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Katen [24]
3 years ago
12

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?
Physics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

f=1.57\times 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=\dfrac{c}{\lambda}

So,

f=\dfrac{3\times 10^8}{0.191}\\\\=1.57\times 10^9\ Hz

So, the frequency of such a system is equal to1.57\times 10^9\ Hz.

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An object with a mass of 2 kg is acted upon by a force of 8 N. What is the resulting acceleration of the object?
Tomtit [17]

Answer:

4 meter per second square is the acceleration

Explanation:

Force=mass x acceleration

8N= 2kg x acceleration

8N/2kg = acceleration

4 meter per second square is the acceleration.

<em>Please mark me as brainliest.</em>

4 0
3 years ago
Which type of scanner uses X-rays, and why?
Mashutka [201]

Answer:

C.  Security baggage scanners, because X-rays either pass through or are absorbed by the objects in the baggage, creating an image of the objects

3 0
3 years ago
Read 2 more answers
A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.
marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

\omega = 25 rpm

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

\omega = 2 \pi f (1)

where

\omega is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

\omega=0.42 rev/s

Keeping in mind that 1 rev = 2\pi rad, the angular speed can be rewritten as

\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42

And re-arranginf eq.(1), we can find the frequency:

f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

v=\omega r

where

\omega is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

\omega = 2\pi \cdot 0.42 rad/s is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

d=vt

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

d=(26.4)(20)=528 cm

Learn more about circular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

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7 0
4 years ago
Read 2 more answers
steve took a train 30 miles north to seattle then he took a different train 25 miles east draw a vector diagram showing steves t
Fed [463]

A = distance traveled by steve in north direction by train = 30 miles

B = distance traveled by steve in east direction by train = 25 miles

R = resultant displacement of the steve relative to starting position

using pythagorean theorem

R = √(A² + B²)

inserting the values

R = √(30² + 25²)

R = 39.05 miles


direction : tan⁻¹(B/A) = tan⁻¹(25/30) = 39.8 deg east of north


5 0
3 years ago
A tennis ball is dropped from 1.16 m above the
sweet-ann [11.9K]

Answer:

Vf = 4.77 m/s

Explanation:

During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 1.16 m

Vf = Final Velocity of Ball = ?

Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)

Therefore, using these values in the equation, we get:

(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²

Vf = √(22.736 m²/s²)

<u>Vf = 4.77 m/s</u>

6 0
3 years ago
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