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inna [77]
3 years ago
13

Appropriate word inside the parentheses.

Physics
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

<em>At constant mass, the  acceleration of an object varies (</em><em>directly</em><em>) with the net external force  applied. That is to say, that an object's acceleration increases as the force applied is (</em><em>increased</em><em>),  but its acceleration decreases if the force applied is (</em><em>decreased</em><em>).</em>

Explanation:

<u>Mechanical Force </u>

According to the second Newton's law, the acceleration of an object varies directly proportional to the external net force applied and inversely proportional to the mass of the object.

If the mass is constant, then the acceleration will vary in the same way as the force does.

Completing the sentences:

At constant mass, the  acceleration of an object varies (directly) with the net external force  applied. That is to say, that an object's acceleration increases as the force applied is (increased),  but its acceleration decreases if the force applied is (decreased).

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Evaluate ( 64800 ms ) 2 to three significant figures and express the answer in Si units. Express your answer using three signifi
Sergeu [11.5K]

Answer:

42.0×10² second²

Explanation:

Here, time is given in milisecond

(64800 ms)²

= 4199040000 ms²

The SI unit is seconds

1 second = 1000 milisecond

1\ milisecond=\frac{1}{1000}\ second

\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2

4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2

42.0×10² second²

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3 years ago
How does this app work
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A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q
Kryger [21]
D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
3 0
3 years ago
An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.
Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

\dfrac{1}{f}=-\dfrac{71}{2522}

f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

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The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0
3 years ago
The fulcrum is between the effort and the load
Inessa05 [86]

Class 1 lever

Explanation:

In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.

  • The effort is the force input and the load is the force output
  • The fulcrum is a hinge between the load and effort.
  • Movement of the effort and load are in opposite directions.
  • There are other classes of lever like the class 2 and 3.
  • They all have different load, fulcrum and effort configurations

learn more:

Load related problems brainly.com/question/9202964

Torque brainly.com/question/5352966

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3 years ago
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