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Monica [59]
3 years ago
15

When observing a specimen in a microscope you are told that the total magnification of the specimen is 630x assuming you are usi

ng a standard occular lens with a magnification of 10x what is the magnification of the objective lens?
Physics
1 answer:
Norma-Jean [14]3 years ago
5 0
<span>Objective Lenses: Usually you will find 3 or 4 objective lenses on a microscope. They almost always consist of 4X, 10X, 40X and 100X powers. When coupled with a10X (most common) eyepiece lens, we get total magnifications of 40X (4X times10X), 100X , 400X and 1000X.</span>
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A student observes that it is hard to hear music underwater in a pool. They state that the sound is always muffled. They
s344n2d4d5 [400]

Answer:

FALSE      

Explanation:

The answer is false.

The speed of the sound in water is  faster when compared to the speed of sound in air. This is because, the particles in air is loosely packed and are far from each other as compared to water or liquid.

The water particles are close to each other than air particles, so water particles are able to transmit the vibrations of the sound faster than the air particles.

Therefore sound waves travels faster in water than in air.

5 0
2 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\&#10;mg=m\frac{v^2}{r}\\&#10;gr=v^2\\&#10;v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
Mary was looking out of the window she saw lightening and then heard thunder a few seconds later explain why she saw lightening
alekssr [168]

Explanation:

It is based upon the fact that " The light travels faster then sound." As the speed of light is faster then the speed of sound, light travels 300,000 km per second and sound travels 1192 km per hour. That is why we observe the lightening first and hear the the sound of thunder later.

        You can do this experiment by yourself. Once you see the lightening start counting the seconds until you hear the sound of thunder.Then divide the seconds by 5, you will find out how many miles away the lightening strike was.

3 0
3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
Two identical 3.0-kg cubes are placed on a horizontal surface in
DedPeter [7]

The magnitude of the force exerted by the left cube on the right cube is 17.64N.

<h3>What is frictional force?</h3>

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.

Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.

From the equilibrium of forces in vertical direction

Normal force N= 2m x g

friction force f = μN =μ(2m)g

From the equilibrium of forces in horizontal direction

F₁ =ma =0

using Newton's third law of motion, we get

F₁  - f =0

F₁  =f = μ(2m)g

Put the values, we get

F₁  = 17.64N

Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.

Learn more about friction force.

brainly.com/question/1714663

#SPJ1

5 0
1 year ago
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