Answer:
V₀y = 0 m/s
t = 2.47 s
V₀ₓ = 61.86 m/s
Vₓ = 61.86 m/s
Explanation:
Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL
<u>V₀y = 0 m/s</u>
For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:
Y = V₀y*t + (0.5)gt²
where,
Y = Height = 30 m
g = 9.8 m/s²
t = time to hit the ground = ?
Therefore,
30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
t² = 30 m/4.9 m/s²
t = √6.122 s²
<u>t = 2.47 s</u>
For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,
V₀ₓ = Xt
where,
V₀ₓ = Initial vertical Velocity = ?
X = Horizontal Distance = 25 m
Therefore,
V₀ₓ = (25 m)(2.47 s)
<u>V₀ₓ = 61.86 m/s</u>
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Due, to uniform motion in horizontal direction:
Final Vertical Velocity = Vₓ = V₀ₓ
Vₓ = 61.86 m/s
Answer:
Explanation:
Time to cover first 100 km = 1 hour.
time remaining = 3.15 - 1 = 2.15 hour .
Time to cover next 42 km = 1 hour .
Time remaining = 2.15-1 = 1.15 hour.
Distance to be covered = 310 - 142
= 168 km
least speed needed = distance remaining / time remaining
= 168 / 1.15
= 146.08 km / h .
The speed change : Δv = 0.41 m/s
<h3>Further explanation</h3>
Given
mass = 5.5 kg
Force = 15 N
time = 0.15 s
Required
the speed change
Solution
Newton 2nd's law
Impulse and momentum
F = m.a
F = m . Δv/t
F.t = m.Δv
Input the value :
15 N x 0.15 s = 5.5 kg x Δv
Δv = 0.41 m/s
Answer:
Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)
Answer:
C) 26.6
Explanation:
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