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Semenov [28]
2 years ago
11

g A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 ra

d/s. What is the constant angular acceleration (in rad/s) of the wheel
Physics
1 answer:
Setler [38]2 years ago
5 0

Answer:

The angular acceleration of the wheel is 15.21 rad/s².

Explanation:

Given that,

Time = 5 sec

Final angular velocity = 96.0 rad/s

Angular displacement = 28.0 rev = 175.84 rad

Let \alpha be the angular acceleration

We need to calculate the angular acceleration

Using equation of motion

\theta=\omega_{i} t+\dfrac{1}{2}\alpha t^2

Put the value in the equation

175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2

175.84=\omega_{i}\times 5+12.5\alpha......(I)

Again using equation of motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

96.0=\omega_{i}+\alpha \times 5

On multiply by 5 in both sides

480=\omega_{i}\times 5+\alpha\times 25....(II)

On subtract equation (I) from equation (II)

480-175.84=\alpha(25-5)

304.16=\alpha\times20

\alpha=\dfrac{304.16}{20}

\alpha=15.21\ rad/s^2

Hence, The angular acceleration of the wheel is 15.21 rad/s².

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<h2>Answer</h2>

Option A that is 8.8 × 10^3 m/s

<h2>Explanation</h2>

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Field-strength = BVqsinΔ

<h2>v = E/B </h2>

Since field are perpendicular so sin90 = 1

           

             v = 4.6/10^4 / 5.2

             v = 8846.15 m /s

The speed at which electrons pass through the selector without deflection = 8846.15 m /s

4 0
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The correct statement about nuclear fusion is the last one: "it produces nearly all the elements that are heavier than helium". Nuclear fusion is the process of combination of two smaller atoms into a larger one. It requires a high activation energy, which is reached in the stars. Elements from He to Iron were formed in the Sun, but larger elements require a higher energy and are formed inside supernovas.

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A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0
sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

5 0
3 years ago
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

  • the peripheral velocity that is directed downward (-V_y) along the y-axis
  • the linear velocity (V_x) that is directed along the x-axis

Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

Also,

-V_y = R* \omega

where \omega(angular velocity) = \frac{d\theta}{dt} = \frac{d}{dt}(8t^4)

-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

3 0
3 years ago
A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium a
nadya68 [22]

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

8 0
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