Which of the following shows the conservation of mass during cellular respiration?

A mule is harnessed to a sled having a mass of 246 kg, including supplies. The mule must exert a force exceeding 1210 N at an an

zimovet [89]

(a) 1800 N

The equation of the forces along the vertical direction is:

$F sin \theta + N - mg = 0$

where

$F sin \theta$ is the component of the applied force along the vertical direction

N is the normal force on the sled

mg is the weight of the sled

Substituting:

F = 1210 N

m = 246 kg

$\theta = 30.3^{\circ}$

We find N:

$N=mg-F sin \theta = (246)(9.8)-(1210)(sin 30.3^{\circ})=1800 N$

(b) 0.580

The equation of the forces along the horizontal direction is:

$F cos \theta - \mu_s N = 0$

where

$F cos \theta$ is the horizontal component of the push applied by the mule

$\mu_s N$ is the static frictional force

Substituting:

F = 1210 N

N = 1800 N

$\theta = 30.3^{\circ}$

We find $\mu_s$ , the coefficient of static friction:

$\mu_s = \frac{F cos \theta}{N}=\frac{(1210)(cos 30.3^{\circ})}{1800}=0.580$

(c) 522 N

In this case, the force exerted by the mule is

$F= 6.05 \cdot 10^2 N = 605 N$

So now the equation of the forces along the horizontal direction can be written as

$F cos \theta - F_f = 0$

where

$\theta=30.3^{\circ}$

and $F_f$ is the new frictional force, which is different from part (b) (because the value of the force of friction ranges from zero to the maximum value $\mu N$ , depending on how much force is applied in the opposite direction)

Solving the equation,

$F_f = F cos \theta = (605)(cos 30.3^{\circ})=522 N$

6 0

3 years ago

A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that osc

Alex777 [14]

Answer:

The period of motion is 0.5 second.

Explanation:

Given;

extension of the spring, x = 3.8 cm = 0.038 m

mass of the object, m = 13 g = 0.013 kg

Determine the force constant of the spring, k;

F = kx

k = F / x

k = mg / x

k = (0.013 x 9.8) / 0.038

k = 3.353 N/m

When the object is replaced with a block of mass 20 g, the period of motion is calculated as;

$T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second$

Therefore, the period of motion is 0.5 second.

3 0

3 years ago

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

4 years ago

Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a

Yanka [14]

Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)² .....(2)

T Cosθ = mg .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ = 0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m

6 0

3 years ago

Newton's first law

Bond [772]

Answer:E. an object will remain in uniform motion unless acted upon by a force

Explanation:

7 0

3 years ago