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goldenfox [79]
2 years ago
8

What is the difference between liquid water and water vapor?

Physics
2 answers:
Nookie1986 [14]2 years ago
5 0

Answer:

D

Explanation:

they are different states of the same pure substance lmk if I'm right I'm 99% sure I am

Vlada [557]2 years ago
3 0
D is the answer honey
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A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
Consider the speed of light in another universe to be only 100 m/s. Two cars are traveling along a highway in opposite direction
olganol [36]

Answer:

-62.45m/s and +62.45m/s

Explanation:

The formula for relativistic speed

This is the speed of A with respect to B

V_{AB}=\frac{V_{A}-V_{B}}{1-\frac{V_{A}V_{B}}{C^2} }

where

V_{A} will be the velocity of person 1: 39m/s

V_{B} will be the velocity of person 2: -31m/s (negative because is travelling in opposite direction)

and C the velocity of light: 100m/s

The velocity of person 1 measured by person 2 is:

V_{AB}=\frac{39m/s-(-31m/s)}{1-\frac{(39m/s)(-31m/s)}{(100m/s)^2}}=62.45m/s

The velocity of person 2 measured by person 1 is:

V_{BA}=\frac{V_{B}-V_{A}}{1-\frac{V_{B}V_{A}}{C^2} }

V_{BA}=\frac{-31m/s-39m/s}{1-\frac{(-31m/s)(39m/s)}{(100)^2} }=-62.45m/s

8 0
3 years ago
Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the
Alex777 [14]
The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire. 

The magnetic field generated by wire 1 is:
B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}=  4 \cdot 10^{-5}T

The magnetic field generated by wire 2 is:
B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T

And so, the resultant magnetic field at the point midway between the two wires is
B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T
8 0
3 years ago
Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it
Oksanka [162]
Coffee mug is cylindrical shape. Therefore, 
Volume of mug = volume of cylinder = πr²h 

r = inner radius of mug. h = height of coffee = 6 cm. 
Density of coffee = Density of water = 1 g/ml 

Therefore, volume of coffee = mass/density = 400/1 = 400 ml = 400 cm³
Volume of coffee = πr²h
400= πr²(6) 
r² = 21.23
r = inner radius of mug = 4.607 cm
6 0
3 years ago
A proton traveling along the x-axis enters a region at x = 0 where the x-component of the electric field is given by E = Ao/x1/2
storchak [24]

.Answer:

The value of the work done is \bf{ 5.29 qA_{0}}.

Explanation:

When a charged particle having charge q is moving through an electric field E, the net force (F) on the charge is

F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

and the work done (W) by the particle is

W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Given, E = \dfrac{A_{0}}{x^{1/2}}.

Substitute the value of electric field in equation (1) and then substitute the result in equation (2).

W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}

7 0
3 years ago
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