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Lelu [443]
4 years ago
9

The density of an object is defined as its mass divided by its volume. Suppose the mass and volume of a rock are measured to be

8 g and 2.8325 cm3.
Physics
2 answers:
Lunna [17]4 years ago
5 0

Answer:

2824.36011 kg/m³

Explanation:

The missing statement is: <em>determine the rocks density in kg/m³</em>

- Data

mass of the rock, m = 8 g

volume of the rock, V = 2.8325 cm³

- Unit conversions

1000 g are equivalent to 1 kg, then 8 g are equivalent to:

1000 g / 8 g = 1 kg / x kg

x = 8/1000 = 0.008 kg

1,000,000 cm³ are equivalent to 1 m³, then 2.8325 cm³ are equivalent to:

1,000,000 cm³ / 2.8325 cm³ = 1 m³ / x m³

x = 2.8325/1,000,000 = 0.0000028325 m³

- Density calculation

Density = m/V

Density = 0.008/0.0000028325

Density = 2824.36011 kg/m³

kicyunya [14]4 years ago
3 0

Answer:

Density of the rock = 2.82436 g/cm³

Explanation:

We are given that:

density = \frac{mass}{volume}

We are also given that:

mass of rock = 8 grams

volume of rock = 2.8325 cm³

Substitute with the given values in the equation to get the density as follows:

Density = \frac{8}{2.8325} = 2.82436 g/cm³

Hope this helps :)

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the young’s modulus of aluminum is 69gpa, of nylon is 3gpa, of tungsten is 400gpa, and of copper is 117gpa. if equal-size sample
olga_2 [115]

Answer:

Least to most elongated: tungsten, copper, aluminum, nylon.

Explanation:

Materials with high Young's modulus are difficult to stretch. σ = Yε and ε = ΔL/L so an object with a high Young's modulus (Y) subject to a certain tensile stress (σ) will have a smaller strain than an object with a smaller Young's 's modulus subject to the same tensile stress. If strain (ε) is smaller, then ΔL will also be smaller.

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2 years ago
a force is applied to an object at rest with a mass of 100 kg. the same force is applied to an object at rest with a mass of 1 k
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4 years ago
The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start
Yanka [14]

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction \mu _s=0.24

Coefficient of kinetic friction \mu _k=0.22

mass of crate m=20\ kg

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maximum static Friction F_s=\mu _sN

N=mg

F_s=0.24\times 20\times 9.8

F_s=47.04\ N

thus applied force is greater than Static friction therefore kinetic friction will come into play

F_k=\mu _kN

F_k=0.22\times 20\times 9.8=43.12\ N

net Force on crate F-F_k=ma

a=\frac{200-43.12}{20}=7.84\ m/s^2

Magnitude of velocity can be obtained by using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

v^2-0=2\times 7.84\times 2

v=\sqrt{31.37}

v=5.60\ m/s                                          

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4 years ago
A mass is suspended from the roof of a lift (elevator) by means of a spring balance. The lift (elevator) is
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8 0
3 years ago
A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distan
OleMash [197]

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Speed of the car, v = 100 km/h = 27.77 m/s

The coefficient of kinetic friction of the tires, \mu_k=0.5

Let f is the net force acting on the body due to frictional force, such that,

-f=ma

a=\dfrac{-f}{m}

a=\dfrac{-\mu _k mg}{m}

a=\mu_k g

a=-0.5\times 9.8

a=-4.9\ m/s^2

We know that the acceleration of the car in calculus is given by :

v.dv=a.dx, x is the stopping distance

\int\limits^0_v {v.dv}=\int\limits^x_0 {a.dx}

\dfrac{v^2}{2}|_v^0=ax

0-(27.77)^2=-2\times 4.9x

On solving the above equation, we get, x = 78.69 meters

So, the stopping distance for the car is 78.69 meters. Hence, this is the required solution.

6 0
3 years ago
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