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lapo4ka [179]
3 years ago
7

When completely filled with water, the beaker and its contents have a total mass of 326.75 g. What volume does the beaker hold?

Use ????=1.00 g/mL as the density of water.
Physics
1 answer:
miv72 [106K]3 years ago
6 0

Answer:

0.003060 cm³

Explanation:

Mass of water = m = 326.75 gram

Density of water = ρ = 1.00 g/mL = 1 g/cm³

density = mass / volume

⇒volume = density / mass

⇒volume = 1 / 326.75

⇒Volume = 0.0030604 cm³

∴ Volume held by beaker = 0.0030604 cm³

Here the combined mass of the beaker and water is given so the volume found will be of the beaker as well as the liquid. But, it can be seen that the volume is so small that subtracting the beaker mass would have negligible effect.

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How long does it take light to travel between earth and the nearest star?
sergejj [24]

Answer:

It takes 8.204 minutes for light to travel from the Sun to Earth

Explanation:

The speed of light is 300000 km/s and the nearest star to the Earth is the Sun which is 149.67 million km. therefore it takes about

147.67×10⁶/300000  = 492.233 seconds for light from the Sun to reach Earth or 8.204 minutes

The other next nearest star is Alpha Centauri which is 4 light years away

5 0
3 years ago
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Quentin is a crime scene investigator. He believes a violent crime may have occurred in the bedroom of an apartment but he sees
Llana [10]

Answer:

Luminol

Explanation:

is a chemical that exhibits chemiluminescence, with a blue glow, when mixed with an appropriate oxidizing agent.

3 0
3 years ago
Keaton is asked to solve the following physics problem:
RideAnS [48]

Answer:

The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"

Explanation:

In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;

<em>1) v=u+at</em>

<em>2)v^2=u^2+2as</em>

<em>3)s=ut+(1/2)(at^2)</em>

the variables are explained below;

u= initial velocity of the body

a=acceleration/Speed of the body

t= time taken by the body while travelling

s= displacement of the body.

Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)

5 0
4 years ago
a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of c
NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}

With the following condition :

  • v = final velocity (m/s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

  • \sf{h_1} = initial height = 45.0 m
  • \sf{h_2} = final height = 10.0 m
  • g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

  • v = final velocity = ... m/s

Step by Step

\sf{v = \sqrt{2 \times g \times \Delta h}}

\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}

\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}

\sf{v = \sqrt{19.6 \times 35}}

\sf{v = \sqrt{686}}

\boxed{\sf{v \approx 26.2 \: m/s}}

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

<h3>See More :</h3>
  • The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625
5 0
2 years ago
A man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s2. What amount of force acted on the ball?
Delvig [45]
Newton taught us that    Force = (mass) x (acceleration)

Force  =  (0.2) x (20) = <em>4 newtons</em> .

Something to think about:  The ball can only accelerate while the club-face
is in contact with it.  Once the ball leaves the club, it can't accelerate any more,
because the force against it is gone.
4 0
3 years ago
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