Answer:
the angle of incident is 40°
Explanation:
NQ is the normal to the mirror, therefore
angle NQA =90°
PQA = 50°
incident angle = NQA - PQA
90°- 50° = 40°
note that the angle of reflection is equal to the angle of incident
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² =
/ r
ω = √(
/ r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a
= g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
In a velocity-time graph, the area under the curve represents the distance.
The distance traveled from 10s to 18 s is

Final Answer: The distance traveled is 40 m from time 10 s to 18 s.