You would use answer 2O g/cm3
Answer:
96%
Explanation:
To find the values of the motor efficiency you use the following formula:
![E=\frac{P_o}{P_i}100](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BP_o%7D%7BP_i%7D100)
P_o: output power = 864J/0.5min=864J/30s=28.8W
P_i: input power = I*V = (3A)(12V) = 36W
By replacing this values you obtain:
![E=\frac{28.8W}{30W}*100=96\%](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B28.8W%7D%7B30W%7D%2A100%3D96%5C%25)
hence, the motor efficiency is about 96%
traslation:
Pentru a găsi valorile eficienței motorului, utilizați următoarea formulă:
P_o: putere de ieșire = 864J / 0.5min = 864J / 30s = 28.8W
P_i: putere de intrare = I * V = (3A) (12V) = 36W
Înlocuind aceste valori obțineți:
prin urmare, eficiența motorului este de aproximativ 96%
The missing diagram is in the attachments.
Answer: X: positive Y: positive
Explanation: Electric field is a vector quantity, which means it can be represented by a vector arrow: the arrow points in the direction of electric field and its length represents the magnitude at a given location. There are another representation of the electric field called electric field lines, <u>in which the line points away from a positively charged source and towards a negatively charged source</u>. This occurs because it follows a pattern, where the lines points in the direction that a positive test charge would have if it is accelerating on the line.
Analyzing the diagram, it can be observed that the lines are pointing away from both of the charged objects. Therefore, both X and Y are <u>positively charged</u>.
Answer:
A)6.15 cm to the left of the lens
Explanation:
We can solve the problem by using the lens equation:
![\frac{1}{q}=\frac{1}{f}-\frac{1}{p}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7Bf%7D-%5Cfrac%7B1%7D%7Bp%7D)
where
q is the distance of the image from the lens
f is the focal length
p is the distance of the object from the lens
In this problem, we have
(the focal length is negative for a diverging lens)
is the distance of the object from the lens
Solvign the equation for q, we find
![\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7B-16.0%20cm%7D-%5Cfrac%7B1%7D%7B10.0%20cm%7D%3D-0.163%20cm%5E%7B-1%7D)
![q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B1%7D%7B-0.163%20cm%5E%7B-1%7D%7D%3D-6.15%20cm)
And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is
A)6.15 cm to the left of the lens