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Damm [24]
3 years ago
13

Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este

0,17kl, să se calculeze:
a) densitatea lichidului;
b) greutatea lichidului.
Physics
1 answer:
Mariana [72]3 years ago
6 0

a) Density of the liquid: 823.5kg/m^3

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

<em>"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate: </em>

<em>a) the density of the liquid; </em>

<em>b) the weight of the liquid."</em>

a)

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

M=m_L + m_V (1)

where

m_L is the mass of the liquid

m_V is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

m_L=M-m_V=175-35=140 kg

The density of the liquid is given by

d=\frac{m}{V}

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = 0.170 m^3 (volume of the liquid, which is equal to the volume of the vessel)

So we get

d=\frac{140}{0.170}=823.5kg/m^3

b)

The weight of a body is given by

F=mg

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

g=9.8 m/s^2 (acceleration due to gravity)

Therefore, its weight is

F=(140)(9.8)=1372 N

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

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6. The image to the right shows a moment of inertia
Trava [24]

The moment of inertia is the rotational analog of mass, and it is given by

the  product of mass and the square of the distance from the axis.

  • The moment of inertia changes as the position of the weight is changed, which indicates that; statement is incorrect

Reasons:

The weight on each arm that have adjustable positions can be considered as point masses.

The moment of inertia of a point mass is <em>I</em> = m·r²

Where;

m = The mass of the weight

r = The distance (position) from the center to which the weight is adjusted

Therefore;

The moment of inertia, <em>I </em>∝ r²

Which gives;

Doubling the distance from the center of rotation, increases the moment of inertia by factor of 4.

Therefore, the statement contradicts the relationship between the radius of rotation and moment of inertia.

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7 0
2 years ago
A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to
nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to  the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

4 0
3 years ago
A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag
VARVARA [1.3K]

Answer:

Induced emf, \epsilon=9.79\times 10^7\ volts

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, \omega=3\ rev/s=18.84\ rad/s

The vertical component of the Earth’s magnetic field is, B=6.5\times 10^5\ T

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

\epsilon=\dfrac{1}{2}B\omega l^2

\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2

\epsilon=9.79\times 10^7\ volts

So, the induced emf between the tip of a blade and the hub is \epsilon=9.79\times 10^7\ volts. Hence, this is the required solution.

5 0
3 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)


8 0
3 years ago
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andrezito [222]

Answer:

i do not know

Explanation:

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3 years ago
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