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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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4 years ago
How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?
salantis [7]

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

Power = \frac {Voltage^{2}}{resistance}

Substituting into the formula, we have;

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3 years ago
When light is directed on a metal surface, the kinetic energies of the photoelectrons a) are random b) vary with the frequency o
jekas [21]

Answer:

b) vary with the frequency of the light

Explanation:

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h = Planck's constat

Then, we can see that K.E is proportional linearly to "v" in the equation above.

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5 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

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r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

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We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

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r_x=v_0t

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3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

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4 years ago
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4 years ago
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