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Some radar systems detect the size and shape of objects such as aircraft and geological terrain. What is the frequency of such a system which can detect objects as small as 19.1 cm?

Answers

Answer 1

Answer:

Answer:

$f=1.57* 10^9\ Hz$

Explanation:

Given that,

A system can detect objects as small as 19.1 cm i.e. 0.191 m. It is the wavelength.

We know that,

Frequency, $f=(c)/(\lambda)$

So,

$f=(3* 10^8)/(0.191)\n\n=1.57* 10^9\ Hz$

So, the frequency of such a system is equal to $1.57* 10^9\ Hz$ .

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Which one defines force?
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(a) Determine the electric field strength between two parallel conducting plates to see if it will exceed the breakdown strength for air (3 ? 106 V/m). The plates are separated by2.98 mm and a potential difference of 5575 V is applied. (b) How close together can the plates be with this applied voltage without exceeding the breakdown strength?

Answers

Answer:

(a) 1.87×10⁶ V/m

(b) 1.12 mm closer

Explanation:

(a)

Electric Field = Electric potential/distance.

E = V/d ................... Equation 1

Where E = Electric Field, V = Electric potential, d = distance.

Given: V = 5575 V, d = 2.98 mm = 0.00298 m.

Substitute into equation 1

E = 5575/0.00298

E = 1.87×10⁶ V/m

(b)

without exceeding the breakdown strength,

make d the subject of equation 1

d = V/E.............. Equation 2

Given: E = 3×10⁶ V/m, V = 5575 V

Substitute into equation 2

d = 5575/3000000

d = 1.86 mm.

the plate will be = 2.98-1.86 = 1.12 mm closer

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft

Answers

I believed the answer is d

Which statements explain why theories change over time? Check all that apply. Scientists change the definition of theory to have their ideas accepted. New information and technology may be developed that influence the theory. Theories change with each new generation of scientists. New experimental methods provide new information. Theories may or may not be supported by new information.

Answers

The theories change over time because new information and technology may be developed that influence the theory. Therefore, option B is correct.

Why theories change over time ?

As new information and perspectives become available, established theories may be changed or refuted. If a new or modified theory explains all the old theory did and then some, scientists are likely to accept it.

When fresh data emerges that refutes an existing theory, scientific theories can be revised or replaced. All scientific theories are founded on observable, testable data, and they can be changed when new information is found that contradicts the present idea.

When certain features of a hypothesis are refuted by fresh experimental data, a theory might be altered in science.

Thus, option B is correct.

To learn more about theories, follow the link;

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Final answer:

Theories change over time due to new information, technology, and the perspectives of new generations of scientists.

Explanation:

Theories change over time for several reasons. First, new information and technology may be developed that influence the theory. For example, advancements in experimental methods can provide new information and lead to the refinement or modification of theories. Additionally, theories may change with each new generation of scientists as they bring new perspectives and ideas. It is important to note that theories may or may not be supported by new information, and scientists do not change the definition of theory to have their ideas accepted.

Learn more about Reasons for changing theories here:

brainly.com/question/34240563

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A 1.7 kg model airplane is flying north at 12.5 m/s initially, and 25 seconds later is observed heading 30 degrees west of north at 25 m/s. What is the magnitude of the average net force on the airplane during this time interval?

Answers

Answer:

Average net force = 0.62 N

Explanation:

We are given;

Mass; m = 1.7 kg

Initial velocity; u = 12.5 m/s

Final velocity; v = 25 m/s

time; t = 25 seconds

Now, we are told that the final velocity was 30° west of North. So, resolving this velocity along the horizontal gives;

v = 25 cos 30°

Now, using Newton's first equation of motion gives;

v = u + at

Where a is acceleration

Plugging in the relevant values gives;

25 cos 30° = 12.5 + 25a

21.6506 - 12.5 = 25a

a = (21.6506 - 12.5)/25

a = 0.3660 m/s²

Now, magnitude of the average net force would be; F = ma

F = 1.7 × 0.366

F ≈ 0.62 N

5. A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with the velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

Answers

Answer:

The recoil velocity of the raft is 1.205 m/s.

Explanation:

given that,

Mass of the swimmer, $m_1=55\ kg$

Mass of the raft, $m_2=210\ kg$

Velocity of the swimmer, v = +4.6 m/s

It is mentioned that the swimmer then runs off the raft, the total linear momentum of the swimmer/raft system is conserved. Let V is the recoil velocity of the raft.

$m_1v+m_2V=0$

$55* 4.6+210V=0$

V = -1.205 m/s

So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.

Answer:

The recoil velocity of the raft would be $v_(r)\approx 1.2(m)/(s)$ (pointing to the left if the swimmer runs to the right)

Explanation:

The problem states thatthe swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.

To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.

Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:

$p_(i)=0$

Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as

$p_(f)=mv-Mv_(r)=0$

The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.

Then, from that previous relation we can clear

$v_(r)=(m)/(M)v=(55)/(210)*4.6(m)/(s)=(253)/(210)(m)/(s)\approx1.2(m)/(s)$

wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).

A truck’s suspension spring each have a spring constant of 769 N/m. If the potential energy of the right front spring is 250 J, how far is the spring compressed?