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3 years ago

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A gas in a sealed container has a pressure of 50 kPa at 27°C. What will the pressure of the gas be if the temperature rises to 8

7°C?

1 answer:

alexgriva [62]3 years ago

4 0

Answer:

the final pressure of the gas is 60 kPa.

Explanation:

Given;

initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa

initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k

final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K

Let the final pressure of the gas = P₂

Apply pressure law;

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} = \frac{50,000 \times 360}{300} = 60,000 \ Pa = 60 \ kPa$

Therefore, the final pressure of the gas is 60 kPa.

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The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.

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3 years ago

Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.

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Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

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2 years ago

How can you tell the difference between and erosion feature and a deposition feature?

Lunna [17]

Deposition moves that have already been eroded, and builds up sediment.
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3 years ago

A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together

Inga [223]

Answer:

The value is $\mu = 0.72$

Explanation:

From the question we are told that

The mass of the first car is $m_1 = 1870\ kg$

the initial speed of the car is $u = 13.5 \ m/s$

The mass of the second car is $m_2 = 2970\ kg$

The distance move by both cars is s = 1.93 m

Generally from the law of momentum conservation

$m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f$

Here $u_2 = 0$ because the second car is at rest

and $v_f$ is the final velocity of the the two car

So

$1870* 13.5+ 0= ( 1870 + 2970 ) * v_f$

=> $v_f = 5.22\ m/s$

Generally from kinematic equation

$v_f^2 = u_2^2 + 2as$

here a is the deceleration

So

$5.22^2 = 0 + 2 *a * 1.93$

=> $a = 7.06 \ m/s^2$

Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as

$F_f = F$

Here F is mathematically represented as

$F = (m_1 + m_2) * a$

$F = (1870 + 2970) * 7.06$

$F =34170.4 \ N$

and

$F_f = \mu * (m_1 + m_2 ) * g$

$F_f = 47432 * \mu$

So

$47432 * \mu = 34170.4$

=> $47432 * \mu = 34170.4$

=> $\mu = 0.72$

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