In a survey, the wording of the questions is important. t or f

How do high-energy electrons from glycolysis and the krebs cycle contribute to the formation of atp from adp in the electron tra

ivanzaharov [21]

Answer

Together with glycolysis, The Krebs cycle, and the electron transport chain release about 36 molecules of ATP per molecule of glucose.The Krebs cycle uses the two molecules of pyruvic acid formed in glycolysis and yields high-energy molecules of NADH and flavin adenine dinucleotide (FADH2), as well as some ATP. The electron transport chain forms a proton gradient across the inner mitochondrial membrane, which drives the synthesis of ATP

5 0

4 years ago

An electrical device in which a long wire is wound into a succession of closely spaced loops around a n insulating core is calle

Iteru [2.4K]

Answer:

Solenoid I think.

Explanation:

6 0

3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

Gemiola [76]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done on the car is equal to its variation in kinetic energy, so:

$W=K_f - K_i$

where in this problem:

W = -36,733 J is the work done on the car (negative because the car is slowing down)

$K_f$ is the final kinetic energy of the car

$K_i=66,120 J$ is its initial kinetic energy

Solving for Kf,

$K_f = K_i + W = 66,120 +(-36,733)=29,387 J$

Now we can find the final speed of the car by writing the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

where:

m = 661 kg is the mass of the car

v is the final speed

K = 29,387 J is the kinetic energy

Solving for v,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

5 0

3 years ago

When astronomers observe the spectra of distant galaxies, they notice that the hydrogen emission lines are shifted noticeably to

Neporo4naja [7]

A; They are moving away from us at high speed.

5 0

3 years ago

So I have these kinematics questions and I’m having a hard time answering it. It says:While riding a hot air balloon, which is s

Ludmilka [50]

a)

Objects moving under free fall have a constant acceleration g, which is the gravitational acceleration, directed downwards:

$g=9.81\frac{m}{s^2}$

Since the balloon is descending at a speed of 2.52 m/s, the initial speed of the phone is 2.52 m/s. To find the speed after 4.00 seconds, use the formula that comes from the definition of acceleration:

$v_f=v_0+gt$

Substitute the value for each parameter:

$\begin{gathered} v_f=2.52\frac{m}{s}+(9.81\frac{m}{s^2})(4.00s) \\ =2.52\frac{m}{s^2}+39.24\frac{m}{s} \\ =41.76\frac{m}{s} \end{gathered}$

b)

From the reference frame of the balloon, the cellphone starts falling from rest, and accelerates with a constant acceleration g.

The equation that describes the distance traveled by an object that starts from rest and moves with constant acceleration g is:

$d=\frac{1}{2}gt^2$

Substitute the values of g and t to find how far is the cell phone below the balloon after 4 seconds:

$\begin{gathered} d=\frac{1}{2}(9.81\frac{m}{s^2})(4.00s)^2 \\ =78.48m \end{gathered}$

c)

If the balloon is rising steadly at 2.52 m/s instead of descending, then the acceleration of the cellphone will be directed in the opposite direction of its initial speed. Then, the initial speed and the acceleration must have opposite signs. Considering the upward direction as positive, we have that the initial speed of the phone is 2.52 m/s and its acceleration is -g. Then:

$\begin{gathered} v_f=(2.52\frac{m}{s})-(9.81\frac{m}{s})(4.00s) \\ =2.52\frac{m}{s}-39.24\frac{m}{s} \\ =-36.72\frac{m}{s} \end{gathered}$

The speed is the modulus of the velocity. Then, the speed of the phone after 4 seconds in this situation, would be 36.72 m/s.

On the other hand, in the reference frame of the rising balloon, the phone still starts moving at rest and accelerating uniformly towards the floor. The answer will be the same, and the phone will be 78.48 m below the balloon after 4 seconds.

Therefore, the answers with the correct amount of significant figures, are:

a)

$41.8\frac{m}{s}$

b)

$78.5m$

c)

$\begin{gathered} a) \\ 36.7\frac{m}{s} \end{gathered}$ $\begin{gathered} b) \\ 78.5m \end{gathered}$

8 0

1 year ago