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2 years ago

14

A balloon was filled to a volume of 2.50 l when the temperature was 30.0∘c. What would the volume become if the temperature drop

ped to 11.0∘c.

1 answer:

Scilla [17]2 years ago

4 0

Answer:

2.34 L

Explanation:

Assuming the pressure inside the balloon remains constant, then we can use Charle's law, which states that for a gas kept at constant pressure, the ratio between the volume of the gas and its temperature remainst constant:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have:

$V_1 = 2.50 L$ is the initial volume

$V_2$ is the final volume

$T_1 = 30.0^{\circ}C+273 = 303 K$ is the initial temperature

$T_2 = 11.0^{\circ}C+273 = 284 K$ is the final temperature

Substituting into the equation and solving for V2, we find the final volume:

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(2.50 L)(284 K)}{303 K}=2.34 L$

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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

Blizzard [7]

<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity $V_{esc}$ is given by the following equation:

$V_{esc}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M$ is the mass of the asteroid

$R$ is the radius of the asteroid

On the other hand, we know the density of the asteroid is $\rho=3.84(10)^{8}g/m^{3}$ and its volume is $V=2.17(10)^{12}m^{3}$ .

The density of a body is given by:

$\rho=\frac{M}{V}$ (2)

Finding $M$ :

$M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})$ (3)

$M=8.33(10)^{20}g=8.33(10)^{17}kg$ (4) This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

$V=\frac{4}{3}\piR^{3}$ (5)

Finding $R$ :

$R=\sqrt[3]{\frac{3V}{4\pi}}$ (6)

$R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}$ (7)

$R=8031.38m$ (8) This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

$V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}$ (9)

Finally:

$V_{esc}=117.626m/s$ This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0

2 years ago

Calculate the force between charges of 5.0 x 10^-8 c and 1.0 x 10^-7 if they are 5.0 feet apart

weqwewe [10]

Answer:

F = 19.375 x 10^-6 N

Explanation:

This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.

It is defined as

F = (ke * q1 * q2)/ r^2

Where,

ke = is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2

q1 = 5.0 x 10^-8 C

q2 = 1.0 x 10^-7 C

r = 5 ft = 1,524 m

F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F = 19.375 x 10^-6 N

6 0

2 years ago

a submarine moving directly upward in the water at constant speed. The weight of the summer and it is 500,000 N. The submarines

Alex787 [66]

The answe would be A

3 0

2 years ago

The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.3x2

ki77a [65]

Answer:

F = 1.65 10⁸ N

Explanation:

In this pressure problem we have to use the definition of pressure

P = dF / dA

dF = P dA

we already have the expression for force and the pressure in a liquid is

P = Po + rho g (H-y)

Where Po is the atmospheric pressure acting on both sides of the dam, whereby its contribution is canceled and (H-y) is the distance from the surface

Let's look for an expression for the area differential

A = xy

dA = dx dy

y = 0.3 x²

x = √(y/0.3)

Let's build our equation with these expressions and integrate between the initial limit where the height is measured from the bottom of the dam y = 0, x = 0 to the upper limit, let's call it H = 200m, x = RA y / 0.3 and F = 0

∫ dF = ∫ (ρ g (H-y) dx dy

-F = ρ g [∫∫ H dy dx - ∫∫ ydy dx]

F = ρ g [∫ H x dy - ∫ x2√2 / 2 dy

Let's evaluate between the limits of integration

F = ρ g [∫ (H (√y /√0.3) dy - ∫ y/0.3 1/2 dy

F = ρ g (H /√0.3 ∫ √y dy - 1 /0.6 ∫ y dy)

Let's do the second integral

F = ρ g (H/√0.3 $y^{3/2}$ ) 2/3 - 1/0.6 y2 / 2)

F = ρ g (2H/3√0.3 $y^{3/2}$ ) - 1/1.2 y2 )

We evaluate at the limits

Y = 0

Y = 38 m

F = 1000 9.8 (2 200/3√3 √38³ - 1/1.2 38²)

F = 9800 (18032 - 1203)

F = 1.65 10⁸ N

5 0

2 years ago

Consider a person standing in a room at 23°C. Determine the total rate of heat transfer from this person if the exposed surface

IRINA_888 [86]

Answer:

The total rate of heat transfer is 138.8894 W

Explanation:

Step 1: Given data

⇒ Temperature in the room Tr= 23° C

⇒ Exposed surface area As= 1.7 m²

⇒ Person's skin temperature Ts= 32 ° C

⇒ heat transfer coefficient h : 3.5 W/m²⋅K

⇒ Emissivity ε = 0.9

⇒ ( Stefan Boltzmann constant σ = 5.67×10^−8 W⋅m^−2⋅K^−4

Step 2: Calculate Qrad

Heat transfer Q = Qrad + Qconv (Thermal properties of people are assumed to be constant. The heat transfer will take place due to both convective and radiation heat loss)

⇒ Qrad = Heat transfer from the person to the surroundings due to radiation

Q = ε*σ*As * (Ts ^4 - Tr ^4 ) + h* As*ΔT

⇒ Qrad =ε*σ*As * (Ts ^4 - Tr ^4)

Qrad = 0.9 * (5.7 * 10 ^ -8) * 1.7 * ((32 + 273.15)^4 -(18 +273.15)^4)

Qrad = 85.3394 W

<u>Step 3</u>: Calculate Qconv

Qconv = The heat transfer from the person to the surroundings due to convection

Qconv = h* As*ΔT

⇒ Q conv = 3.5 W/m²⋅K * 1.7 m² * (32 - 23)

Qconv = 3.5 * 1.7 * 9 = 53.55 W

<u>Step 4: </u>Calculate Q

Q = Qrad + Qconv = 85.3394 + 53.55 = 138.8894 W

The total rate of heat transfer is 138.8894 W

5 0

2 years ago