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Svetllana [295]
3 years ago
6

What is the mass of an object Destiny is 4 grams / milliliters and its volume is 8 ml​

Physics
1 answer:
swat323 years ago
5 0

Answer: mass is 4 g/ ml ×8 ml

Explanation: density = mass/volume ρ = m/V so mass is density × volume

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Name and Title:
Natalka [10]

The variable that is altered by the independent variable is called the dependent variable. the spectrum given out by the hypothetical unknown celestial object.

Name : Christopher Andre

Instructor name : Albert brown

Name of the lab : ELS Laboratory

The term electromagnetic spectrum refers to the range of electromagnetic radiation's frequencies, as well as the wavelengths and photon energies connected to each frequency.

By analyzing the absorption spectra of the planets and moons, the electromagnetic spectrum experiment seeks to identify the components that make up their atmospheres.

From below one hertz to over 1025 hertz, electromagnetic waves are included in the electromagnetic spectrum.

The wavelengths that correlate to the frequency range from tens of thousands of kilometers to a small portion of the size of an atomic nucleus.

Starting at the low frequency (long wavelength) end of the spectrum, each frequency band's electromagnetic waves are referred to by a variety of names.

Hence the dependent variable is the one that changes as a result of the independent variable.

Learn more about electromagnetic spectrum here

brainly.com/question/13803241

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6 0
2 years ago
Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, tra
castortr0y [4]

Answer:

A. Its translational kinetic energy is larger than its rotational kinetic energy.

Explanation:

Given that

Radius = R

Mass = M

We know that mass moment of inertia for the solid sphere

I=\dfrac{2}{5}MR^2

Lets take angular speed =ω

Linear speed =V

Condition for pure rolling , V= ω R

Rotation energy ,RE

RE=\dfrac{1}{2}I\omega^2

RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2

RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2

RE=\dfrac{1}{5}\times MR^2\times \omega^2

RE=\dfrac{1}{5}\times MV^2

RE= 0.2  MV²

The transnational kinetic energy TE

TE=\dfrac{1}{2}MV^2

TE= 0.5 MV²

From above we can say that transnational energy is more than rotational energy.

Therefore the answer is A.

3 0
3 years ago
The transfer of energy when particles of a fluid move from one place to another is called
siniylev [52]

Answer:

Answer below!!!!

Explanation:

Convection is the transfer of thermal energy by particles moving through a fluid.

Hope I Helped!!!

;)

5 0
2 years ago
Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic
Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by \frac{1}{2}kx^{2} where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

                                                              ⇒\frac{1}{2}k(\frac{w}{k})^{2}

                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

4 0
3 years ago
A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t
Shalnov [3]

Answer:

a) yield strength

   \sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa

b) modulus of elasticity

strain calculation

\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046

strain for offset yield point

\varepsilon_{new} = \varepsilon_0 -0.002

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

 \sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa

d) percentage elongation

\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%

e) percentage of area reduction

\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%                            

7 0
3 years ago
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