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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 The Acts were modified several times and ultimately withdrawn when the United States ships were attacked by German submarines.

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A charge of +3.00 μC is located at the origin, and a second charge of -2.00 μC is located on the x-y plane at the point (50.0 cm

romanna [79]

Answer:

0.0567 N

Explanation:

q1 = 3 micro coulomb

q2 = - 2 micro coulomb

OB = 50 cm

AB = 60 cm

By using Pythagoras theorem in triangle OAB

$OA^{2}=OB^{2}+AB^{2}$

$OA^{2}=50^{2}+60^{2}=6100$

OA = 78.1 cm

By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is

$F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}$

F = 0.0885 N

The horizontal component of force is

= F CosФ = $F\times \frac{OB}{OA}$

= 0.0885 x 50 / 78.1 = 0.0567 N

8 0

3 years ago

A 65-cm segment of conducting wire carries a current of 0.35

algol13

The intensity of the magnetic force exerted on the wire due to the presence of the magnetic field is given by
$F=ILB \sin \theta$
where
I is the current in the wire
L is the length of the wire
B is the magnetic field intensity
$\theta$ is the angle between the direction of the wire and the magnetic field

In our problem, L=65 cm=0.65 m, I=0.35 A and B=1.24 T. The force on the wire is F=0.26 N, therefore we can rearrange the equation to find the sine of the angle:
$\sin \theta= \frac{F}{ILB}= \frac{0.26 N}{(0.35 A)(0.65 m)(1.24 T)}=0.922$

and so, the angle is
$\theta=\arcsin(0.921)=67.1^{\circ}$

6 0

3 years ago

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The eagle drops the trout a height of 6.1 m the fish travels 7.9 m horizontaly before hitting the water what is the velocity of

Semenov [28]

The velocity of the eagle is 7.0 m/s

Explanation:

The motion of the fish is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction, where the horizontal velocity of the fish is equal to the horizontal velocity of the eagle

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion, using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where:

s = 6.1 m is the vertical displacement of the fish

u = 0 is the initial vertical velocity of the fish

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t, we find the time of flight of the fish:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(6.1)}{9.8}}=1.12 s$

Now we know that during this time, the fish travels a horizontal distance of

x = 7.9 m

Therefore, the horizontal velocity of the fish is

$v_x = \frac{x}{t}=\frac{7.9}{1.12}=7.0 m/s$

And therefore, this is the initial velocity of the eagle.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

4 years ago

A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat

Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial speed is 3.35 m/s

8 0

3 years ago

When observing a group of children at a daycare center, Emily made the following observations: Five year old children played in

konstantin123 [22]

This question is not about physics science.

The answer is: option <span>a. Five-year-old children have longer attention spans than three-year-old children.

It is the attention ability what let the older children to stay longer in one location instead of being moving between different activities. The younger children who cannot keep their attention long time in a same activity entertain themselves by changing activities.
</span>

5 0

3 years ago

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