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nordsb [41]
2 years ago
12

What is the least count of screw gauge and vernier calliper (9th grade) please help! ​

Physics
1 answer:
nadezda [96]2 years ago
7 0
The least count is the smallest unit of measurement which an instrument can take accurately
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A body of mass 1.0 kg initially at rest slides down an incline plane that is 1.0 m high and 10.0 m long. If the body experiences
Ksivusya [100]

Answer:

answer is

Explanation:

because

6 0
3 years ago
a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav
AURORKA [14]
This next statement is a big deal.  It should be up on a board, surrounded
by flashing red and yellow lights, and hung on the wall of every Science
classroom.   Although we never see it in our daily lives, it's fundamental to
the workings of the universe, and it's also Newton's first law of motion:

<em>Without friction, it doesn't take <u>ANY</u> force to keep a moving object
moving.  </em>
<em>Force is only required to <u>change</u> the object's speed, or to
<u>change</u> the direction </em>
<em>in which it's moving.</em>

The answer to the question is:  On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
same speed.
7 0
3 years ago
Read 2 more answers
A 1,200 kg car travels at 20 m/s. what is it’s momentum ?
WINSTONCH [101]

Answer:c  vx v xv xv x

Explanation:

mlmcmcm

6 0
3 years ago
An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the
Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

(n1)*(sin(theta1))=(n2)*(sin(theta2))

Then we will find the refraction angle relative to the vertical this way:

(n1/n2)*(sin(theta1))=sin(theta2)

(1/1.33)*(sin(45))=sin(theta2)

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

tan(32.12)=(X/30cm)

X=(tan(32.12))*(30cm)

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

tan(32.12)=(24cm/Y)

Y=(24cm/tan(32.12))

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0
2 years ago
When a projectile reaches the highest point the vertical component of the acceleration is:
Mazyrski [523]

Answer:

The acceleration is g.

Taking the upward direction as positive

V = Vy y - 1/2 g t^2

Taking the downward direction as positive

V = -V y + 1/2 g t^2

One can choose either direction as positive, but the acceleration is

the same as g (it is g) while the projectile is in the air.

6 0
3 years ago
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