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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.

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:

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

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♨ 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

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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