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Annette [7]
3 years ago
9

Which parts of a compound microscope magnify objects? a convex and a concave lens two convex lenses two concave lenses one conca

ve and two convex lenses?
Physics
2 answers:
Hoochie [10]3 years ago
6 0
I believe the answer is two convex lenses. A compound microscope has two systems of lenses for greater magnification, the ocular, or eyepiece lens that one looks into and the objective lens, or the lens closest to the object. Both the ocular lens and the objective lens are convex lens.
Inessa [10]3 years ago
3 0

Answer:

two concave lenses

Explanation:

You might be interested in
A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in
VikaD [51]

Answer:

B. - 0.328

Explanation

Potential Energy:<em> This is the energy of a body due to position.</em>

<em>The S.I unit of potential energy is Joules (J).</em>

<em>It can be expressed mathematically as</em>

<em>Ep = mgh........................... Equation 1</em>

<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>

<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>

Substituting these values into equation 1

Ep = 0.00274×12.2×9.8

Ep = 0.328 J.

Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J

The right option is B. - 0.328

<em />

7 0
3 years ago
A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find
Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0
3 years ago
Where is oceanic crust being formed?
Sever21 [200]
Oceanic crust is being formed it is because of the tectonic plates. Tectonic plates work because of volcanic eruption every time it will erupt the tectonic plates underground will have more pressure on magma, then below there will be a volcano thats why their is a volcano in ocean. The heat rises up then it will open, thats what we call divergent bounderies then their will be earthquakes form. Then another plate opens up and the crust is going down, it will stop if the volcano stops erupting because there is more lava left in volcano then it will go underwater and thats why oceanic crust is being formed.

#CarryOnLearning

Butbot na dili na tinuod ga timala ra ko ana
3 0
3 years ago
A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive
Ivahew [28]

Answer:

a

The  x- and y-components of the total force exerted is

           F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

b

 The magnitude of the force is  

            |F_{31 +32}| = 10.25 *10^{-5} N

   The direction of the force is  

         \theta =327.43 ^o   Clockwise from x-axis

Explanation:

From the question we are told that

    The magnitude of the first charge is q_1 = +5.00nC = 5.00*10^{-9}C

      The magnitude of the second charge is q_2 = -2.00nC = -2.00*10^{-9}C

        The position of the second charge  from the first one is  d_{12} = 4.00i \  cm = \frac{4.00i}{100} = 4.00i *10^{-2} m

        The  magnitude of the third charge is q_3 = +6.00nC = 6.00*10^{-9}C

       The position of the third charge from the first one is  \= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} =  (4i + 3j) *10^{-2}m

                |d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m

                |d_{31}| =5 *10^{-2} m

        The position of the third charge from the second  one is

                \= d_{32} = 3j cm = 3j *10^{-2}m

               |d_{32}| =(\sqrt{ 3^2}) *10^{-2} m

               |d_{32}| =3 *10^{-2} m

The force acting on the third charge due to the first and second charge is mathematically represented as

           F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}

 Substituting values

          F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}  \\ \ +  \ \ \ \ \ \ \ \ \   \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3}  * (4i + 3j ) *10^{-2}

            F_{31 +32} = 2.16 *10^{-5} (4i + 3j)  - 12*10^{-5} j

            F_{31 +32} =  (8.64i - 5.52 j) *10^{-5}

The magnitude of     F_{31 +32}  is mathematically evaluated as

            |F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}

             |F_{31 +32}| = 10.25 *10^{-5} N

The direction is obtained as

            tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}

              \theta = tan ^{-1} [-0.63889]

             \theta = - 32.57 ^o

             \theta = 360 - 32.57

            \theta =327.43 ^o

               

                         

5 0
3 years ago
In a perfectly elastic collision between two perfectly rigid objects
ipn [44]

Both the total momentum and the total kinetic energy are conserved

Explanation:

- In a collision between two or more objects, if there are no external forces acting on the system (isolated system), the total momentum of the objects is always conserved. This is called principle of conservation of momentum, and can be written as follows:

mu+MU = mv+MV

where

m, M are the masses of the two objects

u, U are the initial velocities of the two objects

v, V are the final velocities of the two objects

- The total kinetic energy, however, is not always conserved. In fact, we have two types of collision:

1) In a perfectly elastic collision, the total kinetic energy of the objects is conserved. This means that we can write the following equation:

\frac{1}{2}mu^2 + \frac{1}{2}MU^2 = \frac{1}{2}mv^2+\frac{1}{2}MV^2

2) In an inelastic collision, the total kinetic energy of the object is NOT conserved. This means that part of the total kinetic energy is "lost", converted into other forms of energy (mainly thermal energy, due to the presence of frictional forces within the system). The most extreme case is called perfectly inelastic collision, in which the two objects stick together after the collision, and there is the maximum loss of kinetic energy.

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

LearnwithBrainly

7 0
3 years ago
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