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4 years ago

The energy that is released by cellular respiration is the form of

Physics

1 answer:

Y_Kistochka [10]4 years ago

6 0

I believe the energy released in cellular respiration is in the form of ATP.

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Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun

yanalaym [24]

The answer is convection

6 0

3 years ago

During the time interval from 0.0 to 10.0 s, the position vector of a car on a road is given by x(t) = a + bt + ct2, with a = 17

Juli2301 [7.4K]

The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

<h3>Average velocity of the car</h3>

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

<h3>Average velocity</h3>

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1 + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

Learn more about velocity here: brainly.com/question/4931057

#SPJ1

3 0

2 years ago

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

For a bronze alloy, the stress at which plastic deformation begins is 294 MPa and the modulus of elasticity is 121 GPa. (a) What

FromTheMoon [43]

Answer:

a) load in Newton is 96,138 b) 129.314mm

Explanation:

Stress = force/ area (cross sectional area of the bronze)

Force(load) = 294*10^6*327*10^-6 = 96138N

b) modulus e = stress/ strain

Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3

Strain = change in length/ original length = DL/ 129

Change in length DL = 129 * 2.34*10^ -3 = 0.31347

Maximum length = change in length + original length = 129.314mm

7 0

3 years ago

A rectangular piece of cardboard, whose area is 352 square centimeters, is made into an open box by cutting a 2-centimeter sq

Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

$xy=352$

$y=\dfrac{352}{x}$ -------- (1)

The volume of the rectangular piece

$2(x-4)(\dfrac{352}{x}-4)=432$

$x^2-38x+352=0$

$(x-16)(x-22)=0$

x=16,22

Put the value of x in the equation (I)

For x = 16

$y=\dfrac{352}{16}=22$

For x = 22

$y=\dfrac{352}{22}=16$

Dimension of cardboard is 22 m by 16 m

8 0

3 years ago