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I am Lyosha [343]
4 years ago
11

Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose

screen is 65 cm away. What refractive power will these glasses require?
Physics
1 answer:
marusya05 [52]4 years ago
6 0

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= -  .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0
3 years ago
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6 0
3 years ago
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when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv
sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

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A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water a
Kaylis [27]

Answer:

Cp= 0.44 J/g.C

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Explanation:

From energy conservation

Heat lost by metal = Heat gain by water +Heat gain by  calorimeter

Because here temperature of metal is high that is why it loose the heat.The temperature of water and  calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

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By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

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Cp= 0.44 J/g.C

This is heat capacity of metal.

4 0
3 years ago
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