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tankabanditka [31]
3 years ago
7

Im back and need help

Physics
1 answer:
const2013 [10]3 years ago
7 0

Always here to help. Bring it!!!

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A pendulum with a mass of 1 kg and a spring-oscillator with the same mass of 1 kg are both oscillating with a period of 0.5 seco
maria [59]

Answer:

Explanation:

Period of the pendulum will not change as the period is not a function of mass

T = 2π√(L/g)

Period of the spring oscillator will increase with larger mass because period is a function of mass.

T = 2π/ω = 2π/√(k/m) = 2π√(m/k)

7 0
3 years ago
Light waves do not need a medium to travel through. true or false.
melamori03 [73]

Light travels in electromagnetic waves in the form of photons. What do photons travel in? Can a frequency have weight? Carry weight? According to Einstien a photon with energy proportional to its frequency basically explains ultraviolet light(among other things), so does that mean light travels within a "larger" version of itself?

We know that light doesn't need a medium through which to travel because the speed of light is experimentally constant: independent of the movement of the source or detector or the direction in which it travels.

Light contrasts with sound, which travels through the air (or some other material medium). If you're stationary with respect to the air, then the speed of sound is the same in all directions. But if you're moving with respect to the air, the speed of sound will be the same in all directions relative to the air---which means that sound coming up in front of you will seem faster and sound catching up to you from behind will seem slower.

If light were a disturbance in a medium, it would exhibit the same behaviour. But light never does---its speed is the same under all circumstances. So it does you no good to postulate an aether. You can still do it, but it makes the theory more complicated than necessary. The only reason to postulate an aether is that you're uncomfortable with the idea of waves not needing a medium. But our modern understanding of quantum mechanics is that all kinds of particles have a wavelike nature, so, if you accept that matter can travel through empty space, you should have no problem accepting the same for light.

so it's true

6 0
3 years ago
Read 2 more answers
Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At s
Tresset [83]

Answer:

n=0.03928 moles

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

Explanation:

The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs

Volume of oxygen in lungs =V=5*20%= 1 L=1*10^{-3} m^3

Temperature=T=37^oC=273+37=310K

Pressure at sea level = P= 1 atm=1.0125*10^5 Pa

R is universal Gas Constant =8.314 J/mol.K

Formula:

n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol

Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles

8 0
4 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
When the point of reference (the observer) is in motion, it flips the "rules" of vector addition (you ADD the vectors going in o
earnstyle [38]

Answer:

True

Explanation:

4 0
3 years ago
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