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stealth61 [152]
3 years ago
8

You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

happen to have around. With the lens close to your eye and the animal at the lens's focal point, what angular magnification M M do you achieve? Assume your near point is at 25.0 cm.
Physics
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

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The 88-lb force P is applied to the 210-lb crate, which is stationary before the force is applied. Determine the magnitude and d
Marina86 [1]

Answer:

F=-88Ib

Explanation:

From the question we are told that:

Force P=88Ib

Mass of crate M_c=210Ib

Generally the equation for Frictional force F is mathematically given by

Friction\ force (f) = friction\ coefficient\ (u) * Normal\ reaction (N)

F=u*N

with \mu =0.47

F=98.7Ib

Therefore since Static Friction supersedes applied force body remains at rest.

Frictional force =88Ib (negative)

F=-88Ib

5 0
2 years ago
In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equatio
Artist 52 [7]

Answer:

3.626 m/s

Explanation:

v=d/t

1. -0.02/0 = 0 m/s

2. 0.86/0.2 = 4.3 m/s

3. 1.71/0.4 = 4.275 m/s

4. 2.54/0.6 = 4.23 m/s

5. 3.32/0.8 = 4.15 m/s

6. 4.08/1.0 = 4.08 m/s

7. 4.79/1.2 = 3.99 m/s

8. 5.48/1.4 = 3.91 m/s

9. 6.15/1.6 = 3.84 m/s

10. 6.76/1.8 = 3.76 m/s

11. 7.37/2.0 = 3.66 m/s

12. 7.92/2.2 = 3.6 m/s

13. 8.45/2.4 = 3.52 m/s

14. 8.96/2.6 = 3.45 m/s

the mean of these numbers is 3.626

his average velocity ks 3.626 m/s

6 0
2 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will
nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg

So, the mass that is attached to the spring is 1 kg.

6 0
2 years ago
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