I will give 20 points and brainliest

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

The displacement at any given time of an object is x = 6 sin 98t , where the symbols have their usual meanings. i). Proof that t

Fofino [41]

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation, which are represented in the given function as 6 units and 98 rad/s respectively.

<h3>General wave equation for simple harmonic motion</h3>

y = A sinωt

where;

A is amplitude of the motion
ω is angular frequency

<h3>Amplitude of the oscillation</h3>

A = 6 units

<h3>Angular frequency of the wave</h3>

ω = 98 rad/s

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation. Thus, the wave is executing simple harmonic motion.

Learn more about simple harmonic motion here: brainly.com/question/17315536

#SPJ1

5 0

2 years ago

Newton’s Second law of Motion (F = ma) is one of the basis of forces and motion. How is it very similar to his First law and why

koban [17]

Answer:

First law can be deduced from second law.

Acceleration may determine the position and velocity of the system.

Explanation:

When net force is zero, the second law is 0 = ma, or the motion is at constant speed. Thus first law establishes that when there are no forces, the object moves at constant speed, so first law is explained by using the second.

If you determine the acceleration of a system, you may use calculus or kinematic equations to determine velocity and position of the particle and determine how it moves. This is very important in mechanics and engineering, for example, for spacecrafts, forensic situations, etc.

4 0

3 years ago

What is the mass of a dog that weighs 382 N?(unit=kg)

tino4ka555 [31]

Answer:

38.77 kg

Explanation:

The weight of an object is given by:

Weight = mg, g = acceleration due to gravity ( 9.8 m/s square)

and m is the mass.

So, in this case:

Weight of dog is 382 N, so, by putting the values of weight and g into the equation we will get the mass of dog.

382 = mass multiply with 9.8

so, the mass of the dog will be 38.77 kg

4 0

4 years ago

When a light bulb is connected to a 4.5 V battery, a current of 0.16 A passes through the bulb filament. What is the resistance

larisa [96]

Answer:

R = 28.125 ohms

Explanation:

Given that,

The voltage of a bulb, V = 4.5 V

Current, I = 0.16 A

We need to find the resistance of the filament. Using Ohm's law,

V = IR

Where

R is the resistance of the filament

So,

$R=\dfrac{V}{I}\\\\R=\dfrac{4.5}{0.16}\\\\R=28.125\ \Omega$

So, the resistance of the filament is equal to 28.125 ohms.

5 0

3 years ago