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2 years ago

Please help me Why do objects in the sky appear to move and change so much?

Physics

1 answer:

bearhunter [10]2 years ago

6 0

Answer:

It depends on the climate and wind speed at the time.

If the climate is hot and no breeze it will move slower.

If it is cold and very windy it will move rapidly.

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Help, please. I am not sure what to do.

miss Akunina [59]

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

3 0

2 years ago

A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

Mass of the pitcher, m₁ = 50 kg
Mass of the baseball, m₂ = 0.15 kg
Velocity of the ball, u₂ = 35 m/s

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

$u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s$

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0

2 years ago

You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

$f(x,t)=Acos(kx-\omega t)$

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

7 0

3 years ago

A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

ryzh [129]

12414253

Explanation:

8 0

2 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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5 0

3 years ago