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3 years ago

15

A car tire has a pressure of 325 kPa when the temperature is 10 C. If the temperature of the tire rises to 50 C and its pressure

is constant, what is the new pressure?

1 answer:

MAXImum [283]3 years ago

3 0

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Complete the ray diagram and label incident ray, refracted ray, angle of incidence, and angle of refraction

Novay_Z [31]

Answer:

Solution

verified

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(a) The labelled diagram is shown.

(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,

μ=

Speedoflightindiamond

Speedoflightinair

and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is

1/2.42

times the speed of light in vacuum.

Explanation:

a) Draw and label the diagram given :

(i) Incident ray

(ii) Refracted ray

(iii) Emergent ray

(iv) Angle of reflection

(v) Angle of deviation

(v) Angle of emergence

(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?

4 0

2 years ago

In a ________________ wave, such as a sound wave, the particles in the medium vibrate in the same direction that the wave travel

Varvara68 [4.7K]

The answer would be A, transverse.

6 0

2 years ago

Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show

ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = $\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}$

We know that the inductance of the coil is given by the formula:

$L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH$

4 0

2 years ago

What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s

Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity $v$ and at a distance of $r$ from the axis of rotation is given as:

$a=\frac{v^2}{r}$

Here, $v=8\ m/s,r=0.70\ m$

∴ $a=\frac{8}{0.70}=11.43\ m/s^2$

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0

3 years ago

A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a

dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

$A_1v_1 = 18 A_2v_2$

now we know that

$A_1 = 1 cm^2$

$A_2 = 0.4 cm^2$

now from above equation

$1 cm^2 v_1 = 18(0.400 cm^2)v_2$

$\frac{v_2}{v_1} = \frac{1}{18\times 0.4}$

$\frac{v_2}{v_1} = 0.14$

so velocity will reduce by factor 0.14

3 0

3 years ago