All categories

2 years ago

DontBlink1196 is *HEDJL:::A:APSKDL

Physics

1 answer:

agasfer [191]2 years ago

6 0

__________$$$$$$

_________$$____$$

_________$$$__$$$

_________$$_$$_$$

_________$$____$$

_________$$____$$$$$$$

___$$$$$$$$____$$____$$

_$$$$$___$$____$$____$$$$$$

$$$_$$___$$____$$____$$___$$

$$__$$___$$____$$____$$___$$

$$__$$___$$____$$___$$$___$$

$$__$$___$$____$$____$____$$

$$____$$$__$$$$__$$$$___$_$$

$$________________________$$

_$$_______________________$$

_$$$_____________________$$

__$$$$_________________$$$

____$$_________________$

____$$$_______________$$

You might be interested in

Which group has the Same electron shells? A Titanium and Gold B Selenium and Tin C Calcium and Magnesium D Lithium and Boron

Anna11 [10]

im not 100 percent sure but i think its b

6 0

3 years ago

liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i

aksik [14]

$4.80 \times 10^3 \text { seconds }$ long does it take to boil away 2.40 kg of the liquid.

Boiling point of He is $T=4.2 \mathrm{k}$

Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

#SPJ4

3 0

1 year ago

At 20 C, a steel rod of length 40.000 cm and a brass rod

balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

Step 1: identify the given parameters and standard parameters

⇒Steel rod: length of rod = 40.000 cm

Young modulus(Υ) = 20 X 10¹⁰ pa

coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

stress in the rod =?

⇒Brass rod: length of rod = 30.000 cm

Young modulus(Υ) = 9 X 10¹⁰ pa

coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

stress in the rod =?

--------------------------------------------------------------------------------------------

Step 2: calculate the stress in each rod

⇒Steel rod: stress in the rod = -Y*α*ΔT

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

= -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

= -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

Step 3: calculate the new length of each rod

⇒Steel rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

New length = ΔL + L₀ = 0.0192cm + 40cm

New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

ΔL = 2400 X 10⁻⁵ cm = 0.024cm

New length = ΔL + L₀ = 0.024cm + 30cm

New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0

2 years ago

Find the potential energy of a 2 kg ball 15 m in the air.

mart [117]

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

8 0

3 years ago

I need help on the data section of the circuit design lab on Edg.

Arte-miy333 [17]

I hope it's not too late, but here you go

8 0

3 years ago