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Daniel [21]
3 years ago
8

9) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen

, what things could you do to help yourself (before asking the instructor to assist you
Physics
1 answer:
xeze [42]3 years ago
4 0

Answer:

See the answer below

Explanation:

<em>The best thing one can do in this case would be to return the microscope's objective to low power and then </em><em>re-center the specimen </em><em>before switching back to high-dry power.</em>

Most of the time, <u>what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage</u>.  Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.

After doing the above and the specimen still does not come into focus, then the instructor can be called upon.

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Pls answer the 15a answer i cant understand it​
Maksim231197 [3]

Materials required for the experiment of limiting force borne by string:-

  1. String balance
  2. weights
  3. light strings
  4. weight hanger
  5. pan for spring balance
  6. Sand

Steps of procedure for for the experiment of limiting force borne by string:-

  1. First we have to tie a light string to the fixed support and then tie the other end with the weight hanger consists of weight.
  2. Add additional weight to the hanger again and again. And continue the same until the string is broken.
  3. Note down the weight (x) where the string is broken.
  4. Suspend spring balance to a support.
  5. Tie the light string at the end of the balance and at the other end suspend the pan for spring balance.
  6. Now place the weights (x-100 grams) in pan.
  7. Observe the reading in the spring balance.
  8. Add a small amount of sand in the pan by observing the readings.
  9. same is to be done till the string is broken.

Learn more about limiting force here:- brainly.com/question/11371672

#SPJ1

6 0
1 year ago
Which of the following is NOT a way that machines provide a mechanical advantage?
Nonamiya [84]
The following choice that is NOT a way that machines provides a mechanical advantage is to change direction. The correct answer is A. 
4 0
3 years ago
Read 2 more answers
Is it better to wire a house using a series circuit or a parallel circuit?
Svetach [21]

Answer: its better to use parallel because, in parallel connection there will be more advantages than a series connection. and also the electronic devices are wired in series so thats why you should use parralel in house wiring

so its c.

because the parallel is wired through the whole house so if one of the circuits fail youŕe not screwed

Explanation:

5 0
3 years ago
1. A toy car with mass m1 travels to the right on a frictionless track with a speed of 3 m/s. A second toy car with mass m2 trav
Alika [10]

Answer:

v = 3(m1 - 2m2)/(m1 + m2)

Explanation:

Parameters given:

Velocity of first toy car with mass m1, u1 = 3 m/s (taking the right direction as the positive axis)

Velocity of second toy car with mass m2, u2 = -6 m/s (taking the left direction as the negative x axis)

Using conservation of momentum principle:

Total initial momentum = Total final momentum

m1*u1 + m2*u2 = m1*v1 + m2*v2

Since they stick together after collision, they have the same final velocity.

m1*3 + (m2 * -6) = m1*v + m2*v

3m1 - 6m2 = (m1 + m2)v

v = (3m1 - 6m2) / (m1 + m2)

v = 3(m1 - 2m2) / (m1 + m2)

8 0
3 years ago
Read 2 more answers
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.9537 N when separated by
scoray [572]

Answer:

<em>The initial charges on the spheres are  </em>6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

\displaystyle F=\frac{K\ q_1\ q_2}{d^2}

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537

Rearranging

\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}

Solving for q1.q2

\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

\displaystyle q=\frac{q_1+q_2}{2}

Using this second condition:

\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756

\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C

Solving for q2

\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1

Replacing in [1]

\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}

Rearranging, we have a second-degree equation for q1.  

\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0

Solving, we have two possible solutions

\displaystyle q_1=6,796.10^{-6}\ c

\displaystyle q_1=-3.898.10^{-6}\ c

Which yields to two solutions for q2

\displaystyle q_2=-3.898.10^{-6}\ c

\displaystyle q_2=6.796.10^{-6}\ c

Regardless of their order, the initial charges on the spheres are 6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

8 0
3 years ago
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