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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

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The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.

melamori03 [73]

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2 years ago

How many turns should a 10-cm long ideal solenoid have if it is to generate a 1.5-mT magnetic field when 1.0 A of current runs t

Ira Lisetskai [31]

Answer:

N=119.34 turns

Explanation:

The magnetic field of a solenoid is calculated using the formula:

B= µo* $\frac{I*N}{L}$ Equation 1

Where:

B: magnetic field in Teslas (T)

µo: free space permeability in T*m/A

I= Intensity of the current flowing through the conductor in ampere (A)

N= number of turns

L= solenoid length in meters (m)

Data of the problem:

L=10cm= $10*10^{-2}$ , B= 1.5mT= $1.5*10^{-3} T$ ,I=1A

µo= $4\pi *10^{-7} \frac{T*m}{A}$

We cleared N of the equation (1):

N=B*L/ µo*I

N= $(1.5*10^{-3} *10*10^{-2} )/(4\pi *10^{-7} *1$

$N=0.1193*10^{3}$

Answer

N=119.34 turns

3 0

2 years ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

2 years ago

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

Gnesinka [82]

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

$U=mgh$

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

6 0

2 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s