Ask question

Ask question

All categories

3 years ago

5

A model airplane of mass 0.6 kg is attached to a horizontal string and flies in a horizontal circle of radius 6 m, making 1.6 re

volutions every 5 s. (The weight of the plane is balanced by the upward “lift” force of the air on the wings of the plane.) The accelaration due to the gravity is 9.81 m/s2 . Find the speed of the plane. Answer in units of m/s.

1 answer:

nikitadnepr [17]3 years ago

5 0

Answer:

$speed= 12.15\frac{m}{s}$

Explanation:

In this question we have given

mass of airplane=.6 Kg

Radius of horizontal circle,r=6m

Time taken to complete 1.6 revolution=5s

Therefore time taken to complete 1 revolution, t= $\frac{5}{1.6}$

t=3.1 s

We have to find the speed of airplane

We will first find the distance covered by airplane in tracing one circle which is equal to circumference of circle

We know that

Circumference,d = $2 \pi \times r$

or $d=2\times 3.14 \times 6$

$d=37.68 m$

We know that speed= $\frac{d}{t}$ ............(1)

Put value of d and t in equation 1

speed= $\frac{37.68}{3.1}$

$speed= 12.15\frac{m}{s}$

You might be interested in

A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36

Semmy [17]

Answer: the average speed of the rat from the information given above is 0.7m/s

Explanation:

position is given as

x(t) = pt² + qt

finding the diffencial of x(t) with respect to t, we have

d(x(t))/dt = 2pt + q

we substitute the p = 0.36m/s² and q= -1.10 m/s

d(x(t))/dt = 2(0.36)t + (-1.10)

so, at t= 1s

d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s

at t= 4s

d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s

To find the average speed,

average speed = (V1 + V2)/ 2

average speed = (1.78 + (-0.38))/2 = 0.7m/s

5 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

What is potential energy

valkas [14]

Answer:

Potential energy is energy that is stored – or conserved - in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. You can think of it as energy that has the 'potential' to do work.

6 0

3 years ago

Read 2 more answers

To understand the formula representing a traveling electromagnetic wave.

Mumz [18]

Answer:

Explanation:

Time period is the reciprocal of frequency

T = 1/F

F = 1/T

but angular frequency w = 2πF

F = w/2π

The detailed steps is as shown in the attached file

8 0

3 years ago

A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

<u>Where:</u>

<em>N = is the number of turns = 1 </em>

<em>ΔΦ = ΔB*A </em>

<em>Δt = is the time = 0.3 s </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T </em>

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

4 years ago