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3 years ago

A convex lens has a focal length of 0.33 m. The object distance is 0.7 m. What is the image distance?

Physics

1 answer:

dimulka [17.4K]3 years ago

7 0

Answer:

Explanation:

1/v - 1/u = 1/f

given, f = 0.3 m, u = -0.4m

so, 1/v - 1/-0.4 = 1/0.3

or, 1/v = 1/0.3 - 1/0.4 = 1/1.2

v = 1.2 m

now, differentiating 1/v - 1/u = 1/f with respect to t,

-1/v² dv/dt + 1/u² du/dt = 0

or, dv/dt = (v/u)² du/dt

putting, du/dt = 0.01 m/s , v = 1.2 m and u = -0.4 m

so, dv/dt = (1.2/-0.4)² × 0.01

= 0.09 m/s

hence, speed of image with respect to lens is 0.09 m/s .

from formula of magnification

magnification, m = v/u

differentiating with respect to time both sides,

dm/dt = (u dv/dt - vdu/dt)/u²

= (-0.4 × 0.09 - 1.2 × 0.01)/(-0.4)²

= (-0.036 - 0.012)/0.16

= -0.048/0.16

= -0.3 m/s

hence, magnitude of rate of change of lateral magnification is 0.3 m/s

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Read 2 more answers

A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

PIT_PIT [208]

Their velocity afterward is v=3.467 m/s

Explanation:

Given:

Mass of the first football player= 91.5 kg

Initial velocity of the football player 3.73 m/s

Mass of second football player=63.56 kg

Initial velocity of the second football player=3.09 m/s

To find:

Final velocity of both players=?

Solution:

According to the law of conservation of momentum,

Initial momentum =final momentum

mathematically represented as

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...........................(1)

where

$u_1$ =intial velocity of the football player

$u_2$ = inital velocity of second football player

$v_1$ =finall velocity of the first football player

$v_2$ =final velocity of second football player

after tackling , both the football players moves with the same velocity,

so $v_1=v_2=v$

Hence equation (1) becomes

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

now substituting the values,

$v=\frac{(91.5\times+3.73)+(63.5\times3.09)}{(91.5+63.5)}$ $v=\frac{(341.29+196.210)}{155}$

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3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

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1 year ago

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