Help me please i really need this one

I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

2 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

My name is Ann [436]

Answer:

a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C , c) U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

C = k e₀ A / d

Let's calculate

C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

C = 4,012 10⁻¹⁴ F

b) let's look the charge

C = Q / ΔV

Q = C ΔV

Q = 4,012 10⁻¹⁴ 400

Q = 1.6 10⁻¹¹ C

c) The stored energy

U = ½ C ΔV²

U = ½ 4,012 10⁻¹⁴ 400²

U = 3.21 10⁻¹¹ J

4 0

2 years ago

Which phrase desenbes an irregular galaxy ?

Andrews [41]

Answer:

contains many young stars

Explanation:

Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.

However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.

The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.

They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.

7 0

2 years ago

Read 2 more answers

Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t

Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N

Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.

8 0

2 years ago

please tell th answer fast suppose A ball of mass M is thrown vertically upward with initial speed be its speed is continuously

Lisa [10]

Answer:.

the ball would go down and speed of it would not strike so that wouldnt be an example of the conversation momentum

Explanation:

The answer I think

3 0

2 years ago