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2 years ago

i need help please. this is for physics but everything i search for related to this comes up as chemistry

Physics

1 answer:

Annette [7]2 years ago

7 0

The car tyre contains air initially at a pressure of 195 kPa after travelling several km the temperature of the air inside a car tyre rises from 30 to 70°C if the tyre is rigid and does not expand then the new pressure inside the tyre would be 220.74 kPa.

<h3>What is pressure?</h3>

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

By using the Charles law for gases which states that the volume of the gas remains constant then the pressure of the gas is directly proportional to the temperature.

As given in the problem the tyre is rigid and does not expand this means the volume of the tyre remains constant.

The mathematical expression for Charles's law is as follows

P₁/P₂ = T₁/T₂

First, we have to change the temperature from degree Celcius to the kelvin temperature scale

K = 273 + C

where k is the temperature in kelvin and the C is degrees of Celcius

Initially, the temperature was 30° C

T₁ = 273 + 30

T₁ = 303 K

Then after travelling the temperature of the air inside a car tyre rises from 30 to 70°C

T₂= 273+ 70

T₂ =343 K

The car tyre contains air initially at a pressure of 195 KPa

P₁ = 195 kPa

Lets us take the final pressure of the air would be P₂

By substituting the values in the formula

P₁/P₂ = T₁/T₂

195/P₂ = 303/343

P₂ = 220.74 kPa

Thus, the new pressure inside the tyre would be 220.74 kPa.

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AfilCa [17]

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We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

$v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s$

2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2) v$

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

$\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s$

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The impulse exerted on the ball is equal to its change in momentum:

$I=\Delta p$ (1)

The impulse can be rewritten as product between force and time of collision:

$I=F \Delta t$

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

$\Delta p = m\Delta v = m(v_f -v_i)$

So, eq.(1) becomes

$F \Delta t = m(v_f -v_i)$

where:

F = ? is the unknown force

$\Delta t = 0.002 s$ is the duration of the impact

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$v_f = 44 m/s$ is the final velocity of the ball

$v_i = -38 m/s$ is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

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lord [1]

Answer:

4.53482 m/s

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Explanation:

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The linear momentum of the system is conserved

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