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3 years ago

Under the Neutrality Acts, all of the following conditions were agreed to except which?

1 answer:

3 0

Under the Neutrality Act, all of the following conditions were agreed to except America would provide aid to any nation attacked by Axis Powers. The correct option in regards to all the options given in the question is option"a". During the 1930's the United Nations Congress passed the Neutrality Act. It was passed keeping in mind the turmoil Europe and Asia were passing through during that time. Eventually the turmoil led to the starting of World War II. The Acts were modified several times and ultimately withdrawn when the United States ships were attacked by German submarines.

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An 11.0 -W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandes

yarga [219]

The amount or cost that the user of the energy-efficient bulb save during 100h of use will be $0.319.

<h3>How to calculate the cost?</h3>

For the 11.0W bulb, it should be noted that the value will be:

= 11.0 × 100 × (1/1000) × 0.110

= $0.121

The 40W bulb will be:

= 40 × 100 × (1/1000) × 0.110

= $0.44

Therefore, the amount that will be saved will be:

= $0.44 - $0.121

= $0.319

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6 0

2 years ago

Which describes the movement of a fluid during convection?

VikaD [51]

The movement of a fluid during convection is a circular/oval motion since the fluid at the top sinks and the fluid at the bottom rises.

Hope this helps :)

3 0

3 years ago

Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar

kolbaska11 [484]

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 $E = 200000 MPa$

The strain of the bar when subjected to 150 MPa is

$\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075$

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

$\epsilon = \frac{\Delta L}{L}$

$L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm$ or 1.8m

5 0

2 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

2 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

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7 0

3 years ago