Срочно нужно решение задания 5, пожалуйста!!!!!

Answer:

14178 N

Explanation:

5560 Kg * 2.55m/s^2 = 14178 Newtons (Kg*m/s^2)

Answer:

The altitude of the plane is 379.5 m.

Explanation:

Initial horizontal velocity, u = 59.1 m/s

Horizontal distance, d = 521 m

let the time taken by the packet to cover the distance is t.

Horizontal distance = horizontal velocity x time

521 = 59.1 x t

t = 8.8 s

let the vertical height is h .

Use second equation of motion in vertical direction.

Final answer:

The plane must be flying at an altitude of approximately 194.89 m above the pond.

Explanation:

To successfully drop the care package in the small lake, the plane must be flying at a certain altitude above the pond. We can use the horizontal distance and the ground speed of the plane to calculate the time it takes for the package to reach the target location. Then, using the time and the equation for free fall, we can find the altitude above the pond.

First, we calculate the time it takes for the package to reach the target location:

1. Horizontal distance = 521 m
2. Ground speed = 59.1 m/s
3. Time = Horizontal distance / Ground speed = 521 m / 59.1 m/s = 8.81 s

Next, we use the equation for free fall to find the altitude above the pond:

1. Acceleration due to gravity, g = 9.8 m/s^2 (assuming no air resistance)
2. Time for free fall, t = 8.81 s
3. Altitude above the pond = 0.5 * g * t^2 = 0.5 * 9.8 m/s^2 * (8.81 s)^2 = 194.89 m

Therefore, the plane must be flying at an altitude of approximately 194.89 m above the pond in order to successfully accomplish this feat.

Learn more about altitude above the pond here:

brainly.com/question/18722834

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Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²