Ask question

Ask question

All categories

Zigmanuir [339]

4 years ago

9

A ceiling fan is turned on and reaches an angular speed of 120 rev/min in 20 s. It is then turned off and coasts to a stop in an

additional 40 s. The ratio of the average angular acceleration for the first 20 s to that for the last 40 s is which of the following?
a.2
b.-0.5
c.0.5
d.-2

1 answer:

Ilya [14]4 years ago

6 0

Answer:

option D

Explanation:

given,

angular speed of the fan = 120 rev/min

time = 20 sec

After 40 s fan comes to rest

ration angular acceleration of the first 20 s to last 40 s.

angular acceleration for first 20 s

$\alpha_1= \dfrac{\Delta \omega}{t}$

$\alpha_1= \dfrac{120}{20}$

$\alpha_1=6 rev/s^2$

for last 40 s

$\alpha_2= \dfrac{\Delta \omega}{t}$

$\alpha_2= \dfrac{0-120}{40}$

$\alpha_2= -3 rev/s^2$

now, ratio

$\dfrac{\alpha_1}{\alpha_2}=\dfrac{6}{-3}$

$\dfrac{\alpha_1}{\alpha_2}=-2$

hence, the correct answer is option D

You might be interested in

Dominik [7]

Answer:

Option D (N/A-m)(m)(m/s)

Explanation:

(N/A-m)(m)(m/s) is the required dimensional analysis for calculation of emi

3 0

2 years ago

a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calc

kipiarov [429]

A. 0.5 miles an hour
B. 0.2 miles an hour

4 0

3 years ago

Match the chemical formula to the correct name.

11Alexandr11 [23.1K]

Co carbon monoxide
sorry friend i don't know other ones

3 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

Which wavelength of light is capable of penetrating the dust of a nebula?

ivann1987 [24]

When the dust is too thick to penetrate with visible light, such as the Nebula, Radio Waves are used to penetrate the dust. Longer radio waves can completely penetrate the thick cloud cover, allowing scientists to beam radar waves.

8 0

3 years ago