Which example show harassment?

You are woken up in the morning on the farm by a rooster. The rooster is located 30 meters away on top of the 5 meter barn house

algol [13]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = 18.73 \ m/s$

Explanation:

From the question we are told that

The distance of the rooster from farm house is $d = 30 \ m$

The height of the rooster from the ground is $h = 5 \ m$

The angle at which the alarm clock is thrown is $\theta = 53 ^o$

Let the velocity at which the alarm clock is thrown be v

So the horizontal component of v is mathematically represented as

$v_h = v cos \theta$

The distance covered by the alarm clock toward the horizontal direction at velocity v is

$d = v_h$

=> $d = v_h = vcos \theta * t$

=> $vcos \theta * t = 30$

substituting for $\theta$

$v (cos (53)) * t = 30$

=> $v * t = 50$

Considering the motion of the alarm clock in the vertical direction

So the vertical component of v is mathematically represented as

$v_v = vsin \theta$

The distance covered in the vertical direction is mathematically evaluated as follows

From the equation of motion we have

$h = vsin\theta * t - \frac{1}{2} g t^2$

$v* t sin\theta - 4.9t^2 = 5$

Recall $v * t = 50$

So

$50 \ sin \theta -4.9t^2 = 5$

substituting for $\theta$

$50 \ sin (53) -4.9t^2 = 5$

=> $t =\sqrt{7.1289}$

$t =2.673 \ s$

Now

$v * t = 50$

So

$v * 2.673 = 50$

=> $v = 18.73 \ m/s$

3 0

3 years ago

Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations wa

OLga [1]

Answer:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − $\omega$ t) where An is the amplitude f oscillation, $\omega$ is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; $k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f$ where;

$\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency$

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = $\frac{1}{(2/15)}$

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength $\lambda$ = 2m

Transverse speed $v = f \lambda$

$v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s$

Hence, the transverse speed at that point is 15m/s

8 0

3 years ago

When a fixed amount of ideal gas goes through an isobaric expansion

Juliette [100K]

Answer:

its temperature must increase.

Explanation:

An isobaric expansion is a transformation in which the pressure of the gas does not change (isobaric), while the volume increases (expansion).

Since the pressure does not change,

"its pressure must increase."

is a false statement.

The 1st law of thermodynamics is

$\Delta U = Q - W$

where

$\Delta U$ is the change in internal energy of the gas, which is proportional to the change in temperature: $\Delta U \propto \Delta T$

Q is the heat supplied to the gas

W is the work done by the gas, which is given by

$W=p\Delta V$

where p is the pressure and $\Delta V$ is the change in volume. Since the gas is expanding, we can say that $\Delta V>0$ , so the gas does positive work:

$W>0$

This means that the option

"the gas does no work."

is false.

Moreover, from the ideal gas law

$pV=nRT$ (2)

we also know that the temperature of the gas is increasing (because p, the pressure, n the number of moles, and R, the gas constant, are all constant in this process, and since the volume V is increasing, than the temperature T must be increasing also)

So, we know that the option

"its internal (thermal) energy does not change. "

is false.

Finally, in an isobaric expansion, in order to keep the pressure constant heat should be supplied to the system, so

"no heat enters or leaves the gas."

is also wrong

We also said from (2) that the temperature of the gas is increasing, therefore the statement

"its temperature must increase."

is the only correct one.

7 0

4 years ago

How many 60-watt lamps could be connected in parallel on a 15-amp, 120-volt circuit without exceeding 80% of the rating of the c

Alborosie

The total number of lamps is mathematically given as

N=24

<h3>What is the total number of lamps?</h3>

Question Parameters:

How many 60-watt lamps could be connected in parallel on a 15-amp, 120-volt

Generally, the equation for the Power is mathematically given as

P=IV

Therefore

I=P/v

I=60/120

I=0.5A

Since the system does not exceed 80% of the rating

NI=I'(0.8)

N=15*0.8/0.5

N=24

In conclusion, the total number of lamps is

N=24