Focusing on irelevant information can negatively impact the ability to reason and problem solve effectively.

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago

How do I solve this

lesantik [10]

multiply grav pull by mass of astro maybe with a calculator

7 0

4 years ago

The normal eye, myopic eye and old age

yanalaym [24]

Answer:

1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

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

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

$\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}$

f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

1 / f = 1/15 + 1 / 1.5

1 / f = 0.733

f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

f’₀ / f = 1.5 / 1.36

f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

sin θ = y / f

where f is the distance of the lens (eye)

y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

y / f = 1.22 λ / D

y = 1.22 λ f / D

where D is the diameter of the eye

D = 2R₀

D = 2 0.1

D = 0.2 cm

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

\frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

$\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}$

$\frac{1}{f}$ = 0.7066

f = 1.415 cm

therefore the diffraction is

y = 1.22 550 10⁻⁹ 1.415 / 0.2

y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

d = 2y

d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

$\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}$

$\frac{1}{f}$ = 0.733

f = 1.36 cm

diffraction is

y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

y = 4.56 10-6 m

the diffraction diameter is

d_myope = 2y

d_myope = 9.16 10-6 m

$\frac{d_{normal}}{d_{myope}}$ = 9.49 /9.16

\frac{d_{normal}}{d_{myope}} = 1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0

3 years ago

What is the relationship between inertia and mass?

ivann1987 [24]

Answer:

D

Explanation:

The greater the mass, the greater the inertia, and vice versa.

Remark: This means that a more massive object has a greater tendency to resist a change in its state of rest or motion.

8 0

3 years ago