What dose nuclear reactions produce?

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

Tju [1.3M]

Answer:

$_T}=24.57Nm$

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

$_T } = Fr$

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

$_T}=(0.795)*(30.9)$

$_T}=24.57Nm$

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

$_T}=I \omega$

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

$I = mr^2$

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

$_T}=I \omega$

Making angular acceleration the subject of the formula; we have;

$\omega = \frac{_T}{I}$

ω = $\frac{24.57}{707.51}$

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a = ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

5 0

3 years ago

Which molecule has charges that are free to move around?

borishaifa [10]

A polar molecules have charges that are free to move around

6 0

4 years ago

a length 650 cm of thin thread wraps around a cylinder exactly 30 times calculate the circumference and the radius of the cylind

Eduardwww [97]

Answer:

circumference= 65/3 cm = 21.67 cm

radius R = 3.45 cm

Explanation:

To calculate the length of the circumference of the cylinder, we divide 650 cm by 30 (the number of times it wrapped exactly around it)

length of circumference= 65/3 cm = 21.67 cm

now use the formula of the circumference length to find the radius (R):

circumference length = 2 * pi * R

65/3 = 2 * pi * R

R = 65 / (6 pi)

R = 3.45 cm

5 0

3 years ago

How can magnetic levitation be improved?

Ostrovityanka [42]

Https://www.researchgate.net/publication/335238337_A_New_Strategy_for_Improving_the_Tracking_Performance_of_Magnetic_Levitation_System_in_Maglev_Train/fulltext/5d5a958d299bf1b97cf546ba/A-New-Strategy-for-Improving-the-Tracking-Performance-of-Magnetic-Levitation-System-in-Maglev-Train.pdf?origin=publication_detail

The higher the phase margin the more stable is the system and for these tuned parameters, the phase margin is around
. Some researcher given their theory on the phase margin that there are changes of getting sluggish response for larger phase margin but using TLBO algorithm the settling time and as well as peak overshoot of the system shows better response as compared to conventional techniques.

6 0

3 years ago

A 1000-kg car is moving at 40.0 km/hr when the driver slams on the brakes and skids to a stop (with locked brakes) over a distan

liq [111]

Answer:

skid distance = 180 m

Explanation:

given,

mass of the car = 1000 Kg

speed of the car = 40 Km/hr

distance to stop = 20 m

if speed is now = 120 Km/he

distance to stop = ?

from given question we can state that,

Work done by the friction is equal to the change in kinetic energy of the car.

$W= \dfrac{1}{2}mv^2$ ............(1)

And we also know that work is directly proportional to displacement

W = F × x...........................(2)

from the equation (1) and (2)

x ∝ v²

now,

here velocity is increased from 40 km/h to 120 km/h means velocity is increase 3 times

so displacement will be equal to 3² or 9 times

hence, skid distance will be equal to 9 x 20 = 180 m

skid distance = 180 m

8 0

3 years ago