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Tanzania [10]
3 years ago
13

Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller

child hops off to jump straight down into the pool, the bigger child releases herself at the top of the frictionless slide.
(i) Upon reaching the water, how does the kinetic energy of the smaller child compare with that of the larger child?

a.It is greater. b.It is less. c.The two energies are equal.

(ii) Upon reaching the water, how does the speed of the smaller child compare with that of the larger child is?

a.It is greater. b.It is less. c.The two speeds are equal.

(iii) During their motions from the platform to the water, how does the average acceleration of the smaller child compare with that of the larger child? a.It is greater. b.It is less. c.The two accelerations are equal.
d.The First and third are wrong
Physics
1 answer:
IgorC [24]3 years ago
6 0

Answer:

Explanation:

i )

In a conservative field like gravitational field , loss of potential energy  or work done , depends upon the initial and final point and not the manner in which 2 nd point has been reached . Since the initial and final point is same in both the cases of straight and curved path , final velocity will remain same for both of them .

Hence , due to increased mass of larger child , his kinetic energy will be greater .

ii ) Since the initial and final point is same in both the cases of straight and curved path , final velocity will remain same for both of them .

iii ) Smaller child undergo free fall , therefore , he will fall with acceleration g . The larger child falls on curved path . So , he will have only a component of

vertical g at any moment  . hence average acceleration  will be less.

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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
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  • 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
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4 0
2 years ago
A car accelerates from rest at a rate of 6.50 m/s2. Determine thevelocity of the car at t = 4.00 s.
aksik [14]

Answer:

1.62 m/s2

Explanation:

6.50 divided by 4

3 0
2 years ago
How far will a 600 kg boat travel in 10 s if there is a constant 900 N force on it and it starts from rest?
natta225 [31]

Answer:

here

Explanation:

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4 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

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Explanation:

Given that,

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After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

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3 years ago
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