concave <span>ray diagrams were constructed in order to determine the general location, size, orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is a definite relationship between the image characteristics and the location where an object placed in front of a concave mirror. but, convex</span><span>ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by concave mirrors. The ray diagram constructed earlier for a convex mirror revealed that the image of the object was virtual, upright, reduced in size and located behind the mirror. </span>
Answer:
HERE IS YOUR ANSWER
Explanation:
PLEASE MARK MY ANSWER AS BRAINLIEST IF THE ANSWERS ARE CORRECT .
Beacuse of the loose connection of the wire .
Straight
Answer:
<em>Thus, the object is accelerating to the left</em>
Explanation:
<u>The Net Force</u>
The net force is the result of adding all the forces as vectors acting on a body.
Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.
Second Newton's law gives the relation between the net force and the acceleration of the body:
We can see the acceleration is a vector with the same direction as the net force.
The diagram shows two vertical forces and two horizontal forces.
The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.
The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N
Thus, the object is accelerating to the left
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
Answer:
Explanation: I think...
Thermal Energy formula Q = mcΔT
Q = Thermal Energy(J)
m = Mass(kg)
c = Specific Heat(J/kg°C)
ΔT = Change in Temperature(°C)
you have to write the equation based on what you are working on
