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Illusion [34]
3 years ago
9

The overall magnification of a compound microscope with an objective lens magnification of 5 and an eye piece magnification of 3

0 will be
Physics
2 answers:
Katen [24]3 years ago
3 0
Ok the answer to this is quite simple all you do is multiply the objective lens magnification times the eye piece magnification. So multiply 30 times 5. 30 x 5 = 150. So your answer is answer choice C. 150 Hope this helped you. :)
Ann [662]3 years ago
3 0

Answer:

M = 150 times

Explanation:

When two or more lens are used to form image then final magnification of the image is given by the product of magnification due to each lens

so here we can say

M = m_1 \times m_2 \times m_3 ...............

so here we know that

m_1 = 5

m_2 = 30

now the total magnification is given as

M = m_1 \times m_2

M = 5 \times 30

M = 150

so total magnification is 150 times

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A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod. relat
Crank
<span>d.rotating counterclockwise and slowing down This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.</span>
7 0
3 years ago
FILL IN THE BLANKS here on earth, the pull of gravity on a mass of 1 kg is ......... newtons​
Aleks [24]

Answer:

9.8 Newton

Explanation:

At average gravity on earth (conventionally, g=9.80665m/s2),

a kilogram mass exerts a force of about 9.8 newton

I hope this answer helps you

4 0
2 years ago
Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m
inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

     Alf = T R / I (1)

     T = α I / R

Now let's use Newton's second law in the mass that descends

     W- T = m a

     a = (m g -T) / m

The two accelerations need related

     a = R α

    α = a / R

    a = (m g - α I / R) / m

    R α = g - α I /m R

    α (R + I / mR) = g

    α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

4 0
3 years ago
Please send me solution of the question pls​
ELEN [110]

Answer:

20m

Explanation:

P.E=mgh

2000=10×10×h

2000=100h

Divide both side by 100

2000/100=20

4 0
3 years ago
Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .
Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

P_{av}=V_{rms}I_{rms}Cos\phi

As given in the question

CosФ = R / Z

And we know that

V_{rms}=I_{rms}\times Z

So

P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}

P_{av}=I_{rms}^{2}\times R

6 0
3 years ago
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