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Leto [7]
3 years ago
10

A city bus travels along its route from Jackson Street 8 km to Aurora Avenue, where it turns and continues 13 km.

Physics
1 answer:
Elina [12.6K]3 years ago
3 0

im a bit confused on what the question is:(

if its just asking for the total number of km traveled it would be 21 km

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A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagati
Margaret [11]

Answer:

P=2.57\times 10^{-7}\ N/m^2

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

P=\dfrac{2I}{c}

Where

c is speed of light

Putting all the values, we get :

P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2

So, the radiation pressure is 2.57\times 10^{-7}\ N/m^2.

3 0
3 years ago
Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He
umka2103 [35]
<h3>Answer:  130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

  • The initial speed (let's call that Vi) is 8.0 m/s
  • The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
  • This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

  • At 0 seconds, he's going 8 m/s
  • At the 1 second mark, he's slowing down to 8-2 = 6 m/s
  • At the 2 second mark, he's now at 6-2 = 4 m/s
  • At the 3 second mark, he's at 4-2 = 2 m/s
  • Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

  • F = force applied
  • m = mass
  • a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0
3 years ago
describe the motion of objects that are viewed from your reference frame both inside and outside while you travel inside a movin
V125BC [204]

Answer:

The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.

3 0
2 years ago
Why isn't pluto considered a planet anymore?
aivan3 [116]

The answer is. It did not meet the three criteria the IAU uses to define a full-sized planet.

1. It is in orbit around the Sun

2. It has sufficient mass to assume hydrostatic equilibrium

3. It has 'cleared the neighborhood' around its orbit

Pluto has not 'cleared its neighborhood'

3 0
3 years ago
One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F
Alja [10]

Answer:

\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}

f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}

b) Wave speed

(i) Calculate the wavelength

In a  fundamental vibration, the length of the string is half the wavelength.

\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

(b) Calculate the speed s

\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}

\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}

\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

4 0
3 years ago
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