The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only at the surface (i.e. the skin). Since heat loss must be balanced by heat production if an animal is to maintain a constant internal temperature, the relationship between surface area and volume is relevant for physiology. If the surface area of a cube is increased by a factor of 2, by what factor does the volume of the cube change? Give your answer to two significant figures. 1.59

Answers

Answer 1

Answer:

To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

The initial measures,

$\text{Initial Length} = L$

$\text{Initial surface Area} = 6L^2$ (Surface of a Cube)

$\text{Initial Volume} = L^3$

The final measures

$\text{Final Length} = L_f$

$\text{Final surface area} = 6L_f^2$

$\text{Final Volume} = L_f^3$

Given,

$((SA)_f)/((SA)_i) = 2$

Now applying the same relation we have that

$((L_f)/(L_i))^2 = 2$

$(L_f)/(L_i) = √(2)$

The relation with volume would be

$((Volume)_f)/((Volume)_i) = ((L_f)/(L_i))^3$

$((Volume)_f)/((Volume)_i) = (√(2))^3$

$((Volume)_f)/((Volume)_i) = (2√(2))$

$((Volume)_f)/((Volume)_i) = 2.83$

Volume of the cube change by a factor of 2.83

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Answers

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

a. Left support = Right support = 22 kN

b. Left support = 36 kN

Right support = 29 kN

c. Left support force will decrease

Right support force will increase.

d. Right support = 43 kN

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)

0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support, Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)

0.6

Right support = 43 kN

*Which of the following cannot be an example of projectile motion
A. A football flying through the air
B. An apple falling from a tree
C. A pencil rolling on the ground
D.A rocket dropping from its maximum height

Answers

A. football flying through the air

A cause it flying through the also a projectile is a object flying in the air like a arrow for example/also can I get Brainly

Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by magnetic forces. Michelle is a(n)

Answers

Answer:

Explanation:

designer

illusionist

engineer

entrepreneur

salesperson

human

inventor

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) = 514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ / d ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

= 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations : brainly.com/question/4449144

Answer:

$1.7* 10^(-3)$ m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

$w = (D\lambda )/(d)$

$w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))$

$w = 1.7* 10^(-3)$ m

(a) Is the velocity of car A greater than, less than, or the same as thevelocity of car B?
(b) Is the initial position of car A greater than, less than, or equal to the
initial position of car B?
(c) In the time period from t = 0 tot = 1 s, is car A ahead of car B,
behind car B, or at the same position as car B?

Answers

a. ) Is the velocity of car A less than the velocity of car B b. the initial position of car A greater than the initial position of car B c. ahead In the time period from t = 0 tot = 1 s, is car A ahead of car B?.

what is velocity ?

Velocity is the parameter which is different from speed, can be defined as the rate at which the position of the object is changed with respect to time, it is basically speeding the object in a specific direction in a specific rate.

Velocity is a vector quantity which shows both magnitude and direction and The SI unit of velocity is meter per second (ms-1). If there is a change in magnitude or the direction of velocity of a body, then it is said to be accelerating.

Finding the final velocity is simple but few calculations and basic conceptual knowledge are needed.

For more details regarding velocity, visit

brainly.com/question/12109673

#SPJ2

Answer:

a. less than, b. greater than, c. ahead

Explanation:

What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?A. 1.6J
B. 7.9J
C. 15J
D. 20J