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Vedmedyk [2.9K]
3 years ago
10

You have a lens whose focal length is 6.91 cm. You place an object on the axis of the lens at a distance of 11.1 cm from it. How

far is the object\'s image from the lens?
Physics
1 answer:
stellarik [79]3 years ago
7 0

Answer:

4.2591 cm

Explanation:

We have given focal length of the lens f=6.91 cm

Object distance u =11.1 cm

We have to find the image distance that is v

For the lens we know that \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

So \frac{1}{6.91}=\frac{1}{v}-\frac{1}{11.1}

0.1447=\frac{1}{v}-0.09

\frac{1}{v}=0.2347

v=4.2591cm

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A wheelchair ramp is 5.2 m long and 0.8 m high. Calculate the ramp’s mechanical advantage
Arlecino [84]

Answer:

5

Explanation:

62

3 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?
OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0
3 years ago
A student compared some soccer players to the atoms in the gaseous state. Which of the following activities were the soccer play
tensa zangetsu [6.8K]

Answer:

The answer will be <em>D</em>

Explanation:

I choose answer <em>C</em> and got it wrong on the test. Also Liquid atoms vibrate fast and slide past each other. what answer <em>C </em>shows is a gas.

8 0
3 years ago
Read 2 more answers
6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440
MrRissso [65]

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

7 0
4 years ago
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