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2 years ago

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What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per

minute) if the wheel's diameter is 35 cm?

1 answer:

Brilliant_brown [7]2 years ago

4 0

Answer:

$a_{cp}=7.77m/s^2$

Explanation:

The equation for centripetal acceleration is $a_{cp}=\frac{v^2}{r}$ .

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference $C=2\pi r$ and the period T, then we have:

$v=\frac{C}{T}=2\pi r f$

Putting all together:

$a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2$

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Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f

stealth61 [152]

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put 12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

4 0

2 years ago

Which statement describes the relationship between bond strength and the melting and boiling points of a substance? A. As the fo

icang [17]

Answer:

a

Explanation:

4 0

2 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

2 years ago

A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross

Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

$F_D=\frac{1}{2}\rhoC_dAV^2$

Our values are:

$V=30.1m/s$

$C_d=0.45$

$A=3.3m^2$

$\rho=1.2kg/m^3$

Replacing,

$F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2$

$F_D=807.25N$

We need calculate now the energy lost through a time T, then,

$W = F_D d$

But we know that d is equal to

$d=vt$

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

$W=(807.25N)(30.1m/s)(3600s/1hr)$

$W=87.47*10^6J$ (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0

2 years ago

Which of the following are true statements? A. Like charges repel B. Unlike charges repel C. Unlike charges attract or D. Charge

Brums [2.3K]

Answer: The correct answers are (A) and (C).

Explanation:

The expression from electrostatic force is as follows;

$F=\frac{kq_{1}q_{1}}{r^{2} }$

Here, F is the electrostatic force, k is constant, r is the distance between the charges and $q_{1},q_{1}$ are the charges.

The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.

Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.

The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.

Therefore, the true statements from the given options are as follows;

Like charges repel.

Unlike charges attract.

7 0

2 years ago