First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed
, and an accelerated motion on the y-axis, with initial speed
and acceleration
:
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
Therefore:
which has two solutions:
is the time of the beginning of the motion,
is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
Voltage = Current x Resistance
<span>Voltage(?) = 100 x 1.98x10^-4 ohms </span>
<span>Voltage = Current x Resistance </span>
<span>Voltage(?) = 250 x 2.09x10^-4 ohms </span>
<span>Voltage = Current x Resistance </span>
<span>Voltage(?) = 100 x 3.44x10^-4 ohms</span>
Answer:
A unit is represented in kWH or Kilowatt Hour. This is the actual electricity or energy used. If you use 1000 Watts or 1 Kilowatt of power for 1 hour then you consume 1 unit or 1 Kilowatt-Hour (kWh) of electricity.
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved mirror than the centre of curvature.</em>
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During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
1. Is 49.5
2. Is 8.6
4. Is 6.6
Wait I’m not sure
