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Stolb23 [73]
3 years ago
15

Interest groups representing businesses and investors are often among the most successful lobbying groups in foreign policy. Why

would this
be considered a problem?
A. These groups often have interests that are similar to the public interest and they are accountable to the public through voting.
B. These groups' members contribute heavily to many legislators' campaigns and in so doing gain greater access to legislators.
C. These groups work together, promoting a coherent foreign policy agenda.
D. The members of these groups make up the majority of voters in many districts.
Physics
1 answer:
alisha [4.7K]3 years ago
7 0

The members of these groups make up the majority of voters in many districts thus this  be considered a problem.

<u>Option: D</u>

<u>Explanation:</u>

Interest groups play a key role in US politics. Such organizations are made up of wealthy and powerful members who often seek to impose some form of leverage in politicians to promote their goals and agendas. Across the years via many campaigns, they have understood how to speak and manipulate elected leaders and apply leverage to get the kind of legislation that is in their favor. Here the majority of voters in several districts are standing due to group members, as we recognize the interest group belongs to a body in which it uses different methods of lobbying to influence others.

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At the equator earth rotates with a velocity of about 465 m/s.
Dafna1 [17]
The given velocity is 465 m/s.

Part a.
465 \,  \frac{m}{s} =(465 \times 10^{-3} \,  \frac{km}{s})*( 3600 \,  \frac{s}{h} ) = 1674 \,  \frac{km}{h}
Answer: 1674 km/h

Part b.
1674  \frac{km}{h} = (1674 \,  \frac{km}{h})*(24 \,  \frac{h}{day}  ) = 40176 \,  \frac{km}{day}
Answer: 40,176 km/day.

 
3 0
3 years ago
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1) Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way ba
timurjin [86]

Answer:

<h3>B. 19miles</h3>

Explanation:

If Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way back home he out  of gas 6 miles after leaving the supermarket, the distance travel by fred will be the sup of all the distances he covered throughout the journey.

Distance covered by fred = 4miles + 9miles + 6miles

Distance covered by fred = 13miles + 6miles

Distance covered by fred  = 19miles

8 0
3 years ago
During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a
lana66690 [7]
According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
8 0
3 years ago
A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

4 0
2 years ago
If the vertical component of velocity for a projectile is 7.3 meters/second, what is its hang time?
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A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.

So, the hang up time, or the time the object stayed on air is calculated using this equation:

a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
6 0
3 years ago
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