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mash [69]
3 years ago
9

A 6 cm object is 15 cm from a convex lens that has a focal length of 5 cm. What is the distance of the image from the lens, to t

he nearest hundredth?
–33.33 cm
–7.69 cm
7.69 cm
33.33 cm
Physics
2 answers:
cupoosta [38]3 years ago
7 0

Answer:

C.) 7.69 cm

Explanation:

sweet [91]3 years ago
6 0
So we want to know what is the distance d of the object from the lens if the height of the object is h=6 cm, focal length of the lens is f=5 cm and the distance d=15 cm is the distance of the object from the lens. From the formula for the convex lens 1/f=(1/D + 1/d) where D is the distance of the image from the lens we can get D after solving for D: 1/D=(1/f) - (1/d),
1/D=(1/5)-(1/15)=0.2-0,06667=0.13333 so f=1/0.13333=7.500187 cm. Rounded to the nearest hundredth D=7.50 cm.  That is very close to 7.69 cm so the correct answer is the third one. 
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aliina [53]

Answer:

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8 0
4 years ago
A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes
Vadim26 [7]

Complete question

A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes his measurements

a)precise but not accurate

b)accurate but not precise

c)neither precise nor accurate

d)both precise and accurate

Answer:

The measurement is precise but not accurate

Explanation:

A measurement can either be precise or accurate.

  • A Precise measurement describes how close the measured values are to each other.
  • An accurate measurement describes how close a measured value is to the true value.

In this question, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are far from the true value (1.0 kg), therefore the measurement is not accurate.

However, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are close to each other, therefore the measurement is precise.

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4 0
3 years ago
1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Cal
disa [49]

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given:

- The radius of earth r = 6.3781×10^6 m

- The angular speed of earth w = 7.27*10^-5 rad/s

- The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

- The mass of earth m = 5.972 × 10^24 kg

- The inertia of sphere I = 2/5 * m* r^2

Solution:

- The angular acceleration of the earth from rest to w is given by α:

                                α = w / t

                                α = (7.27*10^-5) / (1.57784760 * 10^8)

                                α = 4.60754*10^-13 rad/s^2

- The required torque τ is given by:

                                τ = I*α

                                τ = 2/5 * m* r^2 * α

                           τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

                            τ = 4.47746 * 10^25 N-m

- The power required P to turn the earth to the speed w is:

                           P = τ*w

                           P = (4.47746 * 10^25)*(7.27*10^-5)

                           P = 3.25511*10^21 W

- The energy E required is :

                           E = P / t

                           E = (3.25511*10^21) / (1.57784760 * 10^8)

                           E = 2.06301 * 10^13 J

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Answer:

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Explanation:

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