Which of the following is an example of kinetic energy?

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago

On the weather map, what does the symbol shown below represent?

Evgen [1.6K]

Answer:

wind speed i think where i live?

8 0

3 years ago

The chemical equation provided shows iron rusting to form iron oxide. Use the drop-down menu to choose the coefficients that wil

Brrunno [24]

Answer:

$4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3$

Explanation:

The original equation is:

$Fe + O_2 \rightarrow Fe_2 O_3$

We notice that:

- we have 1 atom of Fe on the left, and 2 atoms of Fe on the right

- we have 2 atoms of O on the left, and 3 atoms of O on the right

Therefore, the equation is not balanced.

In order to balance it, we can add:

- a coefficient 3 in front of $O_2$

- a coefficient 2 in front of $Fe_2 O_3$

So we have:

$Fe + 3 O_2 \rightarrow 2Fe_2O_3$

Now the oxygen is balanced, but the iron it not balanced yet, since we have 1 Fe on the left and 4 on the right. Therefore, we should add a coefficient 4 on the Fe on the left:

$4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3$

3 0

2 years ago

Read 2 more answers

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i

crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

$\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}$

$q=\frac{1}{0.047 cm^{-1}}=+21.3 cm$

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

$h_i = - h_o \frac{q}{p}$

where

$h_i, h_o$ are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

$h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o$

So we notice that:

$|h_i| < |h_o|$ : this means that the image is smaller than the object

$h_i < 0$ : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.

7 0

3 years ago

A very small object carrying -7 μC of charge is attracted to a large, well-anchored, positively charged object. How much kinetic

miss Akunina [59]

Answer:

ΔK.E = 14 nJ

Explanation:

Solution:

- The charge that moves under the influence of an Electric Field produced between a potential difference (V) stores electric potential energy U within that is converted to kinetic energy.

- We will use conservation of energy on the system that contains the charged particle with charge q loses its electric potential energy U as it moves towards positively charged object that converts into a gain in Kinetic energy of the charged particle ΔK.E:

ΔK.E = U

Where,

U = V*q

ΔK.E = V*q

ΔK.E = (7*10^-6)*(2*10^-3)

ΔK.E = 14 nJ

- The gain in kinetic energy is 14 nJ.

7 0

3 years ago