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3 years ago

PLEEEEASE HELP ME PLEASE WILL GIVE BRAINLIEST

Physics

1 answer:

Drupady [299]3 years ago

8 0

C. 7.8 hz I think! Hope this helped ☺️

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What is the product of the cellular respiration process

zlopas [31]

When oxygen is present, cellular respiration produces carbon dioxide, water, and energy in the form of ATP.

When oxygen is not present, cellular respiration produces lactic acid or alcohol.

Hope this helps! :)

4 0

3 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago

Who created the light bulb?

Daniel [21]

There were 2 people
Tomas Edison
Joseph Swan

Tomas Edison invented the first practical incandescent lightbulb.

Sir Joseph Swan is most famous for his role in the development of the first incandescent lightbulb.

5 0

3 years ago

A 4-L pressurecooker has an operating pressure of175 kPa. Initially, one-half of the volume is filled with liquid and the other

Soloha48 [4]

To solve this problem, it is necessary to apply the concepts related to the Energy balance and the mass balance that allow us to find in each state the data necessary to find the total Power of the system through heat exchange.

From the tables of properties of the Water it is possible to obtain at the given pressures the values of the specific volume, the specific energy and the specific enthalpy.

Given these pressures then we have to

$P_1 = 174kPA$

$\Rightarrow v_f = 0.001057m^3/kg\\\Rightarrow v_g = 1.0037m^3/kg\\\Rightarrow u_f = 486.82kJ/kg\\\Rightarrow u_g = 2524.5kJ/kg$

$P_2 = 175kPa \rightarrow$ Saturated vapor

$\Rightarrow v_2 = v_g = 1.0036 m^3/kg\\\Rightarrow v_2 = u_g = 2524.5kJ/kg$

$P_e = 175kPa \rightarrow$ Saturated vapor

$V = 4L$

Considering the process performed, the kinetic and potential energy can be neglected as well as the work involved by specific interactions in the system. Although it is an unstable process it can be treated as a uniform flow process. Considering these expressions we can perform a mass balance for which

$m_{in}-m_{out} = \Delta m_{system} \rightarrow m_e = m_1-m_2$

Similarly through the energy balance you can get that

$E_{in}-E_{out} = \Delta E_{system}$

$Q_{in} -m_eh_e = m_2u_2-m_1u_1 \Rightarrow$ Since $W=ke=pe=0$

The initial mass, initial internal energy, and final mass in the tank are

$m_1 = mf_+m_g = \frac{V_f}{v_f}+\frac{V_g}{v_g}$

$m_1 = \frac{0.002m^3}{0.001057m^3/kg}+\frac{0.002m^3}{1.0036m^3/kg}$

$m_1 = 1.893+0.002= 1.895Kg$

At the same time the internal energy can be defined from the mass in state 1 as,

$U_1 = m_1 u_1$

$U_1 = m_fu_f+m_gu_g$

$U_1 = 1.893*486.82+0.002*2524.5$

$U_1 = 926.6kJ$

The calculation of mass in state 2 can be defined as

$m_2 = \frac{V}{v_2} = \frac{0.004m^3}{1.0037m^3/kg}$

$m_2 = 0.004Kg$

Then from the mass and energy balances,

$m_e = m_1-m_2\\m_e = 1.895-0.004\\m_e = 1.891Kg$

In this way the calculation of the heat of entry would be subject to

$Q_{in} = m_eh_e+m_2u_2-m_1u_1$

$Q_{in} = 1.891*2700.2+0.004*2524.5-926.6$

$Q_{in} = 4188kJ$

Therefore the Power would be given as

$\dot{Q} = \frac{Q}{\Delta t} = \frac{4188kJ}{3600s} = 1.163kW$

Therefore the highest rate of heat transfer allowed is 1.163kW

3 0

3 years ago

A mass-spring-damper system has a mass of 100 kg. Its free response amplitude decays such that the amplitude of the 30th cycle i

Nina [5.8K]

Answer:

the value for the damping constant is c = 5.3408Ns/m and the spring constant is k is 987N/m

Explanation:

Consider the formula for the logarithmic decrement (δ)

δ = $\frac{1}{n} In(\frac{B_1}{B_{n+1}} )----(1)$

Consider the formula for damping ratio (C)

C =δ / √4π² + ² ----------(2)

The given amplitude of the 30th cycle is 20% amplitude of the 1st cycle

The ratio of last peak to first peak is

$\frac{B_{n+1}}{B_1} =20$ %--------------(3)

where

n is the number of cycle

Re arrange eqn (3)

$\frac{B_{n+1}}{B_1} = \frac{100}{5} \\\\\frac{B_{n+1}}{B_1} = 5-----(4)$

The given value for n is 30

substitute 30 for n and eqn (4) in eqn(1)

δ = $\frac{1}{30} In5\\$

$= 0.0536$

substitute 0.0536 for δ in eqn (2)

$C = \frac{0.0536}{\sqrt{39.4784+0.00287} } \\C = 0.0085----(5)$

Consider the formula for damping ratio (C)

C = $\frac{c}{2\sqrt{mk} } ----(6)$

where

c is the damping constant

m is the mass

substitute 100 for m in eqn(6)

$C = \frac{c}{2\sqrt{100k} } \\\\C = \frac{c}{20\sqrt{k} } ----(7)$

compute eqn(5) and eqn(7)

$\frac{c}{20\sqrt{k} } = 0.0085\\\\c = 0.0085(20\sqrt{k} )\\\\c= 0.17\sqrt{k} ----(8)$

The given time to complete 30 cycles in 60s is

period = 60 / 30

= 2s

Formula for spring constant

$k =mw^2_n----(9)$

where

$w_n$ is the undamped natural frequency

consider the formula for damped natural frequency

$w_d=w_n\sqrt{1-C^2}$

rearrange

$w_n = \frac{w_d}{\sqrt{1-C^2} } ----(10)$

substitute eqn (10) to eqn (9)

$k = m(\frac{w_d}{\sqrt{1-C^2} } )\\\\k=\frac{mw^2_d}{1-C^2} ---(11)$

consider the formula for the damped natural frequency

$w_d=\frac{2\pi }{P} ----(12)$

consider eqn (12) in eqn (11)

$k = \frac{m(\frac{2\pi }{P} )^2}{1-C^2}$

substitute 100 for m, 2 for P, 0.0085 for C in the eqn above

$k = \frac{(100)(\frac{2\pi }{2})^2 }{1-(0.0085)^2} \\\\k= \frac{986.96}{0.9999} \\\\k=987N/m$

substitute 987 for k in eqn (8)

$c=0.17\sqrt{987} \\c=0.17(31.4166)\\c=5.3408Ns/m$

Thus, the value for the damping constant is c = 5.3408Ns/m and the spring constant is k is 987N/m

7 0

3 years ago