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2 years ago

Identify the stage in cellular respiration that produces carbon dioxide as a waste product

1 answer:

6 0

During cellular respiration, organisms use oxygen to turn glucose into carbon dioxide, water, and energy in the form of ATP. The process has three stages: glycolysis , the Krebs cycle, and the electron transport chain. Glycolysis in the cytoplasm ), breaks down 1 glucose into 2 pyruvate and 2 ATP. The Krebs cycle (in the mitochondrion's matrix), provides the hydrogen and electrons needed for the electron transport chain. Another 2 are formed here. The electron transport chain (on the inner mitochondrial membrane) forms 32 ATP through oxidative phosphorylation .

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Astronauts are trained for take-off in a high-speed centrifuge of 4.7 m radius that spins in the horizontal plane.

Leokris [45]

Answer:

a) $1.94 \frac{rad}{s}$

b) $9.12\frac{m}{s}$

c) Towards the center of the centrifuge

Explanation:

Becuse the centrifuge rotates in circular motion, there's an angular acceleration tha simulates high gravity accelerations

$a_{rad}=\omega r^{2}$

with r the radius and ω the amgular velocity, in or case $a_rad=3.5g$ so:

$3.5g=\omega r^{2}$ and g=9.8 $\frac{m}{s^{2}}$

solving for ω:

$\omega=\frac{3.5g}{r^2}=\frac{3.5*9.8}{4.2^2}$

$\omega = 1.94 \frac{rad}{s}$

b) Linear speed (v) and angular speed are related by:

$v=\omega r =(1.94)(4.7)$

$v= 9.12\frac{m}{s}$

c) The apparent weigth is pointing towards the center of the circle, becuse angular acceleration is pointing in that direction.

8 0

3 years ago

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago

Scientists have discovered just over _____ different elements with unique properties.

Y_Kistochka [10]

100. (: hope this helps

8 0

3 years ago

Read 2 more answers

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

$\triangle E = 12.79 J$

Explanation:

Sprinters' tendons stretch, $x_s = 43 mm = 0.043 m$

Non athletes' stretch, $x_n = 32 mm = 0.032 m$

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, $E_s = 0.5kx_s^2$

Maximum energy stored in the non athletes, $E_m = 0.5kx_n^2$

Difference in maximum stored energy between the sprinters and the non-athlethes:

$\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J$

4 0

3 years ago

An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

$T=W=mg=(5)(9.81)=49.05N$

With the value of the tension and the velocity you can find the mass density:

$v=\sqrt{\frac{T}{\mu}$

$v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m$

6 0

3 years ago