A test pilot flies with an acceleration of of 5 g . what is the acceleration in meter per second

<u>Metal detectors work by transmitting an electromagnetic field from the search coil into the ground. Any metal objects (targets) within the electromagnetic field will become energised and retransmit an electromagnetic field of their own. The detector’s search coil receives the retransmitted field and alerts the user by producing a target response. metal detectors are capable of discriminating between different target types and can be set to ignore unwanted targets. </u>

1. Search Coil

The detector’s search coil transmits the electromagnetic field into the ground and receives the return electromagnetic field from a target.

2. Transmit Electromagnetic Field (visual representation only - blue)

The transmit electromagnetic field energises targets to enable them to be detected.

3. Target

A target is any metal object that can be detected by a metal detector. In this example, the detected target is treasure, which is a good (accepted) target.

<em>hope this helps PLEASE MARK AS BRAINLIEST:)</em>

8 0

3 years ago

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

2 years ago

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be $\sqrt{10^3}}$ years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

$1year * \frac{360degrees}{31.62years} = 11.38 degrees$

or about 12 degrees.

6 0

3 years ago