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laila [671]
3 years ago
9

The magnification is always +ve for​

Physics
1 answer:
Nikitich [7]3 years ago
6 0

Please complete your question and then send

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Explain, using numbered steps, how you could find the volume of a small, irregularly shaped rock.
dmitriy555 [2]
<span>1. Get a graduated cylinder. 2. Fill the graduated cylinder to a known amount of water. Record the amount of water in the cylinder. 3. Place rock into the graduated cylinder 4. Measure the new volume of the graduated cylinder with the rock in it. 5. Take the difference of the new volume and the old volume and that is the volume of the rock.</span>
8 0
3 years ago
Atmospheric pressure and density at the top of Mount Everest ​
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The summit of Mount Everest has an average pressure around 30 kPa. ... A barometer also measures variations in atmospheric pressure. As altitude increases, the air becomes thinner, the density of air decreases, and the pressure of the air decreases as well.

3 0
4 years ago
The air speed of a plane is defined as its velocity with respect to the surrounding air, or in other words how fast the plane wo
amid [387]

Answer:

3

Explanation:

because

7 0
3 years ago
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

<h3>45 J</h3>

Hope this helps you

7 0
3 years ago
a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
4 years ago
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