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3 years ago

Define relative density

Physics

1 answer:

weeeeeb [17]3 years ago

5 0

Hello There!

Relative density is another word for specific gravity. Relative density is comparing the density of one thing to the density of another. Most of the time, you look at the density of water so we look at the density of an object relative to the density of water.

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right

V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

g: Gravitational Acceleration

θ: Angle with the vertical

N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

m*g*cos(θ) - 0 = m*vf^2 / R

g*cos(θ) = vf^2 / R

vf^2 = R*g*cos(θ)

vf^2 = 1.45*32.2*cos(34)

vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

ΔK.E = ΔP.E

0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))

vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

vi^2 = 22.744

vi = 4.77 ft/s

4 0

3 years ago

In an isolated system, Bicycle 1 and Bicycle 2, each with a mass of 10 kg,

dimulka [17.4K]

Answer: 20 kgm/s

Explanation:

Given that M1 = M2 = 10kg

V1 = 5 m/s , V2 = 3 m/s

Since momentum is a vector quantity, the direction of the two object will be taken into consideration.

The magnitude of their combined

momentum before the crash will be:

M1V1 - M2V2

Substitute all the parameters into the formula

10 × 5 - 10 × 3

50 - 30

20 kgm/s

Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s

7 0

2 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

Rama09 [41]

Please elaborate more on your question so I can help you

7 0

2 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

2 years ago

You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in

Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

Volume = $\frac{\text{mass}}{\text{density}}$

So, V = $\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}$

= $5.181 \times 10^{-8} m^{3}$

Now, we will calculate the area as follows.

Area = $\frac{\text{volume}}{\text{length}}$

= $\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}$

= $2.15 \times 10^{-11} m^{2}$

Formula to calculate the resistance is as follows.

R = $\rho \frac{l}{A}$

= $\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}$

= $2.71 \times 10^{6} ohm$

Thus, we can conclude that the resistance of given wire is $2.71 \times 10^{6} ohm$ .

4 0

2 years ago

Define relative density​

Define relative density