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aniked [119]
3 years ago
5

If the near-point distance of the jeweler is 22.0 cm, and the focal length of the magnifying glass is 7.70 cm, find the angular

magnification when the diamond is held 5.68 cm from the magnifying glass. Assume the magnifying glass is directly in front of the jeweler's eyes.
Physics
1 answer:
Strike441 [17]3 years ago
5 0

Answer: The angular magnification M = 2.1

Explanation:

If the jeweler holds the diamond at the focal point of the magnifying glass, its image will be produced infinitely far away.

This gives an angular magnification of M = N/f

Since the diamond is held 5.68 cm from the magnifying glass,

Then, M = (N - 5.68)/f

Where N = 22.0 cm and f = 7.7 cm

M = (22 - 5.68)/7.7

M = 2.1

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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If
mart [117]

Answer:  29.50 m

Explanation: In order to calculate the higher accelation to stop a train  without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction  multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

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5 0
3 years ago
Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

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Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

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r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

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