Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Answer: Look where the points are.
Explanation:
Yes potential increases while kinetic decreases
Answer:
<em>11.06m/s²</em>
Explanation:
According to Newtons second law of motion
Given
Mass m = 17kg
Fm = 208N
theta = 36 degrees
g = 9.8m/s²
a is the acceleration
Substitute
208 - 0.148(17)(9.8)cos 36 = 17a
208 - 24.6568cos36 = 17a
208 - 19.9478 = 17a
188.05 = 17a
a = 188.05/17
a = 11.06m/s²
<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>
The only graph that accurately depict the given motion is graph D.
The given parameters;
- initial position of the man = 0
- direction of the man's first displacement = backward
- time of first motion, t₁ = 6 seconds
- velocity of this first displacement = v₁
- time without any motion (<em>zero movement</em>) = 6 seconds
- direction of the second displacement = forward
- velocity of second displacement = 2v₁
Let the acceleration of the first displacement = a
Acceleration of the second displacement = 2a
From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.
The only options with initial motion towards the negative direction are;
The difference between graph B and D;
- in graph B there is a uniform motion for 6 seconds
- in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).
Thus, the only graph that accurately depict the given motion is graph D.
Learn more here: brainly.com/question/21095906
