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MAVERICK [17]
3 years ago
14

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?
Physics
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.
5 0
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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5 0
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

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=> 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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