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Leokris [45]
2 years ago
7

An automobile tire is filled to a pressure of 240.0 kPa early in the morning when the temperature is 15.0C. After the car is dri

ven all day over hot roads, the tire temperature is 18.0C. Estimate the new pressure.
Physics
1 answer:
yuradex [85]2 years ago
5 0

Answer:

<h2> 242.5kPa</h2>

Explanation:

According to one of the gas laws, \frac{P1}{T1} =\frac{P2}{T2}

Given P1 = 240.0kPa, T1, 15.0°C, T2 = 18.0°C, P2 = ?

Substituting this values into the equation, we have;

\frac{240}{15+273} =\frac{P2}{18+273}\\\frac{240}{288} =\frac{P2}{291}\\

Cross multiplying we have;

288P2 = 240*291

P2 = \frac{240*291}{288} \\P2 = 242.5kPa

The new pressure is 242.5kPa

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
Please re-explain the following phrases in terms of momentum
zheka24 [161]
A. An object in motion will remain in motion unless acted upon by an outside force : The momentum of an object is constant unless an outside force acts on the object.

B. Force is defined as mass times acceleration : the rate of change of the momentum of a particle is proportional to the force F acting on it, hence the force is equal to <span>mass times acceleration.

C. </span>For every action there is an equal and opposite reaction : <span>to every action force there is an equal and opposite reaction force. </span>
8 0
2 years ago
A 5kg bucket hangs from a ceiling on a rope. A student attaches a spring scale to the buckets handle and pulls horizontally on t
7nadin3 [17]
I don’t know what the angle is in your diagram so I used the angle from the vertical.

6 0
2 years ago
At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature
Margaret [11]

The molar latent enthalpy of boiling of iron at 3330 K is  ΔH = 342 \times 10^3 J.

<u>Explanation:</u>

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

                      d ln p = (ΔH / RT^2) dt

                   (1/p) dp = (ΔH / RT^2) dt

                    dp / dt = p (ΔH / RT^2) = 3.72 \times 10^-3

                  (p) (ΔH) / (8.31) (3330)^2 = 3.72 \times 10^-3

                          ΔH = 342 \times 10^3 J.

8 0
2 years ago
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A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t
choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25
8 0
2 years ago
Read 2 more answers
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