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3 years ago

A or B??????????????

Physics

2 answers:

Cerrena [4.2K]3 years ago

8 0

Answer:

Explanation:

Keep playing is probably not the answer.

Neporo4naja [7]3 years ago

7 0

I’m pretty sure it’s a

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Cho hệ thống thùng lắc có mô hình tại vị trí đang xét như hình vẽ

professor190 [17]

Answer:

<h2>I can't understand this language!!!</h2>

4 0

3 years ago

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A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular

AURORKA [14]

Answer:

A) ω₂ = 28 rev/min

B) ω_3 = 28 rev/min

Explanation:

A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).

Thus;

I₁ = ((1/12)ML²) + 2(mr²)

Where;

M is mass of rod = 0.03 kg

L is length of rod = 0.4m

m is mass of small rings = 0.02 kg

r is radius of small rings = 0.05 m

Thus;

I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)

I₁ = 0.0021 kg.m²

Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;

I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)

I₂ = 0.0036 kg.m²

From conservation of angular momentum, we know that:

I₁ω₁ = I₂ω₂

We are given ω₁ = 48 rev/min.

Thus; plugging in the relevant values;

0.0021 x 48 = 0.0036ω₂

0.1008 = 0.0036ω₂

ω₂ = 0.1008/0.0036

ω₂ = 28 rev/min

b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.

Thus,

I₂ω₂ = I_3•ω_3

So,0.0021 x 48 = 0.0036ω_3

ω_3 = 28 rev/min

7 0

3 years ago

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input pi

AfilCa [17]

Answer:

The correct answer is "3899.92 N".

Explanation:

The given values are:

Force,

$F_{app}=58 N$

Ratio,

$\frac{R_2}{R_1}=8.2$

As we know,

Area, $A=\pi r^2$

or,

⇒ $\frac{F_2}{F_1} =\frac{A_2}{A_1}$

On substituting the value of "A", we get

⇒ $\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}$

⇒ $\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}$

On applying cross-multiplication, we get

⇒ $F_2=F_1\times (\frac{r_2}{r_1} )^2$

On substituting the given values, we get

⇒ $=58\times (8.2)^2$

⇒ $=58\times 67.2$

⇒ $=3899.92 \ N$

6 0

3 years ago

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the

kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

$\theta=tan^{-1}(\frac{v_y}{v_x})$ (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

$v_x=v_o$

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

$v_y^2=v_{oy}^2+2gy$ (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

$v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}$

Finally, you use the equation (1) to find the angle:

$\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°$

The angle below the horizontal is 12.60°

7 0

3 years ago

Which of the following is a small star that has reached the end of its stellar evolution?

Akimi4 [234]

D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

Hope this helps!

3 0

3 years ago

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