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Brrunno [24]
3 years ago
8

A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦

Physics
1 answer:
bija089 [108]3 years ago
5 0
At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
pV=nRT
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and R=0.082 atm L K^{-1} mol^{-1} the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
n= \frac{pV}{RT}= \frac{(1.00atm)(20.0L)}{(0.082 LatmK^{-1}mol^{-1})(273 K)}=0.89 mol
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What do you mean by reflection of sound​
Bogdan [553]

Answer:

Reflection of sound waves also leads to echoes. Echoes are different than reverberations. Echoes occur when a reflected sound wave reaches the ear more than 0.1 seconds after the original sound wave was heard. ... Reflection of sound waves off of curved surfaces leads to a more interesting phenomenon.

8 0
3 years ago
a cliff diver drops from rest to the water below. how many seconds does it take for the diver to go from 0 to 60. that is to go
aleksandr82 [10.1K]
U = 0, initial vertical velocity
v = 60 mph = 88 ft/s

Ignore air resistance and take g = 32 ft/s².
It t = time to attain 60 mph, then
(88 ft/s) = (32 ft/s²)*(t s)
t = 88/32 = 2.75 s

Answer: 2.75 s
6 0
3 years ago
5. If 100N effort is required to lift a load of 300N in a lever of 1m and distance between fulcrums to load is 30 cm then, calcu
pishuonlain [190]

effort distance =?

effort =100N

load =300N

lever =1m

load D =30cm

ma=?

vr=?

efficiency =?

we know that

effort distance =l ×ld

= 300×30

9000

ma= l ×e

= 100×300

= 30000

vr = ld ×ed

= 1×9000

=9000

efficiency =(ma÷vr)/100

(30000÷9000)×100

3.33×100

333

5 0
3 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
(Write the answer in fair test) You have to investigate whether surface area affects how fast some water evaporates
Damm [24]

Answer:

Evaporation increases with an increase in the surface area

If the surface area is increased, then the amount is of liquid that is exposed to air is larger. More molecules can escape with a wider surface area.

Explanation:

7 0
3 years ago
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