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Andrews [41]
2 years ago
12

A baseball bat could be considered a uniform rod. It ha a mass of 0.97 kg and a length of 97 cm. If a player accelerates it from

rest to 2.8 rev/s in 0.20s, how much torque is applied?
Physics
2 answers:
grigory [225]2 years ago
5 0

Answer:

The torque is 26.75 N-m.

Explanation:

Given that,

Mass = 0.97 kg

Length = 97 cm

Time = 0.20 s

Angular speed \omega= 2.8 rev/s = 2.8\times2\pi\ rad/s

We need to calculate the moment of inertia of rod at one end

Using formula of the moment of inertia

I=\dfrac{Ml^2}{3}

Put the value into the formula

I=\dfrac{0.97\times(0.97\times10^{-2})^2}{3}

I=0.3042\ kg m^2

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\Delta \omega}{\Delta t}

Put the value into the formula

\alpha=\dfrac{2.8\times2\pi}{0.20}

\alpha=87.964\ rad/s^2

We need to calculate the torque

Using formula of torque

\tau=I\times\alpha

Put the value into the formula

\tau=0.3042\times87.964

\tau=26.75\ N-m

Hence, The torque is 26.75 N-m.

sukhopar [10]2 years ago
4 0

Answer:

Applied torque, T = 26.74 N-m

Given:

Mass of rod, M = 0.97 kg

Length of the rod, L = 97 cm = 0.97 m

Change in angular velocity, \Delta \omega = 2.8 rev/s = 2.8\times 2\pi = 17.59 rad/s

time, t = 0.20 s

Solution:

Now, we know that torque is given by:

T = I\alpha                          (1)

where

I = moment of inertia of the rod

\alpha  = angular acceleration

Using perpendicular axis formula to calculate the moment of inertia:

I = \frac{ML^{2}}{12}                    

Now, the base ball bat will be held from one end, therefore,  the moment of inertia about the axis at the end is given by:

I = \frac{ML^{2}}{12} + \frac{ML^{2}}{4} = \frac{ML^{2}}{3}

I = \frac{0.97\times (0.97)^{2}}{3} = 0.304 kg-m^{2}

Now, angular acceleration can be calculated as:

\alpha = \frac{\Delta \omega}{t} = \frac{17.59}{0.20} = 87.95 rad/s^{2}

Now, calculation of torque using eqn (1):

T = 0.304\times 87.95 = 26.74 N-m

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In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of
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Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

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3 years ago
A plece of titanium has a mass of 67.5g and a volume of 15cm<br> What is the density?
sveta [45]

Answer:

4.5g/cm^3

Explanation:

Here, Mass(m)=67.5g

         Volume(v)=15cm^3

Now, According to formula,

        Density(p)=m/v

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8 0
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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!        

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a) since force = mass * acceleration  

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similarly b) f = 0

7 0
3 years ago
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