A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a while since you've checked. You can't find a tire gauge in the car, but you do find the owner's manual and a ruler. Fortunately, you've just finished taking physics, so you tell your friend, "I don't know, but I can figure it out." From the owner's manual you find that the car's mass is 1000 kg . It seems reasonable to assume that each tire supports one-fourth of the weight. With the ruler you find that the tires are 13 cm wide and the flattened segment of the tire in contact with the road is 13 cm long.

Answers

Answer 1

Answer:

21 psi

Explanation:

The weight of the car is:

W = mg

W = 1000 kg * 9.8 m/s²

W = 9800 N

Divided by 4 tires, each tire supports:

F = W/4

F = 9800 N / 4

F = 2450 N

Pressure is force divided by area, so:

P = F / A

P = (2450 N) / (0.13 m × 0.13 m)

P ≈ 145,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 21 psi

Answer 2
Final answer:

The pressure in the car's tires can be calculated using physics principles. By calculating the weight of the car and dividing it by the area of contact each tire has with the ground, the pressure is found to be approximately 21.46 psi.

Explanation:

To determine the tire pressure, we'll use the following physics concept: Pressure = Force / Area. The force in this case would be the weight of the car distributed among the four tires and the area would be the contact area of the tire with the ground.

First, we calculate the weight of the car by using the formula Weight = mass x gravity. Given that the mass of the car is 1000 kg and acceleration due to gravity is 10 m/s^2 (approx), the weight would be 10000 N (Newtons). Since each tire supports a quarter of the car's weight, the force each tire experiences is 2500 N.

Then, we calculate the area of contact the tire has with the ground. The flattened section would approximate a rectangle with length and width equal to 13 cm each, so Area = length x width = 0.13 m x 0.13 m = 0.0169 m^2.

Finally, we calculate the pressure by dividing the force by the area. Pressure = Force / Area = 2500 N / 0.0169 m^2 ≈ 148,000 Pa (Pascal).

Please note that 1 Pascal (Pa) = 0.000145 psi, so the pressure in psi would be approximately 21.46 psi, which is a bit lower than recommended by the tire manufacturer.

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Related Questions

Most of the funding for research comes from the federal government or ? And is provided to Principal Investigators (PIs) through the organizations for which they work.

Answers

Final answer:

The federal government and industry are the main sources of funding for research, providing funding to Principal Investigators (PIs) through their organizations. The U.S. economy has increasingly relied on industry-funded research, but the government still plays a significant role in funding research.

Explanation:

The federal government is one of the main sources of funding for research, along with industry. The government provides funding to Principal Investigators (PIs) through the organizations they work for. Over time, the U.S. economy has relied more heavily on industry-funded research and development (R&D). However, the government still plays a significant role in funding research, especially in areas where private firms are not as active.

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Most funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through their organizations.

Explanation:

Most of the funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through the organizations for which they work. The federal government, through agencies such as the National Institutes of Health (NIH) and the National Science Foundation (NSF), provides grants for research in various fields. Industry-funded research, on the other hand, is supported by companies and private entities who invest in research and development projects.

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The probable question can be: Complete the following sentence. Most of the funding for research comes from the federal government or ______ and is provided to Principal Investigators (PIs) through the organizations for which they work.

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Consider two copper wires with the same cross-sectional area. Wire A is twice as long as wire B. How do the resistivities and resistances of the two wires compare? Check all that apply. Check all that apply. Wire A and wire B have the same resistance. Wire A has twice the resistance of wire B. Wire A and wire B have the same resistivity. Wire B has twice the resistivity of wire A. Wire B has twice the resistance of wire A. Wire A has twice the resistivity of wire B. SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Your answer indicates that you need to review resistivity and resistance. Provide Feedback Next Incorrect. Incorrect; Try Again; 4 attempts remaining. Feedback. Your answer indicates that you need to review resistivity and resistance. End of feedback. g

Answers

Answer:

Wire A has twice the resistance of wire B.

