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2 years ago

A cat is fed a diet designed for dogs. Which amino acid is most likely to be deficient in this cat's diet?

Engineering

1 answer:

nalin [4]2 years ago

3 0

Answer:

Your cat may also be a risk of a taurine deficiency if he or she enjoys sampling your dog's food. Dog food doesn't contain taurine, as dogs can create their own supply of the amino acid. Although your cat may like the taste, a steady diet of dog food is a bad idea

hope this helps

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What are the three most common types of relearn procedures?

WITCHER [35]

Answer:

The three types of relearn procedures are auto relearn, stationary and OBD.

Explanation:

In TPMS system, after the direct service like adjustment of air pressure, tire rotation or replacement of sensors etc, is performed then maximum vehicle often needs TPMS system relearn that needs to be performed.

For performing these relearn procedure, there are mainly three types:

auto relearn
stationary relearn
OBD

After applying the relearn process, the TPMS system will again be in proper function.

8 0

4 years ago

Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers

4vir4ik [10]

Answer:

2132hp ed enregia

Explanation:

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6 0

4 years ago

The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and

garik1379 [7]

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, $x_1$ = 39 mm

$x_2$ = 225 mm

a). From the work energy principal,

Work forces = kinetic energy

$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$

$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$

$9.75= 7.5(V^2_2-0.08^2)$

$1.3= V^2_2-0.08^2$

$V_2=1.14\ m/s$

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of $V_2$ in (1),

$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$

$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$

$98.43x - 100(4x^2+0.156x)+0.048=0$

$98.43x - 400x^2-15.6x+0.048=0$

$82.83x - 400x^2+0.048=0$

$400x^2- 82.83x-0.048=0$

x = 0.20 m

4 0

3 years ago

The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.

emmasim [6.3K]

Answer:

The pressure inside the cooker is 1.0804 atm.

Explanation:

Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.

For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.

So, Standard conditions:

T₁ = 373.15 K

P₁ = 1 atm

Given ,

The water boils at Temperature = 130 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So, the temperature, T₂ = (130 + 273.15) K = 403.15 K

To find pressure inside the cooker (P₂) :

Applying Amontons's Law as:

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

So,

$P_2=\frac {P_1}{T_1} \times {T_2}$

$P_2=\frac {1 atm}{373.15 K} \times {403.15 K}$

$P_2=1.0804 atm$

Thus, The pressure inside the cooker is 1.0804 atm.

6 0

3 years ago

The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] h

marshall27 [118]

Answer:

ΔT = 62.11°C

Explanation:

Given:

- Brass Shell:

Inner Diameter d_i = 32 mm

Outer Diameter d_o = 39 mm

E_b = 93 GPa

α_b = 15.1*10^-6 / °C

- Ceramic Core:

Outer Diameter d_o = 32 mm

E_c = 310 GPa

α_c = 3.2*10^-6 / °C

- Unstressed @ T = 8°C

- Total Length of the cylinder L = 160 mm

Find:

Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

Solution:

- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:

б_b = - 60 MPa

- The force equilibrium can be written as:

б_b*A_b + б_c*A_c = 0

Where, б_b is the stress in core

A_b is the cross sectional area of the shell

A_c is the cross sectional area of the core

б_b*pi*( d_o^2 - d_i^2) / 4 + б_c*pi*( d_i^2) / 4 = 0

б_b*( d_o^2 - d_i^2) + б_c*( d_i^2) = 0

б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)

Plug in the values:

б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)

б_c = 29.121 MPa , б_b = - 60 MPa

- The total strains in both brass shell and ceramic core is given by:

ξ_b = α_b*ΔT + б_b / E_b

ξ_c = α_c*ΔT + б_c / E_c

- The compatibility relation is:

ξ_b = ξ_c

α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c

ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b

ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )

Plug in values and solve:

ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )

ΔT = 62.11°C

8 0

3 years ago