In which orbitals would the valence electrons for selenium (Se) be placed?

1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The len

igomit [66]

Answer:

a) $\alpha = -0.233 rad/s^{2}$

b) the rotational acceleration will remain the same, $\alpha = -0.233 rad/s^{2}$

c) When r = 36 m, $a_{c} = 157.25 m/s^{2}$

When r = 18 m, $a_{c} = 78.63 m/s^{2}$

d) When r = 36 m, $a_{c} = 39.31 m/s^{2}$

When r = 36 m, $a_{c} = 19.66 m/s^{2}$

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60 = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

$\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}$

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, $a_{c} = w^{2} r$

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

$a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}$

At the halfway point, r = 18 m

$a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}$

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

$a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}$

When r = 18 m

$a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}$

8 0

3 years ago

Why friction made it difficult to discover Newton’s first law of motion.

Amiraneli [1.4K]

Mandatory friction makes it difficult for a free body of mass to move with constant speed.

5 0

3 years ago

What is the current in the 10.0 , resistor?

In-s [12.5K]

Answer:

B. 12.0A

Explanation:

The circuit shown in the image is a parallel circuit which mean that the voltage of the source is equal across each resistance in the circuit because they are directly connected to the terminals of the voltage source.

The voltage across the 10 Ohms resistor is 120V

So, let's proceed to calculate the current

By using Ohm's Law

$I=\frac{V}{R} \\I= \frac{120.0V}{10.0Ohm} \\I= 12.0A$