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siniylev [52]
3 years ago
10

Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the ele

ctric field due to only one of the positive charges is about 18 V/m. What is the magnitude of the total electric field due to both charges at this location
Physics
1 answer:
yKpoI14uk [10]3 years ago
6 0

Answer:

the magnitude of the total electric field is 25 V/m

Explanation:

Given data

point above = 0.5 m

charges = 18 V/m

to find out

the magnitude of the total electric field

solution

the magnitude of the total electric field is 25 V/m because number is less than    the twice magnitude of field by every charge and  the horizontal component in electric field by every charge cancel out and vertical component in field added together

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What does addition of two vectors give you?
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A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them
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Answer:

B. There is a direct proportion between the mass and force listed in the table.

Explanation:

From the table, the values of force increases with increase in the value of mass.

if 5kg=25 N

Finding the contant of proportionality k;

k=25/5=5

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3 years ago
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2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t
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Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

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Answer:

The last one makes the most sense as they combine two like things that are easy to visualize

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