Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N
Given:
The mass of the truck is m1 = 3162 kg
The speed of the truck is v1i = 12 m/s in East
The mass of the parked car is m2 = 510 kg
The speed of car is v2i = 0 m/s
The speed of car after collision is v2f = 24 m/s in East
To find the speed of the truck after collision.
Explanation:
The final velocity of the truck will be
Thus, the speed of the truck after collision is 8.129 m/s
(a) The speeds of the tips of both rotors; main rotor <u>178.3 m/s</u> and tail rotor <u>218.4 m/s</u>.
(b) The speed of the main rotor is <u>0.52</u> speed of sound, and the speed of the tail rotor is <u>0.64</u> speed of sound.
<h3>Linear speed of main motor and tail rotor</h3>
v = ωr
where;
- ω is the angular speed (rad/s)
- r is radius (m)
v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)
v(main rotor) = 178.3 m/s
v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)
v(tail rotor) = 218.4 m/s
<h3>Speed of the rotors with respect to speed of sound</h3>
% speed (main motor) = 178.3/343 = 0.52 = 52 %
% speed (tail motor) = 218.4/343 = 0.64 = 64 %
Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.
Learn more about linear speed here: brainly.com/question/15154527
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Answer:
See the answers below
Explanation:
We can solve both problems using vector sum.
a)
Let's assume the forces that help the diver dive as positive downward, and the forces that oppose upward, as negative
![F_{resultant}=100+30-85+900\\F_{resultant}=845[N]](https://tex.z-dn.net/?f=F_%7Bresultant%7D%3D100%2B30-85%2B900%5C%5CF_%7Bresultant%7D%3D845%5BN%5D)
The drag force is horizontal d this way in the horizontal direction we will only have the drag force that produces the water stream.
![F_{drag}=50[N]](https://tex.z-dn.net/?f=F_%7Bdrag%7D%3D50%5BN%5D)
b)
Let's assume the forces that propel the rocket upwards as positive and forces like the weight of the rocket and other elements as negative forces.
![F_{resultant}=960+7080-7700\\F_{resultant}=340 [kN]](https://tex.z-dn.net/?f=F_%7Bresultant%7D%3D960%2B7080-7700%5C%5CF_%7Bresultant%7D%3D340%20%5BkN%5D)
