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Delicious77 [7]
3 years ago
5

The air pressure inside the tube of a car tire is 430 kPa at a temperature of 13.0 °C. What is the pressure of the air, if the t

emperature of the tire increases to 59.5 °C? Assume that the volume of the tube doesn't change.
Physics
1 answer:
KengaRu [80]3 years ago
7 0

Answer:

The pressure of the air is 499.91 kPa.

Explanation:

Given that,

Initial pressure = 430 kPa

Temperature = 13.0+273=286 K

Final temperature = 59.5+273=332.5 K

We need to calculate the final pressure

Using relation of pressure and temperature

At constant volume,

\dfrac{P'}{P}=\dfrac{T'}{T}

\dfrac{P'}{430}=\dfrac{332.5}{286}

P'=\dfrac{332.5}{286}\times430

P'=499.91\ kPa

Hence,The pressure of the air is 499.91 kPa.

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The engine in your car is sometimes called: A. A 2-stroke engine
Juli2301 [7.4K]

Answer:A

Explanation:

Engines in car are 4 stroke engine . A 4-stroke engine is internal combustion engine which derives its power by four piston strokes . Internal combustion means combustion takes inside the engine i.e. is in cylinder.

There are process in 4 stroke engine

  • Intake: Intake of air
  • Compression:compression of intake air to a high pressure
  • Combustion:Fuel is injected and burned to get power
  • Exhaust:removal of exhaust gases after combustion    
7 0
3 years ago
The terms intrusive and extrusive are used to describe which one of the three rock groups
tino4ka555 [31]

Answer:

IGNEOUS ROCKS

Explanation: Igneous rocks are those rocks that solidify from magma.

Igneous rock is divided into two ,they are:

1. Intrusive

Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.

2 Extrusive igneous rock is also known as known as volcanic rocks

8 0
3 years ago
Birdman is flying horizontally at a
zavuch27 [327]

Answer:

X=92.49 m

Explanation:

Given that

u= 21 m/s

h= 97 m

Time taken to cover vertical distance h

h= 1/2 g t²

By putting the values

97 = 1/2 x 10 x  t²          ( g = 10 m/s²)

t= 4.4 s

The horizontal distance

X= u .t

X= 21 x 4.4

X=92.49 m

3 0
3 years ago
show the trajectory of a body projected with an initial velocity vat an angle of departure thita is a paraboler​
Jobisdone [24]

Answer:

Let me look up a couple of things regarding this question.

Explanation:

Then I will get back to you.

5 0
3 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
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