If a star contains a certain element, it can be determined by studying the star’s

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago

the focal length of a simple magnifier is 8.50 cmcm . assume the magnifier to be a thin lens placed very close to the eye.

Igoryamba

When the object is at the focal point the angular magnification is 2.94.

Angular magnification:

The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.

Here we have to find the angular magnification when the object is at the focal point.

Focal length = 6.00 cm

Formula to calculate angular magnification:

Angular magnification = 25/f

= 25/ 8.5

= 2.94

Therefore the angular magnification of this thin lens is 2.94

To know more about angular magnification refer:: brainly.com/question/28325488

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5 0

1 year ago

Which force is stronger electrical or gravitational?

Alinara [238K]

Gravitational force is stronger

8 0

3 years ago

A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the bl

KengaRu [80]

Answer:

$a) \ d_2=d_1\\b) \ d_2=4d_1$

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

$E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}$

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

6 0

4 years ago

Enumerate two ways that you practice to control manage noise pollution

motikmotik

1. Sound insulation at construction sides.
2. Using silencers in automobiles and replacing old noisy machines with new quitter machines or using lubricants

4 0

2 years ago