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3 years ago

You make the following measurements of an object 42kg and 22m3 what would the objects density be

2 answers:

4 0

The density of the object is approximately 1.91 kg per m³.

42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship $density = mass / volume$ to solve for the object's density.

42 kg $\div$ 22 m³ $\approx$ 1.91 kg per m³.

sammy [17]3 years ago

4 0

<h2>Answer: </h2>

Density = 1.909 kg/m3

<h3>Explanation: </h3>

Given data:

Mass of the object = 42 kg

Volume of the object = 22 m3

Density of the object = ?

solution

Density = mass/volume

Density =42/22 = 1.909

Density = 1.909 kg/m3

Hence the density of the object is 1.909 and its unit is kg/m3 which is system international unit of density.

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A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made

anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

$B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }$ ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

$B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }$

B = 6.99 x 10⁻⁶ T

3 0

2 years ago

She left the cubes in the water for three hours which of the following describes a heat flow that took place during those three

quester [9]

Answer:

veffvevfevve

Explanation:

3 0

2 years ago

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The distance from the Earth to the Sun is 92 868 000 miles.

ANEK [815]

Your answer can be either 92 900 000 or 9.29e+7

6 0

2 years ago

Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

$L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K$

The view factor between two coaxial parallel disks would be

$\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33$

$\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75$

Then the view factor between base to top surface of the cylinder becomes $F_{12} = 0.26$ . From the summation rule

$F_{13} = 1-0.26$

$F_{13} = 0.74$

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

$\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}$

$\dot{Q_3} = 2\dot{Q_{13}}$

$\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)$

$\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)$

$\dot{Q_3} = 780.76W$

Therefore the rate heat radiation is 780.76W

5 0

2 years ago

Summarize Newton's three laws of motion

Sedaia [141]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

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2 years ago