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2 years ago

What is the main source of fuel for cellular respiration?

Physics

2 answers:

Tamiku [17]2 years ago

6 0

The primary source of fuel for cellular is glucose, but oxygen is also utilized. Hope this helps! :)

OleMash [197]2 years ago

3 0

The main source of fuel for cellular respiration is glucose

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A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e = v_c/m + vc_e

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

5m*vs_e - 4m*(v_c/m) = 5m*vc_e

vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

vc_e = 3760 - 0.8*(90)

vc_e = 3,688 km/h

5 0

3 years ago

Read 2 more answers

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

2 years ago

The amount of energy which is transported past a given area of the medium per unit of time is known as the ________ of the sound

Aliun [14]

The term that describes the amount of energy transported past a given area of the medium per unit time would be "intensity." In addition, the formula for computing intensity would be:

Intensity = Energy / (Time * Area)

It can be implied that the wave would be more intense when its energy transfer rate gets increased and vibration amplitudes also increases.

6 0

2 years ago

Which factors describe a gas

Alekssandra [29.7K]

Answer:

B. changing shape and changing volume

Explanation:

*no definite shape (takes the shape of its container)

*no definite volume

7 0

3 years ago

Determine the magnitude of the electrostatic force on a 0.06000 C charged object when it is placed in an electric field of magni