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3 years ago

12

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]3 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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Question 2 (1 point)

mixer [17]

Answer:

The initial vertical velocity is zero, u = 0 m/s

Explanation:

Given;

height of the table, h = 0.55 m

horizontal distance traveled by the tennis, x = 0.12 m

Apply the following kinematic equation;

h = ut + ¹/₂gt²

where;

u is the initial vertical velocity = 0, since the tennis ball rolled off the edge of a table.

h = ¹/₂gt²

The time to fall from the vertical height is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(0.55)}{9.8} }\\\\t = 0.335 \ s$

The initial horizontal velocity of the tennis is given by;

x = vₓt

vₓ = x / t

vₓ = (0.12) / (0.335)

vₓ = 0.358 m/s

Therefore, the initial vertical velocity is zero, u = 0 m/s and initial horizontal velocity, vₓ is 0.358 m/s

3 0

3 years ago

Which of these best explains why people on Earth cannot see the entire shape of the Milky Way

BartSMP [9]

I believe the answer is A.

Since the Earth is in the Milky Way and not outside it, we cannot see the exact shape of it. Physicists have been able to track and graph the movements of the planets accurately for thousands of years, but that does not mean we know the shape of the entire solar system.<span />

5 0

3 years ago

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

stiv31 [10]

Answer:

Magnetic field, $B=2\times 10^{-4}\ T$

Explanation:

Given that,

Velocity of electron, $v=5\times 10^7\ m/s$

It enters a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, $E=10^4\ V/m$

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, $F_m=qvB\ sin\theta$ .......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

$qvB\ sin\theta=qE$

$B=\dfrac{E}{v}$

$B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}$

B = 0.0002 T

or

$B=2\times 10^{-4}\ T$

Hence, this is the required solution.

3 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

Does an increase in velocity necessarily mean an increase in acceleration?

Vinil7 [7]

We know, acceleration = final velocity - initial velocity / time
Here, if velocity is increasing, then,
Final velocity > initial velocity, in that case, acceleration is also increasing, as it is directly proportional to velocity

In short, Your Answer would be "Yes"

Hope this helps!

3 0

3 years ago

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