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stiv31 [10]
3 years ago
13

A technician fills a tank with a liquid to a height of 0.20 m. The tank is cylindrical with radius 0.10 m. The mass of the liqui

d is 1.0 kg. What is the density of the liquid in 160 kg/m3
Physics
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

159.2 kg/m^3

Explanation:

The mass of the liquid is

m = 1.0 kg

The volume of the cylindrical tank is given by

V=\pi r^2 h

where

r = 0.10 m is the radius

h = 0.20 m is the heigth

Substituting,

V=\pi (0.10 m)^2 (0.20 m)=6.28\cdot 10^{-3}m^3

So now we can find the density of the liquid:

\rho = \frac{m}{V}=\frac{1.0 kg}{6.28\cdot 10^{-3} m^3}=159.2 kg/m^3

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Read the scenario.
Fofino [41]

Answer:

displacement = 2 m west

Explanation:

The displacement of an object is a vector connecting the final point of the motion of the object to the initial point, and its magnitude is equal to the length of the vector. So the magnitude of the displacement is basically the distance (measured in a straight line) between the final point and the starting point.

In this problem, we have:

- initial position of the acorn: 0 m

- final position of the acorn: 2 m west

So, the displacement has a magnitude of

d = 2 m - 0 m = 2 m

And the direction is west, since the final position is west compared to the initial position.

5 0
4 years ago
A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
3 years ago
The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?
Strike441 [17]

Answer:

Option B. N2(g) + 3H2(g) → 2NH3(g)

Explanation:

When nitrogen react with hydrogen, they form a product as shown below:

N2+ H2 → NH3

We need to balance the equation. This is illustrated below:

There are 2 atoms of nitrogen on the left side and 1 atom on the right side. To balance it, put 2 in front of NH3 as shown below:

N2+ H2 → 2NH3

Now, There are a total of 6 atoms of Hydrogen on the right side and 2 atoms on the left side side. This can be balanced by putting 3 in front of H2 as shown below:

N2+ 3H2 → 2NH3

Now we see clearly that the equation is balanced as we have equal numbers of atoms of N and H on both sides of the equation

4 0
3 years ago
Read 2 more answers
A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what
Fudgin [204]

Answer:

v = 40 [m/s].

Explanation:

Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

P=m*v\\

where:

m = mass [kg]

v = velocity [m/s]

P =20*10\\P =200 [kg*m/s]

Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.

v = P/m\\v = 200/5\\v = 40 [m/s]

3 0
3 years ago
CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!
BARSIC [14]
<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.  

Now, in this case we are on Earth, so the gravity value is 9.8\frac{m}{s^{2}}  

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely  true on Earth, unless the following condition is fulfiled:

If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at 9.8\frac{m}{s^{2}} and hit the ground at approximately the same time.  


5 0
4 years ago
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