A tow truck pulls a 2,000-kilogram car with a net force of

A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without

kari74 [83]

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

7 0

3 years ago

Which restricted basin has the coolest temperatures?

sattari [20]

Hudson Bay is the restricted basin that has the coolest temperatures

Hudson Bay is a restricted basin which remains frozen or is dominated by ice over the summer solstice and through- out much of the high-sun season. This basin experiences a harsh continental climate.

The average annual temperature in almost the entire bay is around 0 °C (32 °F) or below. In the extreme northeast, winter temperatures average as low as −29 °C or −20.2 °F. The region of this basin has very low year-round average temperatures.

This basin starts freezing up by early November, and the northern part of the basin is typically entirely iced over by the end of the month.

correct answer is Hudson bay

learn more about basin :

brainly.com/question/11871406?referrer=searchResults

#SPJ4

3 0

2 years ago

Explain why mess extinctions occur and why it make sense to measure geologic time between mass extinctions

Nastasia [14]

Mass extinction occur from natural disasters, such as a n asteroid hitting earth or a volcano errupting and spread ash everywhere.

It makes sense to measure geologic time between mass extinctions because after each mass extinction, there is almost no life left and the few left have to repopulate, which may lead way to new mutations and new varieties of plants and animals.
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3 0

3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea

Semmy [17]

Answer:

${\rm K} = 2.4\times 10^{-19}~J$

Explanation:

The electric field inside a parallel plate capacitor is

$E = \frac{Q}{2\epsilon_0 A}$

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

$F = qE = \frac{qQ}{2\epsilon_0 A}$

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

$C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60$

Plugging this identity into the force equation above gives

$F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q$

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is $1.6 \times 10^{-19}$

Therefore, the kinetic energy is $2.4\times 10^{-19}$

8 0

4 years ago