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2 years ago

What is focusingHelp me please

Physics

1 answer:

sveticcg [70]2 years ago

4 0

Focusing means you’re paying particular attention to something

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Estimate the number of atoms in 1 cm^3 of a solid

dexar [7]

Avogadro's number: 6.02 x 10^23 atoms is present in 1mol of a solid (i.e. 22, 400 cm3)

Hence, in 1 cm3, 6.02 x 10^23 /22400 atoms is present = 2 x 10 ^ 19 atoms.

7 0

2 years ago

0.22 L of pancake syrup has a mass of 33 g.

katrin2010 [14]

Answer:

a. 150 g/L

b. 75 g

c. 120 mL

Explanation:

a. 33g/0.22L=150 g/L

b. 33g/0.22L=150 g/L

150 g/L*0.5L=75g

c. 0.22L/33g=0.006667L/g

0.006667L/g*18g=0.12L

0.12L*1000=120mL

6 0

2 years ago

1. What do you notice or wonder about ultrasound imaging? (black and white pictures)

a_sh-v [17]

Answer:

the shape how it involve into a picture

Explanation:

3 0

2 years ago

Two students make the following claims:

antiseptic1488 [7]

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

$F=m.g$

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

$W=F.s\cos\theta$

where:

$s=$ displacement of the object

$\theta=$ angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

In case of free-fall the angle between the force is and the displacement is zero.
In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

6 0

3 years ago

Calculate the constant acceleration a in g’s which the catapult of an aircraft carrier must provide to produce a launch velocity

goblinko [34]

The solution to the problem is as follows:

First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.

So now write down all the values you know:

Vfinal = 275.7 ft/s

Vinitial = 0 ft/s

distance = 299ft

Now just plug in Vf, Vi and d to solve

Vf^2 = Vi^2 + 2 a d

BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.

5 0

2 years ago

What is focusingHelp me please​

What is focusingHelp me please