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3 years ago

A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

Physics

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

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For these pictures is more or less friction needed?

antiseptic1488 [7]

Answer:

8: More

9: More

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11: Less

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3 years ago

A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th

Andrew [12]

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2 V² = g x 2r + 1/2 v²

V² = g x 4r + v²

V² = 9.8 x 4 + 8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T = mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

3 0

4 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected

V125BC [204]

Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e

q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)

=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.

3 0

3 years ago

Cells & Cellular Processes

never [62]

Answer:

one or more cells

Explanation:

I need to have 20 characters to submit lol

7 0

3 years ago