Wire A and wire B have the same resistivity.

Explanation:

- Resistivity is a property of a material, that tells how much is the material able to oppose to the flow of current through it. The value of the resistivity of a wire depends on the material only: this means that two wires made of the same material have same resistivity. Since both wire A and B here are made of copper, they have the same resistivity.

- Resistance of a wire instead is given by

[tex]R=\rho \frac{L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

Here, the two wires have same resistivity and same cross-sectional area, while wire A is twice as long as wire B (so, L for A is twice the value of L for B): therefore, the resistance of wire A will be twice that of wire B.

Fuel cells have been developed that can generate a large amount of energy. For example, a hydrogen fuel cell works by combining hydrogen and oxygen gas to produce water and electrical energy. If a fuel cell can generate 10.0 kilowatts of power and the current is 15.8 amps, what is the voltage of the electricity?
A.
0.63 volts
B.
158volts
C.
633 volts
D.
158,000 volts
E.
5.8 volts

Answers

Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.

10,000 W / 15.8 A = V

632.9, or 633.

The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.

What is electric power?

Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.

The given data in the problem is;

V is the voltage = ? Volt (V)

Electric current (I)= 15.8 amps (A)

P is the power =10.0 kilowatts =10⁴ watt

The formula for the power is given as;

[tex]\rm P= V I \\\\\ 10^4= V \times 15.8 \\\\ V=632.9 \ V[/tex]

The voltage of the electricity will be 63.29 V.

Hence, option B is correct.

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In a laboratory, you determine that the density of a certain solid is 5.23×10−6kg/mm3. Convert this density into kilograms per cubic meter.Notice that the units you are trying to eliminate are now in the denominator. The same principle from the previous parts applies: Pick the conversion factor so that the units cancel. The only change is that now the units you wish to cancel must appear in the numerator of the conversion factor.

Answers

Answer:

[tex]5.23\cdot 10^3 kg/m^3[/tex]

Explanation:

The density of the solid is

[tex]d = 5.23\cdot 10^{-6}kg/mm^3[/tex]

we want to convert it into kg/m^3. We must note that:

[tex]1 m^3 = 1 m \cdot 1 m \cdot \1m =1000 mm\cdot 1000 mm \cdot 1000 mm=1\cdot 10^9 mm^3[/tex]

Therefore, the conversion can be done as follows:

[tex]d=5.23\cdot 10^{-6} \frac{kg}{mm^3} \cdot (1\cdot 10^9 \frac{mm^3}{m^3}) =5.23\cdot 10^3 kg/m^3[/tex]

____ is undeniably accepted by scientists all over the world as the primary language of science.

Answers

Answer:

Latin

Explanation:

In order for the scientists to have a common and official name for a particular thing that can be understood by every scientist in the world, a single language has been established for the purpose. The language chosen is the Latin language. The official scientific names are given in this language, so it is a necessity for the scientists to know and understand this language. The terms that are commonly used are regional, and they come in many different languages, which is why this language has been chosen. Occasionally, the ancient Greek language is used as well, though much less than the Latin.

Answer: English

Explanation: I would say the Answer is English but I’m not Sure!

Juan and Anita each lift an identical stack of books onto a table, but Anita places the books on a table twice as high as Juan. Therefore, her actions involve twice as much__

Answers

Answer:

Work

Explanation:

Anita exerted the same amount of force but for twice the distance.  So she did twice as much work.

Which of he following explains the significance of the observations that some microbes can kill or inhibit the growth of other microbes?
A.) It led to the development of the first antibiotic
B.) It led to the development of the first antiviral
C.) It led to the development of the Theory of Evolution
D.) It led to the development of the Germ Theory of Disease

Answers

Some microbes can kill or inhibit the growth of other microbes led to the development of the first antibiotic.

