Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as
Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
The percentage error in his experimental value is -51.97%.
<h3>What is percentage error?</h3>
This is the ratio of the error to the actual measurement, expressed in percentage.
To calculate the percentage error of the student, we use the formula below.
Formula:
- Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1
From the question,
Given:
- Calculated value = 4.15 g/cm
- accepted value = 8.64 g/cm
Substitute these values into equation 1
- Error(%) = (4.15-8.64)100/8.64
- Error(%) = -4.49(100)/8.64
- Error(%) = -449/8.64
- Error(%) = -51.97 %
Hence, The percentage error in his experimental value is -51.97%.
Learn more percentage error here: brainly.com/question/5493941
