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3 years ago

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A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

ective, the antenna is a circular aperture through which the microwaves diffract. What is the diameter of the radar beam at a distance of 30 km

1 answer:

My name is Ann [436]3 years ago

5 0

Answer:

915m

Hope this helps.

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Your friend says that if Newton’s third law is correct, no object would ever start moving. Here is his argument: You pull a sled

m_a_m_a [10]

Answer:

Explanation:

You pull a sled exerting a 50 N force on it , sled also exerts a force on you . These forces are action and reaction force , as per third law of Newton . These two forces are equal and opposite . But they do not act on the same object so they do not cancel each other . They act on different objects , one on the sledge and the other on you . Due to force on sledge , sledge moves in the direction of force or towards you . You will start moving in opposite direction if frictional force of ground is nil or less .

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3 years ago

A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.

horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

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3 years ago

Read 2 more answers

You are given four resistors, 2 ohms, 3 ohms, 5 ohms, and 10 ohms. Your friend say you can connect them so you obtain an equival

AlexFokin [52]

If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

The equivalent circuit is,

R equivalent for the series connection is,

$\begin{gathered} Req(S)=2+3+5=10ohms \\ Now,\text{ } \\ Req\text{ }for\text{ }10\text{ }ohms\text{ }and\text{ }10\text{ }ohms\text{ }is, \\ \frac{1}{Req}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ So, \\ Req=5\text{ ohms} \end{gathered}$

The equivalent resistance is 5 ohms.

So your friend is saying true.

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1 year ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

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3 years ago

A track athlete jumps over a hurdle. What is the equal and opposite force for the force of the athlete pushing on the ground as

Jet001 [13]

Answer: Normal Force of the ground pushing the athlete up.

Explanation: I did this on Khan Academy and got the answer right.

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3 years ago