Answer: Option A

Explanation:

Antibiotics are the substance that restricts the bacteria growth and replication. These are designed against microbes and it targets bacterial infections in or on a human body. Bacteria's comes under the microbe category that cause harm to the human body.  

The substance that targets, restricts and kills microbial cells includes antibiotics, antiseptics and antiviral. The antibiotics that we use today can be produced in labs and they can be found in nature also. Hence, antibiotics are the one that kills or restricts the microbe's growth.

X-rays with an energy of 265 keV undergo Compton scattering from a target. If the scattered rays are deflected at 41.0° relative to the direction of the incident rays, find each of the following. (a) the Compton shift at this angle _________nm (b) the energy of the scattered x-ray __________keV (c) the kinetic energy of the recoiling electron ___________keV

Answers

Answers:

(a) Compton shift

The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex]     (1)

Where:

[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]h=4.136(10)^{-15}eV.s[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c=3(10)^{8}m/s[/tex] the speed of light in vacuum.

[tex]\theta=41\°[/tex] the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are deflected at [tex]41\°[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(41\°))[/tex]   (2)

[tex]\Delta \lambda=\lambda' - \lambda_{o}=5.950(10)^{-13}m[/tex]   (3)

But we are asked to express this in [tex]nm[/tex], so:

[tex]\Delta \lambda=5.950(10)^{-13}m.\frac{1nm}{(10)^{-9}m}[/tex]  

[tex]\Delta \lambda=0.000595nm[/tex]  (4)

(b) the energy of the scattered x-ray

The initial energy [tex]E_{o}=265keV=265(10)^{3}eV[/tex] of the photon is given by:

 [tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex]    (5)

From this equation (5) we can find the value of [tex]\lambda_{o}[/tex]:

[tex]\lambda_{o}=\frac{h.c}{E_{o}}[/tex]    (6)

[tex]\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{265(10)^{3}eV}[/tex]    

[tex]\lambda_{o}=4.682(10)^{-12}m[/tex]    (7)

Knowing the value of [tex]\Delta \lambda[/tex] and [tex]\lambda_{o}[/tex], let's find [tex]\lambda'[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}[/tex]

Then:

[tex]\lambda'=\Delta \lambda+\lambda_{o}[/tex]  (8)

[tex]\lambda'=5.950(10)^{-13}m+4.682(10)^{-12}m[/tex]  

[tex]\lambda'=5.277(10)^{-12}m[/tex]  (9)

Knowing the wavelength of the scattered photon [tex]\lambda'[/tex]  , we can find its energy [tex]E'[/tex] :

[tex]E'=\frac{h.c}{\lambda'}[/tex]    (10)

[tex]E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{5.277(10)^{-12}m}[/tex]    

[tex]E'=235.121keV[/tex]    (11) This is the energy of the scattered photon

(c) Kinetic energy of the recoiling electron

If we want to know the kinetic energy of the recoiling electron [tex]E_{e}[/tex], we have to calculate all the energy lost by the photon in the wavelength shift, which is:

[tex]K_{e}=E_{o}-E'[/tex]  (12)

[tex]K_{e}=265keV-235.121keV[/tex]  

Finally we obtain the kinetic energy of the recoiling electron:

[tex]E_{e}=29.878keV[/tex]  

Answer:

The first one:

the energy of the scattered x-ray

The answer for last on:

Kinetic energy of the recoiling electron

A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle? A) 1.56 × 106 N⋅m2/C B) 6.60 × 105 N⋅m2/C C) 1.42 × 105 N⋅m2/C D) 5.49 × 104 N⋅m2/C E) 4.23 × 104 N⋅m2/C

Answers

Answer:

[tex]6.60\cdot 10^5 Nm^2/C[/tex]

Explanation:

The electric flux through the rectangle is given by

[tex]\Phi = E A cos \theta[/tex]

where

E is the electric field strength

A is the area of the rectange

[tex]\theta[/tex] is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

[tex]A=2.50 m \cdot 5.00 m=12.5 m^2[/tex]

and the angle is

[tex]\theta=65.0^{\circ}[/tex]

so, the electric flux is

[tex]\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C[/tex]

4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 106 J/kg . Answer in units of J. 006 (part 2 of 3) 10.0 points Find the amount of heat added to the water to accomplish this process. Answer in units of J.

