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3 years ago

15

When the bells are charged up, the left bell acquires a positive charge, the right bell a negative charge. When this occurs, the

nut that is suspended between the two bells experiences a torque (it lines itself up with a line between the two bells) but it doesn't experience a net force. What does this tell us about the charge on the nut

1 answer:

MAXImum [283]3 years ago

5 0

Answer:

Thus simply tells us that The nut has no net charge and so therefore, There will be a negative charge on the left side, and an equal positive charge on the right side

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To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

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If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?

REY [17]

Answer:

The car's angular speed is $\frac{rad}{s}$ .

Explanation:

Angular velocity is usually measured with $\frac{radians}{sec}$ , so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω) is given by:

$V = \omega R$

Using the values from the question, we get:

$2 \frac{m}{s} = \omega (0.4m)$

$\omega = 5 \frac{rad}{s}$

Therefore, the car's angular speed is $5 \frac{rad}{s}$ .

Hope this helped!

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3 years ago

Scientists use a substance called Iodine-131 to treat cancer. As Iodine-131 undergoes radioactive decay and emits beta particles

olganol [36]

Answer:

I think it's strong I'm not to sure I'm sorry if it's wrong

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A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T .

kupik [55]

Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

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3 years ago

A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The

Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

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3 years ago