The range of the projectile is 188 m
Explanation:
The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
The path of a projectile is the combination of these two motions: see figure in attachment.
In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.
We have:
t = 5.0 s (time of fligth of the projectile)
and the horizontal velocity is constant, and it is given by

where
is the initial velocity
is the angle of projection
Substituting,

And therefore, the range of the projectile is:

Learn more about projectile motion:
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<u>Explanation</u>
- The relationship between the strength of a bond (single vs double vs triple) and its wave-number on an IR spectrum as the bond strength increases the wave number increases.
STRENGTH OF BONDS TRIPLE>DOUBLE>SINGLE
WAVE NUMBER SINGLE>DOUBLE>TRIPLE
- wave number for single bond is greatest because it has greatest bond frequency among the three( more the frequency greater is the wave number).
Answer:

Explanation:
As we know that Far sighted person has near point shifted to 80 cm distance
so he is able to see the object 80 cm
now the distance of lens from eye is 2 cm
and the person want to see the objects at distance 10 cm
so here the image distance from lens is 80 cm and the object distance from lens is 8 cm
now from lens formula we have



I’m assuming you’re supposed to calculate the resultant force?
425N (right) -300N (left)
=125 N to the right
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
<h3>
Heat released by the water when it freezes</h3>
The heat released by the water when it freezes is calculated as follows;
Q = mcΔФ
where;
- m is mass of water
- c is specific heat capacity of water
- ΔФ is change in temperature = Фf - Фi
when water freezes, the temperature, Фf = 0 °C
Q = 82 x 4200 x (0 - 0)
Q = 0
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
Learn more about heat flow here: brainly.com/question/14437874