Answers

1. 408.4 J

The work done by a gas is given by:

[tex]W=p\Delta V[/tex]

where

p is the gas pressure

[tex]\Delta V[/tex] is the change in volume of the gas

In this problem,

[tex]p=1.01\cdot 10^5 Pa[/tex] (atmospheric pressure)

[tex]\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3[/tex] is the change in volume

So, the work done is

[tex]W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J[/tex]

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

[tex]Q = m \lambda_v[/tex]

where

m is the mass of the water

[tex]\lambda_v = 2.26\cdot 10^6 J/kg[/tex] is the specific latent heat of vaporization

The initial volume of water is

[tex]V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3[/tex]

and the water density is

[tex]\rho = 1000 kg/m^3[/tex]

So the water mass is

[tex]m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg[/tex]

So, the amount of heat added to the water is

[tex]Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J[/tex]

A small laser used as a pointer produces a beam of red light 5 mm in diameter, and has a power output of 5 milliwatts. What is the magnitude of the electric field in the laser beam?

Answers

Answer:

434.0 V/m

Explanation:

The power output of the laser is:

[tex]P=5 mW = 0.005 W[/tex]

while the radius of the beam is

[tex]r=\frac{5 mm}{2}=2.5 mm = 0.0025 m[/tex]

so the cross-sectional area is

[tex]A=\pi r^2 = \pi (0.0025 m)^2=2.0\cdot 10^{-5} m^2[/tex]

So the intensity of the laser beam is

[tex]I=\frac{P}{A}=\frac{0.005 W}{2.0\cdot 10^{-5} m^2}=250 W/m^2[/tex]

The intensity of a laser beam is related to the magnitude of the electric field by

[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

Solving the formula for E, we find

[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(250 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=434.0 V/m[/tex]

Light passes through a single slit. If the width of the slit is reduced, what happens to the width of the central bright fringe? (a) The width of the central bright fringe does not change, because it depends only on the wavelength of the light and not on the width of the slit. (b) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes smaller. (c) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. (d) The central bright fringe becomes narrower, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger. (e) The central bright fringe becomes narrower, because the angle that locates the first dark fringe on either side of the central bright fringe becomes smalle

Answers

Answer:

(c) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.

Explanation:

The formula that gives the angle of the first minimum of the diffraction pattern from a single-slit is

[tex]sin \theta = \frac{\lambda}{w}[/tex]

where

[tex]\lambda[/tex] is the wavelength of the light

w is the width of the slit

We see that the angle is inversely proportional to the width of the slit: therefore, if the width of the slit is reduced (so, w is decreased), the angle that locates the first minimum [tex]\theta[/tex] increases, and so the central bright fringe becomes wider.

Final answer:

Reducing the width of a single slit through which light passes leads to a wider central bright fringe. This occurs because the narrowed slit causes greater divergence of light rays, leading to a larger angle for locating the first dark fringe.

Explanation:

The phenomenon being discussed here pertains to physics, specifically light diffraction and interference. When light passes through a single slit, it exhibits a diffraction pattern with a central bright fringe flanked by smaller, dimmer fringes on either side. This pattern is influenced by the width of the slit and the wavelength of the light source.

For the scenario provided in the question, light passes through a single slit that is then narrowed. The correct answer is (c): The central bright fringe becomes wider because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.

This is explained by the fact that as the slit narrows, the rays of light diverge more upon exiting the slit, due to the wave nature of light. The larger spread of these rays leads to a larger angle for locating the first dark fringe, which subsequently results in a wider central bright fringe.

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