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kolezko [41]
3 years ago
12

If a gas at 25.0 °C occupies 5 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 4 atm?

Physics
1 answer:
inessss [21]3 years ago
8 0

Answer:

1.25L

Explanation:

Applying Gay-Lussac's Law which states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.

p_1v_1=p_2v_2\\\\5l\times 1.00atm=4.0atm\tiimes x\\\\x=5/4=1.25L

Hence the volume at 4.0atm is 1.25L

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A falling skydiver has a mass of 105 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resis
LenKa [72]
F-free = m*g - F_air = m*a

F_air = 1.2 * m

a= (105 kg * 9.8 m.s^2 - 5*105) / 105 kg

a = 9.3 m/s

Hope this helps
7 0
3 years ago
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si
solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0
3 years ago
Newton's first law equations like velocity and stuff like that
Klio2033 [76]

Answer:

Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force. ... An object sliding across a table or floor slows down due to the net force of friction acting on the object.

Explanation:

please give me a heart

4 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
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