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3 years ago

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A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass of 1.56 kg. Calculate it

s density, specific weight and specific gravity.

1 answer:

Kazeer [188]3 years ago

3 0

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = $\frac{m}{V}$

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = $\frac{1,56Kg}{1,767145L}$ = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) / φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

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In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

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

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

3 years ago

When creating any map, what four basic elements should you include?

melisa1 [442]

Title,Scale,Date of Publication,North Arrow (Legend,Location Information,and Source of Information)

5 0

3 years ago

Read 2 more answers

A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

lisabon 2012 [21]

Answer:

$\alpha =10.93radian/sec^2$

Explanation:

We have given given the final angular velocity $\omega _{final}=13.5rad/sec$

And $\omega _{initial}=22rad/sec$

Displacement $\Theta =13.8radian$

We have to find the angular acceleration $\alpha$

According to law of motion $\omega _{final}^2=\omega _{initial}^2+2\alpha \Theta$

So $13.5^2=22^2+2\times \alpha \times 13.8$

$\alpha =-10.93radian/sec^2$

In question we have tell about magnitude only so $\alpha =10.93radian/sec^2$

4 0

3 years ago

A car travels 554 miles south in 10.4 hours . what is the velocity of the car ? show all steps

kirill115 [55]

Answer:

<h3>The answer is 53.27 mi/hr</h3>

Explanation:

To find the velocity covered by the car we use the formula

$v = \frac{d}{t} \\$

where

d is the distance

t is the time

From the question

d = 554 miles

t = 10.4 hrs

We have

$d = \frac{554}{10.4} \\ = 53.2692307...$

We have the final answer as

<h3>53.27 mi/hr</h3>

Hope this helps you

4 0

3 years ago

Two children (each having a mass of 60 kg) are standing on the edge a merry-go-round (mass of 140 kg) as it spins with an angula

poizon [28]

Answer:

The angular velocity after the children jump off is approximately 1.4 rad/s

Explanation:

The given parameters are;

The masses of each child, m₁, and m₂ = 60 kg

The mass of the merry-go-round, m₃ = 140 kg

The initial angular velocity, $\omega_i$ = 0.75 rad/s

The angular velocity after the children jump off = $\omega_f$

According to the principle of conservation of angular momentum

The angular momentum = I × ω

The moment of inertia, I = m × R²

The total initial angular momentum = $I_i \times \omega_i = m_i \times R^2 \times \omega_i$

The total angular momentum after the children jump off = $I_f \times \omega_f = m_f \times R^2 \times \omega_f$

The initial mass, $m_i$ = m₁ + m₂ + m₃ = 60 kg + 60 kg + 140 kg = 260 kg

The final mass, $m_f$ = m₃ = 140 kg

According to the principle of conservation of linear momentum, we have;

$I_i \times \omega_i$ = $I_f \times \omega_f$

Therefore;

260 kg × R² × 0.75 rad/s = 140 kg × R² × $\omega_f$

∴ $\omega _f$ = (260 kg × R² × 0.75 rad/s)/(140 kg × R²) = 1.39285714 rad/s. ≈ 1.4 rad/s

The angular velocity after the children jump off, $\omega _f$ ≈ 1.4 rad/s.

7 0

3 years ago