Answer:
a. 12.57 m/s b. 39.5 m/s² c. Her centripetal force is four times her weight.
Explanation:
a. What is Missy's linear speed on the rotor?
Missy's linear speed v = 2πr/T where r = radius = 4.0 m and T = time it takes to complete one revolution = 2.0 s
So, v = 2πr/T
= 2π(4.0 m)/2.0 s
= 4π m/s
= 12.57 m/s
b. What is Missy's centripetal acceleration on the rotor?
Missy's centripetal acceleration, a = v²/r where v = linear velocity = 12.57 m/s and r = radius = 4.0 m
a = v²/r
= (12.57 m/s)²/4.0 m
= 158.01 m²/s² ÷ 4.0 m
= 39.5 m/s²
c. If her mass is 50-Kg, how is the centripetal force compare to her weight?
Her centripetal force F = ma where m = mass = 50 kg and a = centripetal acceleration = 39.5 m/s².
Her weight W = mg where m = mass = 50 kg and g = acceleration due to gravity = 9.8 m/s².
So, comparing her centripetal force to her weight, we have
F/W = ma/mg
= a/g
= 39.5 m/s² ÷ 9.8 m/s²
= 4.03
≅ 4
So her centripetal force is four times her weight.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
Answer:
alternating mountain ranges and valley.
Hope u find this helpful
The cat has two directions of motions:
The horizontal motion = Dx = 2.2 m
The vertical motion = Dy = -1.3 m (negative sign indicates that the cat is falling)
a = 9.8 m/sec^2
Vy = zero (since you are not moving up)
From the laws of motion:
<span>Dy = Vyt + 0.5ayt^2
</span>-1.3 = 0(t) + 0.5(-9.8)t^2
<span>t = 0.52s
</span>
Then, again using the laws of motion (but for the horizontal direction this time)
Dx = Vxt
<span>2.2 = Vx0.52 </span>
<span>Vx = 2.2/0.52 </span>
<span>= 4.23 m/s
</span>
<span>Therefore the cat's speed when it slid off the table is 4.23 m/s horizontally.</span>
