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OleMash [197]
4 years ago
14

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

A = 23° above the x axis, and the angle θB = 53° below the x axis.
1) What is the magnitude of the car’s acceleration?
2) What is the x component of the car’s acceleration when it is at point A
3) What is the y component of the car’s acceleration when it is at point A
4) What is the x component of the car’s acceleration when it is at point B
5) What is the y component of the car’s acceleration when it is at point B
Physics
1 answer:
Fantom [35]4 years ago
6 0

Centripetal acceleration of car is given by formula

a_c = \frac{v^2}{R}

now plug in the values in this

a_c = \frac{23^2}{210}

a_c = 2.52 m/s^2

Part b)

At position A we have

x component of acceleration is given as

a_x = -a_c cos23

a_x = -2.32 m/s^2

Part c)

At position A we have

y component of acceleration is given as

a_y = -a_c sin23

a_y = -0.98 m/s^2

Part d)

At position B we have

x component of acceleration is given as

a_x = -a_c cos53

a_x = -1.52 m/s^2

Part e)

At position B we have

Y component of acceleration is given as

a_y = a_c sin53

a_y = 2.01 m/s^2

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Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

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4 years ago
Rays traveling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus?
abruzzese [7]

Answer:

True

Explanation:

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When a light ray travelling through focus of a concave mirror then after reflection the light ray reflect out through the mirror and  go parallel to principle axis.

Therefore, rays travelling parallel to the principle axis of a  concave mirror will reflect out through the mirrors focus.

It is true.

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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

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1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

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