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Hoochie [10]
2 years ago
13

9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large

tank measuring 2 m x 1 m x 20 cm. Density of water is 1000 kg/m³ or 1.0 g/cm³. 3. Find the volume of a lump of softwood whose mass is 120 g. Density of softwood is 0.6 gcm-3 or 600 kgm-³. -3​
Physics
1 answer:
My name is Ann [436]2 years ago
4 0

Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. density=\frac{mass}{volume}

1000=\frac{mass}{2 * 1 * 0.2}

1000*0.4=mass

400kg = mass

3. density=\frac{mass}{volume}

0.6=\frac{120}{volume}

volume=\frac{120}{0.6}

volume= 200cm

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3. Label the parts of each wave.<br> Please help me!!!
lisabon 2012 [21]

Answer:

D is the wavelength

A is the crest

C is amplitude

B is trough

G is wavelength

H is compression

I is rarefaction

4 0
3 years ago
The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const
polet [3.4K]

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

Power = \frac{Energy}{time}

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, s = ut + 0.5at^{2}

The drill head starts from rest, u = 0 m/s

5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

6 0
3 years ago
2. Using examples to describe each term, differentiate
melisa1 [442]

Answer: find the answer in the explanation

Explanation:

Average velocity is the average value of magnitude of initial velocity and final velocity.

If U = initial velocity and V = final velocity, then average velocity can be expressed as

Average velocity = ( U + V )/2

A vehicle who takes 60 minutes to cover 30 miles north and then 30 miles south and end up at the same place, has an average speed of 60 miles divided by 60 minutes, or 1 mile per minute.

Instantaneous velocity is the distance covered in a specific direction per time taken. Instantaneous velocity can be expressed as

Velocity = displacement/ time.

Uniform velocity occurs when we have a constant velocity. That is, when velocity does not change. When a vehicle travels in equal distances in equal intervals of time.

7 0
3 years ago
PLEASSEE HELP MEEE WITHH THESEEE
DanielleElmas [232]

Answer:

Q1: (a)

1cm \:  = 10 {}^{ - 2}m

so

45 \times  {10}^{ - 2} m

Q2:

1km = 10 {}^{3} m \\ 6378km \times  {10}^{ 3}

this is the radius in meter

diameter = 2r

in this i don't know which choice is the right answer

Q3:

1gram = \:  {10}^{ - 3} kg

so

250 \times  {10}^{ - 3}  = 2.5 \times  {10}^{ - 1} kg

again there is no correct answer

7 0
3 years ago
A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17
Korolek [52]

<span>Given:

Mass of cart: 15kg</span>

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

 

So, the solution would be like this for this specific problem:

 

<span><span>1)    </span>W(by applied force) = F(applied) x s x cosθ <span>
=>W(a) = 12 x 14 x cos17* = 160.66 J </span></span>
<span><span>
2)    </span>By F(net) = F(applied) - F(friction) <span>
=>As v = constant => a = 0 => F(net) = 0 
=>F(applied) = F(friction) 
<span>=>W(friction) = -F(friction) x s
=>W(friction) = -F(applied) x s 
=>W(f) = -12 x 14 x cos17* = -160.56 J </span></span></span>
<span><span>
3)    </span>0, as the displacement is perpendicular to Force </span>
<span><span>
4)    </span>0, as the displacement is perpendicular to Force</span>  
<span>
To add, the force that is applied<span> to an object by a person or another object is called the applied force.</span></span>
8 0
3 years ago
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