Given that
Velocity of missile (v) = 20 m/s ,
Angle of missile (Θ) = 53°
Determine , Vertical component = v sin Θ
= 20 sin 53°
= 15.97 m/s
Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.
Answer:
a = 1.16 m/s²
Explanation:
In order to find the acceleration of the ball we will use 3rd equation of motion.
2as = Vf² - Vi²
where,
a = acceleration = ?
s = displacement = 21.9 m
Vf = Final Velocity = 7.14 m/s
Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)
Therefore, using the values, we get:
2a(21.9 m) = (7.14 m/s)² - (0 m/s)²
a = (50.97 m²/s²)/(43.8 m)
<u>a = 1.16 m/s²</u>
Answer:
The focal length of the lens should be -51.5 cm (a concave lens).
Explanation:
The purpose of the lens is to make objects at 48.5 cm appear at the healthy near point. The healthy near point is 25.0 cm.
We use the lens formula
where <em>f</em> = focal length, <em>u</em> = object distance and <em>v</em> = image distance.
In this case, <em>u</em> = 48.5 cm and <em>v</em> = -25.0 cm.
<em>v</em> is negative because the image is virtual an not real. (Here, we are using the real-is-positive sign convention)
The negative sign indicates the lens is concave.
