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ruslelena [56]
3 years ago
8

How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad

/ s 2?
Physics
1 answer:
Rzqust [24]3 years ago
7 0

initial angular speed is given by 33.3 rpm

w_0 = 2\pi \frac{33.3}{60}

w_0 = 3.49 rad/s

final angular speed is given by 78 rpm

w_f = 2\pi \frac{78}{60}

w_f = 8.17 rad/s

now by using kinematics we will have

w_f = w_0 + \alpha * t

8.17 = 3.49 + 2.15 * t

t = 2.17 seconds

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drek231 [11]

The advantage of using a solar cooker is that it is Eco-friendly and the disadvantage is that it can be used only under certain conditions.

<h3><u>Explanation:</u></h3>

A solar cooker is used for cooking food without having to use electricity or gas. Instead, the appliance uses heat from the sun to cook food. It is used widely in by people who travel in remote areas or go on trips. But the appliance has limitations of its own too.

ADVANTAGES

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DISADVANTAGES

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8 0
3 years ago
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o
umka2103 [35]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

F = qv \times B = qvBsin(\theta)     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

K = \frac{1}{2}mv^{2}

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

3 0
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