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zhuklara [117]
3 years ago
12

When comparing the proper position for spotting different exercises, you can conclude that, in spotting any exercise, it is impo

rtant to
Physics
2 answers:
Hitman42 [59]3 years ago
8 0
It is important to work with a partner pay attention to your partner  always ask if he needs help never take your eye off of him because in a second he could let go of that bench press because its too heavy and he could seriously be injured.
dezoksy [38]3 years ago
7 0

Answer:The answer is C

Explanation:

Just took the test

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The motion map shows an object’s position and velocity at given times. A long black arrow right labeled x. Above that are 5 gree
Digiron [165]

The map be changed <u>by adding vectors that are all the same length and placing them above the current top row of vectors</u>

<h3>Further explanation</h3>

A motion map represents the position, velocity, or acceleration of an object at various time readings

a small point or dot describes the position of the object. the object's velocity / acceleration is represented by an arrow / vector.

In this motion map, The arrows (vectors) are all the same length (same speed) and point in the same direction (direction to the right) for t = 0 (origin) to t = 4s

At t = 5s, the object has slowed down, and at t = 7s and 8s the object stops

At t = 9s-11 s, the object accelerates to the left

So the position of the next moving object will be above the previous movement, and the length of the same vector shows the same speed / acceleration

So that motion maps show constant acceleration after the object changes direction, it can be done by adding vectors that are all the same length and placing them above the current top row of vectors

<h3>Learn more</h3>

The initial velocity of the bird before the gust of wind

brainly.com/question/13207873

Keywords: motion map,  constant acceleration, dot, arrows, object,length, vectors, row

#LearnwithBrainly

5 0
3 years ago
Read 2 more answers
Why does a rolling sphere slows down?​
Iteru [2.4K]
There’s frictional force acting on the sphere, which causes it to gradually slow down, and eventually come to a stop.
4 0
3 years ago
A race car drives one lap around a race track that is 500 meters in length.
pychu [463]

Explanation:

Check out the picture I drew for a minute before reading this...

B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.

A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

3 0
2 years ago
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0
2 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
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