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lianna [129]
3 years ago
9

all of the following methods are ways to correct a run-on sentence except ,placing a semicolon between the compete sentences.put

ting a comma in between the complete sentences.adding a conjunction between the complete sentences.
Physics
1 answer:
olchik [2.2K]3 years ago
6 0
All of the following methods are ways to correct a run-on sentence except: Putting a comma in between the complete sentences. 

One of the uses of semi-colon is to separate two complete or independent sentences. A conjunction can also be used such as: and, however and but, to connect two complete sentences and transform them into a compound sentence.
You might be interested in
Which object represents a negatively charged particle? which object represents a positively charged molecule? which object repre
kari74 [83]

The answers to your questions are as written below:

  • The objects that represents a negatively charged particle is : Image B
  • The object that represents a positively charged molecule is : Image A
  • The object that represents an uncharged molecule is : Image C
  • The object the will not move when in an electric fied is : Image C

<h3>Different types of charges molecules</h3>

A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.

A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.

Hence we can concude that the answers to your questions are as listed above.

Learn more about electric charges :brainly.com/question/857179

#SPJ4

attached below is the missing image

8 0
2 years ago
28. Identify whether the following objects are in
Marianna [84]

Answer:

a. A baseball after it has been  hit - not in free fall

b. A rock that is thrown in the  air - not in free fall

c. The moon - free-fall

d. A paper airplane - not in free fall

e. A bird flying - not in free fall

Explanation:

  1. The free-fall is defined as the falling of an object due to the action of gravity. The object is not experiencing any other force neglecting the air resistance.
  2. If an object is in free-fall, the direction of its motion is directed towards the center of the earth. It does not have a horizontal component of velocity.
  3. If the body is under free-fall, but a centripetal force acts on it where it is equal to the gravitational force at that point. The object will have two components of velocity along the tangential line, perpendicular to the radius of the orbit.

a. A baseball after it has been  hit - not in free fall according to point 1 & 2.

b. A rock that is thrown in the  air - not in free fall according to point 1.

c. The moon - free-fall according to point 3.

d. A paper airplane - not in free fall according to point 1 & 2.

e. A bird flying - not in free fall according to point 1 & 2.

7 0
2 years ago
A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0
2 years ago
Estimate how long the sun would last if it were merely a huge fire that was releasing chemical energy. Assume that the sun begin
stepan [7]

Answer:

≅ 17000 years or 1.7 x 10⁴ years

Explanation:

time= total energy/power

=  (10⁸J/kg)(2x10³⁰ kg) / 3.8 x 10²⁶ J/s

 = 526,315,789,473 s

=  16689 years

≅ 17000 years or 1.7 x 10⁴ years

7 0
2 years ago
julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a
aliya0001 [1]
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 
Time to height 
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s 
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m 
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m 
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s 
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s           (ANSWER)
8 0
3 years ago
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