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nekit [7.7K]
3 years ago
7

N an experiment, researchers want to determine if the __________ variable causes changes in the __________ variable.

Physics
2 answers:
chubhunter [2.5K]3 years ago
7 0

C

independent; dependent

wlad13 [49]3 years ago
4 0

Answer:

It's C. independent . . . dependent

Explanation:

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A negative charge is at rest at the origin of an axis system. Location x is at coordinate point (2m,3m) while location y is at (
Sergeu [11.5K]
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.

Location ' x ' is  √(2² + 3²) = √13 m  from the charge.

Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.

The magnitude of the E-field is the same at both locations.

The direction is also the same at both locations ... it points toward the origin.


5 0
3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

5 0
3 years ago
What is the name and symbol of the element in the second row and fourteenth column of the periodic table? Hint: Review your peri
Papessa [141]
The answer is Carbon. Beyond being the only element listed here that is located in the second row, it is also in the fourteenth column of the table if you count from left to right. Hope this helps!
4 0
3 years ago
Fossils are dated using radioactive...<br><br> a. hydrogen <br> b. helium<br> c. carbon
OlgaM077 [116]

Answer:

carbon

Explanation:

7 0
3 years ago
Read 2 more answers
A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg
iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

F=\frac{k*q1*q2}{d^{2}}

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0
3 years ago
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