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boyakko [2]
2 years ago
14

A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field

created by the positive test charge also specify the direction of the electric field with respect to the force

Physics
1 answer:
Sophie [7]2 years ago
8 0

Explanation:

direction of electric field is same as that of force experienced by the test charge

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A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of
djverab [1.8K]

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

F_c=ma_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN

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3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
adoni [48]

Answer:

B) R1 = 6 V and R2 = 6V

Explanation:

In series, both resistors will carry the same current.

that current will be I = V/R = 12 / (10 + 10) = 0.6 A

The voltage drop across each resistor is V = IR = 0.6(10) = 6 V

6 0
3 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

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Answer:

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Explanation:

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mart [117]

Answer: 3.92 N.

Explanation:

Your box weighs 400g, or 0.4kg. In order to lift it, you need to overcome the force of gravity. F = ma, and acceleration due to gravity is -9.8 m/s^2. So gravity acts on the box with a force of 0.4 kg * -9.8 m/s^2 = -3.92 N. A force of +3.92 N is required to overcome this.

5 0
3 years ago
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