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DochEvi [55]
2 years ago
6

An object of height 2.2 cm is placed 5.1 cm in front of a diverging lens of focal length 19 cm and is observed through the lens

from the other side 50% Part (a) How many centimeters beyond the lens does the observer locate the object's image? Grade Summa Deductions Potential ry 0% 100% cosO asin0a Sin Submissions Attempts remaining:J5 (5% per attempt) detailed view cotana acosO acotan0sinh0 atan cosh0 tanh cotanhO Degrees Radians END DELI CLEAR Submit Hint I give up! Hints: deduction per hint. Hints remaining: Feedback: 296 deduction per feedback. là 50% Part (b) What is the height of the image, in centimeters?
Physics
1 answer:
monitta2 years ago
3 0

Answer:

a)   i = -4.02 cm , b)     h’= 1,576 cm

Explanation:

a) The constructor equation is

              1 / f = 1 / i + 1 / o

Where f is the focal length, i and o are the distance to the image and the object

Let's clear the distance to the image

             1 / i = 1 / f - 1 / o

             1 / i = 1 / -19 - 1 / 5.1

             1 / i = 0.2487

              i = -4.02 cm

b) let's use the expression of magnification

              m = h’/ h = - i / o

              h’= - h i / o

              h’= 2.2 4.02 /5.1

              h’= 1,576 cm

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Answer:

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Explanation:

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7 0
3 years ago
Wavelength multiplied by frequency equals
Kruka [31]
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4 0
3 years ago
which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20
MA_775_DIABLO [31]
<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
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Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
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Plug in the values in equation (A):

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7 0
3 years ago
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Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
2 years ago
Besides gravity, what factor keeps the moon and Earth in orbit?
earnstyle [38]

Answer:

interna

Explanation:

please mark as brainllest

8 0
2 years ago
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