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ahrayia [7]
3 years ago
7

Calculate the wavelength in centimeters of radar energy at a frequency of 10 GHz. What is the frequency in gigahertz of radar en

ergy at a wavelength of 25 cm
Physics
1 answer:
konstantin123 [22]3 years ago
7 0

Explanation:

(a) Frequency of radar energy, f = 10\ GHz =10^{10}\ Hz

The relation between frequency and wavelength is given by :

c=f\lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{10^{10}}

\lambda=0.03\ m

or

\lambda=3\ cm

(b) If wavelength, \lambda=25\ cm=0.25\ m

c=f\lambda

f=\dfrac{c}{\lambda}

f=\dfrac{3\times 10^8}{0.25}

f=1.2\times 10^9\ Hz

or

f = 1.2 GHz

Hence, this is the required solution.

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3 years ago
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n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It
Pani-rosa [81]

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

v_x = RT\\\\T = \frac{v_x}{R}

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

v_x = horizontal speed = 9 m/s

Therefore,

T = \frac{9\ m/s}{40\ m}

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4 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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People often use simple machines like pulleys, levers, and ramps because they say the machine “makes the work easier.” Which of
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I think the answer is D.
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How are the direction of the net force on an object and the direction of the object’s acceleration related?
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They're the same direction.

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