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levacccp [35]
3 years ago
15

Due to the friction between a tire and the road surface, the temperature of the air inside a tire increases as one drives. If th

e tire pressure is 37 psi when the car is in the driveway at 25° C, What is the pressure of the tire when the temperature increases to 50° C?
Physics
1 answer:
inysia [295]3 years ago
5 0

Answer:

40.10 atm .

Explanation:

Here volume of tire is assumed to be nearly constant .

Modified gas law ,

P₁ / T₁ = P₂ / T₂

37 / ( 273 + 25 ) = P₂ / ( 273 + 50)

37 / 298 = P₂ / 323

P₂ = 37 x 323 / 298

= 40.10 atm .

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Are plants carnivores<br>Ture or False​
monitta

Answer:

False they are producers and they don' eat anything only produce and provide

Explanation:

3 0
2 years ago
Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
Makala arrives late to class and misses the first few minutes of the lecture. These are the notes she takes:
lukranit [14]

Answer:

B. Quantum Mechanics

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Quantum mechanics is the foundation of all quantum physics including quantum chemistry, quantum field theory, and quantum information science.

4 0
2 years ago
Read 2 more answers
A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate
GaryK [48]

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\

Solving for Y given  X=2m/s we get

\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s

8 0
3 years ago
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