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12345 [234]
2 years ago
12

The focal length of david's lens is 60 cm. if rebecca stands in front of david at a distance of do and david perceives the posit

ion of rebecca at di, what does do equal if the magnification is -0.40?
Physics
1 answer:
Makovka662 [10]2 years ago
5 0

if rebecca stands in front of david at a distance of do and david perceives the position of rebecca at di, di will be +84 cm

<h3>What is focal length ?</h3>

How strongly light converges or diverges depends on an optical system's focal length, which is the inverse of optical power. A system with a positive focus length is said to converge light, whereas one with a negative focal length is said to diverge light.

focal length = +60 cm

magnification m = -0.40

focal length being positive an magnification negative.

given lens is a convex lens.

for a lens

m = di/do and 1/f = (1/di) - (1/do)di

= -0.4do1/f = (1/-0.4do) - 1/do do

= -210 cmdi = -0.4 * -210

di = +84 cm

To learn more about focal length go to - brainly.com/question/25779311

#SPJ4

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The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer
zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

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Lifted acceleration = 5.0\ m/s^2

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

F = ma

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in m/s^2

Now, we substitute m = 20 kg and a = 5.0\ m/s^2 in Newton's second law,

F = 20\times 5.0

F = 100\ m/s^2

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

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A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p
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Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

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= 3² + 2 x 1.837 x 9.8

= 9 + 36

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6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
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