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iren [92.7K]
2 years ago
8

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.
Physics
1 answer:
Kazeer [188]2 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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Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0
2 years ago
What statements accurately describe sunspots? Check all that apply.
polet [3.4K]

Answer:

sunspots are storms on the Suns surface

Sunspots are marked by intense magnetic activity

Sunspots produce solar flares and hot gassy ejections.

Sunspots can affect Earth’s climate.

Explanation:

I just did this lesson

6 0
2 years ago
Calculate the kinetic energy of a motorcycle of mass 60kg travelling at a velocity of 40km/h​
ELEN [110]

Answer:

1848.15J

Explanation:

KE =1/2 mv^2

Mass = 60kg, velocity =40km/h =11.11m/s

Hence

KE =30 x(11.1)^2 /2 = 1848.15J

5 0
2 years ago
One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has
vovangra [49]

Answer:

The mass of the cube is 420.8 kg.

Explanation:

Given that,

Length of edge = 38.9 cm

Density \rho= 7.15 \times10^{3}\ kg/m^3

We need to calculate the volume of cube

Using formula of volume

V = 38.9^3

V=0.058863\ m^3

We need to calculate the mass of the cube

Using formula of density

\rho = \dfrac{m}{V}

m = V\times\rho

m =0.058863\times7.15 \times10^{3}

m=420.8\ kg

Hence, The mass of the cube is 420.8 kg.

7 0
2 years ago
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
2 years ago
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