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Aleksandr-060686 [28]
3 years ago
12

Identify the region in the image where lighter elements, such as hydrogen or helium, are more likely to be found in a liquid or

gaseous state.
Physics
2 answers:
otez555 [7]3 years ago
4 0

Answer:

Further out (Jupiter to Pluto)

Explanation:

Annette [7]3 years ago
3 0

Lower in an arrangement

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Does the moon fall partly into earth's shadow when it is n "full"?
murzikaleks [220]
Yes, <span> the moon fall partly into earth's shadow when it is in its full size</span>
5 0
3 years ago
A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?
Scrat [10]

Answer:

{ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

♨ Question :

  • A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ \underbrace{ \sf{Required \: Answer : }}

☄ Given :

  • Initial velocity ( u ) = 0
  • Final velocity ( v ) = 60.0 m/s
  • Time ( t ) = 7.50 s

☄ To find :

  • Acceleration ( a )

✒ We know ,

\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) =  \frac{Final velocity ( v )  - Initial velocity ( u)}{t} }}}}

Substitute the values and solve for a.

➛ \sf{a =  \frac{60.0 - 0}{7.50}}

➛ \sf{a =  \frac{60.0}{7.50}}

➛ \boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}

---------------------------------------------------------------

✑ Additional Info :

  • When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
  • When a moving object comes in rest , in the case , final velocity ( v ) = 0
  • If the object is moving with uniform velocity , in the case , u = v.
  • If any object is thrown vertically upwards in the case , a = -g
  • When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

4 0
2 years ago
Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of
Fudgin [204]

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San  Francisco i.e 400 mi

considering ,

Alan's speed v_A=50mph

Beth's speed v_B=60mph

->For Alan:

The time required t_A= d/v_A= 400/50 => 8h

-> For beth:

The time required t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

3 0
2 years ago
Two soccer players kick a soccer ball back and forth along a straight line. The first player kicks the ball 16 m to the right to
Oliga [24]

Answer:

a

The total distance is d_t =19.1 m

b

The displacement is

  D_t = 12.9m

Explanation:

From the question we are told that

        Distance traveled by the ball for first player d_f = 16m to the right

        Distance  traveled by the ball for second player d_s = 3.1 m to the left

       

The total distance traveled by the ball is mathematically represented as

                   d_t = d_f + d_s

Substituting values

                  d_t = 16 + 3.1

                  d_t =19.1 m

The displacement is mathematically represented as

              D_t  = d_f -d_s

This is because displacement deal with direction and from the question we are told that right is positive and left is negative

          Substituting values  

                D_t = 16 -3.1

                 D_t = 12.9m

     

5 0
3 years ago
An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
Zigmanuir [339]

Answer:

(A) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )

\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )

\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

7 0
3 years ago
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