All categories

2 years ago

An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

Physics

1 answer:

miss Akunina [59]2 years ago

4 0

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

$X = V . \sqrt{\frac{2y_0}{g}}$

Using the given values:

$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

x = $6.7*10^2$ m

You might be interested in

If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these

insens350 [35]

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

$\beta = 10Log(\frac{I}{I_0})$

Where,

I = Intensity of Sound

$I_0$ = Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

$I= 15I_0$

And the total mutual intensity is 100 dB, so we should

$\beta = 100-10*log(\frac{15I_0}{I_0})$

$\beta = 100-10*log(15)$

$\beta = 100-11.76$

$\beta = 88.23dB$

Therefore each one of these engines produce D. 88dB.

5 0

2 years ago

The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the

Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

$E = \dfrac{kq}{r^2}$

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

$q= \dfrac{Er^2}{k}$

now,

$q= \dfrac{180000\times 0.015^2}{9\times 10^9}$

q = 4.5 x 10⁻⁹ C

q = 4.5 nC

the charge on the object is equal to 4.5 nC

8 0

2 years ago

Read 2 more answers

The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point

aleksley [76]

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

6 0

2 years ago

Read 2 more answers

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

8 0

2 years ago

Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with

mr_godi [17]

A challenge scientists face with this process is the use of ultrathin iron oxide, to pull protons off water and produce hydrogen gas, which itself is a poor electrical conductor.

5 0

1 year ago

Read 2 more answers