Answer:
C. The block and the sphere would have the same weight.
Explanation:
If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.
Answer:
v = 2 v₁ v₂ / (v₁ + v₂)
Explanation:
The body travels the first half of the distance with velocity v₁. The time it takes is:
t₁ = (d/2) / v₁
t₁ = d / (2v₁)
Similarly, the body travels the second half with velocity v₂, so the time is:
t₂ = (d/2) / v₂
t₂ = d / (2v₂)
The average velocity is the total displacement over total time:
v = d / t
v = d / (t₁ + t₂)
v = d / (d / (2v₁) + d / (2v₂))
v = d / (d/2 (1/v₁ + 1/v₂))
v = 2 / (1/v₁ + 1/v₂)
v = 2 / ((v₁ + v₂) / (v₁ v₂))
v = 2 v₁ v₂ / (v₁ + v₂)
Answer:
Technician A says that this is the normal operation of the ETC self -test is the correct answer.
Explanation:
An engine control unit (ECU), also widely referred to as an engine control module (ECM), is a type of electronic control device that controls an internal combustion engine with a series of actuators to ensure maximum engine performance.
It achieves so by reading values from a multitude of sensors within the engine bay, translating data using multidimensional feedback maps (the so-called lookup tables) and modifying the actuators.
Mechanically fixed and dynamically regulated by mechanical and pneumatic means, air-fuel combination, ignition time, and idle speed were before ECUs.
As soon as the system gets battery voltage, after ignition is turned, the efi computer makes a self-test of all the actuators and sensors, included the ETC.
Answer:
<em>The penny will hit the ground at 6.39 seconds</em>
Explanation:
<u>Free Fall</u>
The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:
Where 
, and t is the time.
Solving for t:
Using the value y=200:
t=6.39 sec
The penny will hit the ground at 6.39 seconds
Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E = 
indicates that the field is E for r = 2m
E = 
(1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C
