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3 years ago

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an object 8.25 cm from a lens creates a virtual image of magnification 2.40 what is the focal length of the lens (mind your minu

s sign) (unit=cm)

1 answer:

murzikaleks [220]3 years ago

5 0

Answer:

Focal length of the lens is 14.14 cm

Explanation:

As we know that the image formed due to lens is virtual image

so here we have

M = 2.40

now we have

$\frac{d_i}{d_o} = 2.40$

now we have

distance of object is 8.25 cm

so we have

$d_i = 8.25(2.40) = 19.8 cm$

now by lens formula

$\frac{1}{d_i} - \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{-19.8} - \frac{1}{-8.25} = \frac{1}{f}$

$f = 14.14 cm$

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

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3 years ago

The downward pull of an object due to gravity is the objects _______?

elena-s [515]

The downward pull of an object due to gravity is the object’s weight.

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3 years ago

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A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

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3 years ago

List five situations in which temporary wiring would be used

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an emergency

building a test circuit

temp joining components

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You are getting bored as you are stuck at home, and want to find out the effective 'spring constant' of random objects. You find

Luda [366]

Given Information:

Mass of sock = 0.23 kg

Stretched length of sock = x = 2.54 cm = 0.0254 m

Required Information:

Spring constant = k = ?

Answer:

Spring constant = k = 88.82 N/m

Explanation:

We know from the Hook's law that

F = kx

Where k is spring constant, F is the applied force and x is length of sock being stretched.

k = F/x

Where F is given by

F = mg

F = 0.23*9.81

F = 2.256 N

So the spring constant is

k = 2.256/0.0254

k = 88.82 N/m

Therefore, the spring constant of the sock is 88.82 N/m

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3 years ago

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