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STALIN [3.7K]
3 years ago
7

A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r

ope can withstand before breaking is 11.0 N, what is the maximum angular speed of the ball
Physics
1 answer:
Murrr4er [49]3 years ago
4 0

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l =  4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s

Let \omega is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s

So, the maximum angular speed of the ball is 2.1 rad/s.

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an
LUCKY_DIMON [66]

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

\Phi_B=BA=\mu_oNIA

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

M=\mu_o N_1 N_2 \frac{A_2}{l}

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

\epsilon_1=M\frac{dI_2}{dt}

by replacing you obtain:

\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV

the emf is 20mV

7 0
3 years ago
Read 2 more answers
What part of the crane is used to lift and lower the load?
Flura [38]
I think the answer is b.boom
7 0
3 years ago
Read 2 more answers
A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori
yarga [219]
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.


First, the vertical component of tension (Tsin theta) is equal to the weight of the object. 
 T * sin θ = mg =</span> 1.55 * 9.81 <span>
 T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind. 
 T * cos θ =  13.3 

Tan θ = sin  </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82. 

T then is equal to 20.20 N
4 0
3 years ago
What is the SI unit for intensity?
Allisa [31]

Answer:

The SI unit of intensity is the watt per square meter/metre (W/m^2.)

Explanation:

Intensity is equal to the power transferred per unit area. Since power is measured in watts (W) and 1 W = 1 J/s, then intensity can be viewed as how fast energy goes through a certain area.

In physics, intensity is often used when studying light, sound, or other phenomena that involve waves or energy transfer. (With waves, the power value is taken as the average power transfer over the wave's period.)

8 0
3 years ago
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0
3 years ago
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