This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

7 0

2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

6 0

1 year ago

A. Why is the stratosphere considered a "stable" layer in the atmosphere?

vovangra [49]

Answer:

A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .

D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

E. Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

7 0

2 years ago

A radioactive source has a half life of 80s.

Burka [1]

Lets make the original number of nuclides at the start is 100.

If 7/8 of 100 is decayed, that means 87.5 decayed.

$\frac{7}{8} \times 100 = 87.5$

And there is 1/8 left of the number of nuclide 100. Which is 12.5

$100 - 87.5 =12.5$

$\frac{1}{8} \times 100 = 12.5$

How many Half lifes passed for 100 to become 12.5 is 3 Half-Lives.

$100 \div 2 \div 2 \div 2 = 12.5$

Each Half-Life is 80 seconds so there is 240 seconds

$3 \times 80 = 240$ The answer is that it takes 240 seconds.

6 0

2 years ago