnitude (the size of a force) and opposite in direction. This slows down the ball

Explanation:

you can use friction friction (a force applied on the ball in the opposite direction to its motion), but the direction of its motion will remain the same.

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Please help, I will give brainylest
Anna [14]

Answer:

Explanation:

Divide both of the like fractions, then multiply the answers even if their iregular.

5 0
2 years ago
At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements
Gnom [1K]

Given that:

  • At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

     Initial volume of    Final volume of    Mass of water   Density (g/mL)

     burette (mL)        burette   (mL)       dispensed (g)

 

Sample 1      2.33                     7.34                   5.000               -----

Sample 2      7.34                    12.37                 5.025                -----

Sample 3      12.37                   18.50                6.112                  -----

Sample 4      18.50                  24.57               6.064                 -----

Sample 5     24.57                  31.31                6.720                  -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

  • The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

\mathbf{density = \dfrac{mass}{volume}}

\mathbf{density = \dfrac{5.01 g}{5.000 ml}}

density = 0.998004 g/ml

Sample 2:

\mathbf{density = \dfrac{5.025 g}{5.03ml}}

density = 0.999006 g/ml

Sample 3:

\mathbf{density = \dfrac{6.112 g}{6.13ml}}

density = 0.997064 g/ml

Sample 4:

\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}

density = 0.999012 g/ml

Sample 5:

\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}

density = 0.997033 g/ml

Thus, the average density for all the samples is:

\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 +   0.999012  + 0.997033  )}{5}}

= 0.998024

∴

The percentage error for the two densities measurement is:

=\dfrac{ (experimental \  value -theoretical  \ value)\times 100 }{theoretical  \ value}

Given that the theoretical value = 0.997044 g/ml

Then;

\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0
2 years ago
WILL GIVE BRAINLIEST Which effect is one likely result of a forest fire? a. extinction b. adaptation c. speciation d. forced mig
MAVERICK [17]

Answer:

d. forced migration

Explanation:

Certain hazardous occurrences affect living organisms in their natural habitat. One of those occurrences is forest fire. Forest fire or vegetation fire is an uncontrollable break out of fire in a vegetation, affecting the inhabitants of the area.

The occurrence of a forest fire will lead to a forced migration of organisms from their natural habitat. Animals and other mobile organisms will be forced to leave behind their devastating habitat and migrate to a less threatened area in order to survive.

4 0
3 years ago
Read 2 more answers
A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what
bulgar [2K]

Answer:

The mass of KF in the mixture is 2.77 gms.

Explanation:

Given;

Total weight of mixture (LiF+KF)=5.32gms

Let, mass of KF in the mixture = x gms

⟹ mass of LiF in mixture =(5.32-x)gms.

We know that :

Atomic weight of F=19gms.

Atomic weight of Li =7gms.

Atomic weight of K = 39 gms.

moles=mass/(molecular weight)

Thus, moles of KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

Thus,

moles of F in KF=moles of KF=x/58 ---(1)

moles of F in LiF =moles of LiF= (5.32-x)/26---(2)

From (1) & (2),

Total moles of Fluorine

=(x/58)+((5.32-x)/26)

Hence,

total weight of Fluorine in sample = moles*Atomic weight

=((x/58)+((5.32-x)/26))*19gms.

=3.12 gms.---(given)

Now, solving the equation for x,

26x +(5.32*58)-58x

=3.12*58*26/19

22x=308.56-247.62

x=60.94/22

=2.77 gms. (Answer)

Thus, the mass of KF in the mixture is 2.77 gms.

8 0
3 years ago
Which diagram shows a pair of electrons that have opposite spins?
Nastasia [14]

It's B ------------------------------------

7 0
3 years ago
Read 2 more answers
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