Elements that give up electrons easily are called _______

What does wave speed have to do with all SONAR AND RADAR technologies?

Elena L [17]

Explanation:

1) Radar uses radio waves, which are a type of electromagnetic energy. Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects. Sound waves are a type of acoustic energy. Because of the different type of energy used in radar and sonar, each has its own applications.

2)Radar systems operate using radio waves primarily in air, while sonar systems operate using sound waves primarily in water (Minkoff, 1991). Despite the difference in medium, similarities in the principles of radar and sonar can frequently result in technological convergence.

4 0

3 years ago

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A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo

Alex

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

Speed = $\frac{distance}{time taken}$

Since time is unknown, we make it the subject of the expression;

time = $\frac{distance }{speed}$ = $\frac{400}{600}$

time = 0.67s

Learn more:

Speed brainly.com/question/10048445

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5 0

3 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

3 0

3 years ago

During which two time intervals does the particle undergo equal displacement?

san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, $d_1=\dfrac{1}{2}\times 2\times 10=10\ m$

Area under BC, $d_2=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under CD, $d_3=\dfrac{1}{2}\times 2\times 7=7\ m$

Area under DE, $d_4=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under EF, $d_5=\dfrac{1}{2}\times 2\times 3=3\ m$

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0

3 years ago

A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per

GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0

3 years ago

Elements that give up electrons easily are called ________