Which of the following statements uses a connotative word to evoke the most negative feeling?

A spherical gas-storage tank with an inside diameter of 9 m is being constructed to store gas under an internal pressure of 1.50

lutik1710 [3]

Answer: 33 mm

Explanation:

Given

Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m

Internal pressure of gas, P(i) = 1.5 MPa

Yield strength of steel, P(y) = 340 MPa

Factor of safety = 0.3

Allowable stress = 340 * 0.3 = 102 MPa

σ = pr / 2t, where

σ = allowable stress

p = internal pressure

r = radius of the tank

t = minimum wall thickness

t = pr / 2σ

t = 1.5*10^6 * 4.5 / 2 * 102*10^6

t = 0.033 m

t = 33 mm

The minimum thickness of the wall required is therefore, 33 mm

6 0

3 years ago

A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.

babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

$p=\frac{F}{A}$

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

$p\propto \frac{1}{A}$

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

8 0

3 years ago

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z

sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So, 4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0

3 years ago

Please answer these questions for me! I am giving MAX points! Brainliest!

guajiro [1.7K]

I think you have them all marked correctly

7. They carry hereditary material from parent

8. 50 percent

1. Nutrition

2. X-linked dominant

If Im wrong Im sorry. its been like 4 years since i took biology but i remember some things

3 0

3 years ago

Read 2 more answers