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3 years ago

Which kind of electromagnetic waves has a wavelengths greater than 10 cm?

Physics

1 answer:

bonufazy [111]3 years ago

3 0

Radio waves
Hope this helps

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A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K).

mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

$P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa$

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

$P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa$

(c) Same steps as in part (a)

$P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa$

ΔP=9.164-9.14

ΔP=0.0245kPa

8 0

3 years ago

A person travels by car from one city to another with different constant speeds between pairs of cities. she drives for 50.0 min

liraira [26]

I think it's 45 miles. Don't know for sure though

3 0

3 years ago

Which kind of air front forms when a warm air mass meets the area of a cooler air mass? A. stationary front B. cold front C. war

zaharov [31]

This would be a cold front that forms when a warm air mass meets the area of a cooler air mass, although it should be noted that this doesn't always feel "cold".

3 0

3 years ago

Read 2 more answers

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

3 years ago

How do you relate mas and acceleration?

Shtirlitz [24]

Answer:

yes i relate mass but not acceleration

3 0

3 years ago