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inysia [295]
2 years ago
6

HELP ASAP

Physics
1 answer:
Sladkaya [172]2 years ago
8 0
Answer:
I think the answer is
C) iron nails are attracted towards all materials
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the scores of players on a golf team are shown in the table. the teams combined score was 0 what was travis's score?
Alona [7]

Answer:

what table?

Explanation:

3 0
3 years ago
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Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
The refractive index for glycerine is n_g=1.473, while for air it is n_a = 1.00.

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
3 years ago
A capacitor consists of two metal surfaces separated by an insulating layer. A new capacitor has no charge on either of its surf
mihalych1998 [28]
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>

When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a  measure of the ability of a capacitor  to store energy. </em>For any capacitor, the capacitance is  a constant defined as:

C=\frac{Q}{V_{ab}}

To maintain C constant, if Q increases V also increases.

On the other hand, the potential energy U can be expressed as:

U=\frac{Q^{2}}{2C}

In conclusion, as Q increases the potential energy also increases.

5 0
3 years ago
A 2 kg book is pushed from rest to a final velocity of 3 m/s. The book travels 2 m. How much force was the book pushed with
Studentka2010 [4]

Explanation:

first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s

now using the 3rd law of motion

v^2=u^2+2as

0=9+2a2

a= -9/4m/s^2

now force=mass×accelration

=2kg×(-9/4)m/s^2

=4.5 N

4.5 newton force applied on the book!

✌️:)

6 0
3 years ago
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