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Law Incorporation [45]
2 years ago
11

If a balloon is heated, what happens to the pressure of the air inside the balloon if the volume remains constant?.

Physics
1 answer:
Molodets [167]2 years ago
8 0

Answer:

P V = N R T    is the ideal gas equation

P2 / P1 = T2 / T1

The pressure will increase.

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Why is the nitrogen cycle considered a closed system?
Svetradugi [14.3K]

B nitrogen stays in order so it can't change movement

8 0
3 years ago
Read 2 more answers
A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of
Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

8 0
2 years ago
Newton
vodomira [7]

Given parameters:

Mass on earth  = 50kg

Unknown:

Mass on planet Xenon = ?

Weight on planet Xenon = ?

Mass is the amount of matter contained in a particular substance.

Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.

         Weight  = mass x acceleration due to gravity

Planet Xenon has half the gravitational force of Earth.

 This translated gives \frac{9.8}{2}   = 4.9m/s²

Now, mass is always the same every where if the amount of matter in a substance does not change.

In this problem, mass = 50kg on planet xenon.

Weight =  mass x acceleration due to gravity  = 50 x 4.9  = 245N

The weight on Xenon is 245N and the mass is 50kg

4 0
3 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel
BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

t=\frac{6.22}{2}=3.11 s

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

Solving the  formula, we find

s=\frac{1}{2}(9.8)(3.11)^2=47.4 m

7 0
3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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