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FrozenT [24]
4 years ago
12

In example 20.3 in the text, the net force on a 1.0 nC charge located between two 10 nC charges is calculated. How would the ans

wer change if the two 10 nC charges were replaced by two -30 nC charges, leaving the 1.0 nC charge the same
Physics
1 answer:
VashaNatasha [74]4 years ago
7 0

Answer:

The forces experienced by the middle particle are attractive, and the net force will remain the same (0) if and only if the distances of the sides particles to the middle particle are the same.

Explanation:

In example 20.3 the forces experienced by the middle particle are repulsive because all the particles are positive, for the case in which the particles on the sides are replaced for negative charge particles the forces experienced by the middle particle are attractive. Regarding the net force, because we don't know the distances we can not give a definitive answer, what we can say is that if the distances from the middle particle to the sides particles are the same the net force is zero for both cases (remain unchanged).  

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A ball is kicked off of a roof at 23 m/s [R 25° U]. What is the height of
Ymorist [56]

Answer:

Explanation:

Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:

v_{0y}=v_0sin\theta and filling in:

v_{0y}=23sin(25) which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:

v_{0y}=9.7m/s

a = -9.8 m/s/s

t = 2.8 seconds

Δx = ?? m

The one-dimensional motion equation that utilizes all of these variables is

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 9.7(2.8)+\frac{1}{2}(-9.8)(2.8)^2 I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have

Δx = 27 + \frac{1}{2}(-9.8)(2.8)^2 and then to the right of the + sign and to 2 significant digits we have

Δx = 27 - 38 so

Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.

6 0
3 years ago
ANSWER QUICK 30 POINTS
labwork [276]

Answer:

Gravity controls the movement of the planets around the sun, holds together stars grouped in galaxies, and galaxies grouped in clusters.  The Universal Law of Gravitation depends on two things.  First it depends on mass of each object  and the second factor is the distance between two objects.  If the mass of one object is Larger, the gravitational pull towards it will be larger  and the smaller distance, the larger the gravitational pull will be between the objects. Therefore the Larger planets have more moon and the inner planets have less.

Explanation:

6 0
3 years ago
a runner makes one lap around a 200m track in 25s, what is the runners (a) average speed and (b) average velocity
TEA [102]

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5 0
3 years ago
What factors will increase the amount of energy by kinetic
marissa [1.9K]

Gravitational energy is the potential energy associated with gravitational force. If an object falls from one point to another inside a gravitational field, the force of gravity will do positive work on the object and the gravitational potential.

Found this on cha-cha.com

4 0
3 years ago
A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving
Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

m_av_a + m_sv_s = 0

Where m_a = 92kg is the mass of the astronaut, m_s = 1200kg is the mass of the satellite, v_s = 0.14 m/s is the speed of the satellite. We can calculate the speed v_a of the astronaut:

v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0
3 years ago
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