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2 years ago

Part 1: Fill in the SI unit for each of the following measurements. 1. Time: 2. Length: 3. Mass: 4. Temperature:

Physics

1 answer:

Alik [6]2 years ago

3 0

Answer:

1. Time: Second

2. Length: Meter

3. Mass: Kilogram

4. Temperatur: Kelvin

Explanation:

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A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0

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Answer:

Explanation:

Given

mass of box $m=2\ kg$

speed of box $v=1.9\ m/s$

distance moved by the box $x=10\ cm$

coefficient of kinetic friction $\mu _k=0.66$

Friction force $f_r=\mu_kN$

$f_r=0.66\times mg$

$f_r=0.66\times 2\times 9.8=12.936 \N$

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

$\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2$

$\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2$

$3.61-1.2936=0.005\times k$

$k=463.28\ N/m$

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A skydiver prepares to jump out of a plane. Explain how gravity and air resistance will affect the motion of the skydiver before

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Las condiciones iniciales de un gas son 3000 cm3

slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

$PV=nRT$

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

$n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles$

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

$T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C$

hence, T'=92.70°C

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