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3 years ago

10

The revolving nosepiece of a compound microscope is used to: a. move the condenser up or down b. change the objective lens c. ad

just the slide to move the object into view d. change eyepiece lenses

1 answer:

Jlenok [28]3 years ago

6 0

Answer:

Option B: change the objective lens

Explanation:

The revolving nosepiece is one of the parts of a microscope. Its responsibility is to hold the objective lenses.

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Just after two identical point charges are released when they are a distance D apart in outer space, they have an acceleration a

Darya [45]

Answer:

d) 4a

Explanation:

r = Distance

Electrostatic force is given by

$F=\dfrac{kq_1q_2}{r^2}$

$ma=\dfrac{kq_1q_2}{r^2}$

It can be seen that the force is inversely proportional to distance

$a\propto \dfrac{1}{r^2}$

If the distance is reduced

$\dfrac{a_1}{a_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{a_1}{a_2}=\dfrac{(\dfrac{d}{2})^2}{d^2}\\\Rightarrow \dfrac{a_1}{a_2}=\dfrac{1}{4}\\\Rightarrow a_2=4a_1$

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The answer is d) 4a

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4 years ago

Motion and force of physics. Plz answer.

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Explanation:

a. " for every action there is an equal and opposite reaction".

b. The electric fan does not stop moving just after the switch turns off because of rational inertia force.

c. The force applied to first vehicle is 120N.

d. In my view it doesn't support the law of conservation of momentum. Momentum of 1 and 2 object before the collision is equal to the total momentum of two object after collision.

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3 years ago

A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the accele

ElenaW [278]

(a) $1.23 m/s^2$

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

$d = ut + \frac{1}{2}at^2$

Since u=0 (the block starts from rest), it becomes

$d=\frac{1}{2}at^2$

So by solving the equation for a, we find the acceleration:

$a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2$

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

$W_p = mg sin \theta$

where

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

$\theta=33.0^{\circ}$ is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

$F_f = - \mu mg cos \theta$

where

$\mu$ is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

$W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma$

Solving for $\mu$ , we find

$\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50$

(c) 12.3 N

The frictional force acting on the block is given by

$F_f = \mu mg cos \theta$

where

$\mu = 0.50$ is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=33.0^{\circ}$ is the angle of the incline

Substituting, we find

$F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N$

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

$v^2 - u^2 = 2ad$

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

$v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s$

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Answer:

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