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Alexus [3.1K]
3 years ago
6

Plzzzzz help me swear will mark u brainiest

Physics
2 answers:
GarryVolchara [31]3 years ago
6 0
Vroom vroom b.............
photoshop1234 [79]3 years ago
4 0

Answer:

B

Explanation:

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7. An object can be accelerating even if its speed is
steposvetlana [31]

Answer:

I think that the answer might be d

4 0
3 years ago
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Consider helium at 350 K and 0.75 m3/kg. Using Eq. 12-3, determine the change in pressure corresponding to an increase of (a) 1
Alex777 [14]

Answer:

a) (dP)_{v} = 9.692 kPa

b) (dP)_{T} = -9.692 kPa

c) dP = 0 Pa

Explanation:

The specifies equation is :

dz = (\frac{\delta z}{\delta x}) _{y} dx + (\frac{\delta z}{\delta y}) _{x} dy

Note that:

dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV

1% increase in temperature at specific volume:

dT = \frac{0.01}{1} *350\\dT = 3.5 K

a) Change in pressure of helium at constant volume:

(dP)_{v} = \frac{R}{v} dT

R = 2.0769 kJ/kg-K

dT = 3.5 K

v = 0.75 m³/kg

(dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa

b)

dv = (1%/100%) *0.75

dv = 0.0075 m³/kg

Change in pressure of helium at constant temperature:

(dP)_{T} = \frac{-RT}{v^{2} } dv

R = 2.0769 kJ/kg-K

T = 350 K

v = 0.75 m³/kg

dv = 0.0075 m³/kg

(dP)_{T} = \frac{-(2.0769*350)}{0.75^{2} } *0.0075\\(dP)_{T} = -9.692 kPa

c) The change in pressure of helium :

dP = (dP)_{v} + (dP)_{T}

dP = 9.692 - 9.692

dP = 0

5 0
4 years ago
List and describe the four phases of matter. Give a common example of each.
Marrrta [24]
Solid
example book
liquid
example water
gas
example helium or hydrogen
plasma
eg. florescent lights or plasma TVs
4 0
4 years ago
How far will a car travel it is traveling at 60 mph for 2 hours
Molodets [167]
120 miles because of a car is going at 60 miles per 1 hour, 2 hours would result in 120 miles
3 0
3 years ago
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laser produces a coherent beam of light that does not spread (diffract) as much in comparison to light from other sources, like
Fiesta28 [93]

Answer:

The diameter of the aperture is 0.321 m

Explanation:

The expression for the angle with wavelength is equal to:

\theta _{R} =sin^{-1} (\frac{1.22\lambda }{d} ) (eq. 1)

Where

λ = wavelength

d = diameter of the lens

The expression for the angle of the satellite is:

\theta _{m} =tan^{-1} (\frac{delta-x}{h} ) (eq. 2)

Where

h = distance between the Earth and the Moon

Like small angles:

sin\theta _{R} =\theta _{R}  \\tan\theta _{m} =\theta _{m}

Matching both equations and clearing Δx:

delta-x=h(\frac{1.22\lambda  }{d} ) (eq. 3)

Where

λ = 633 nm = 633x10⁻⁹m

For the width of the central maxima is equal:

w=2*delta-x\\delta-x=\frac{w}{2}

Where

w = 1.85 km

Replacing:

delta-x=\frac{1.85}{2} = 0.925km

From equation 3, the diameter of the aperture is:

d=h(\frac{1.22\lambda }{delta-x} )

h = 3.84x10⁵km = distance between the Earth and the Moon

Replacing:

d=3.84x10^{5}km *(\frac{(1.22)*(633x10^{-9}m) }{0.925km} )=0.321m

3 0
3 years ago
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