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3 years ago

Juan lives 100 m away from Bill.What is Juan's averege speed if he reaches Bill's home in 50 s?

1 answer:

7 0

Juan lives a hundred miles away from Bill then the average speed that he reaches Bill's Home in 50 seconds means that Juan lives 50 seconds away from bill because 50 + 50 seconds equals 100 seconds so Juansorry my bad Juan lives 50 miles away from Belle but he probably how I must have run to get to bill in 50 seconds .

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E= (500.0lm) 4 (100)

drek231 [11]

Answer:

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3 years ago

Describe the difference between distance, position, and displacement

nata0808 [166]

Answer:

Explanation:

Position is the location of the object (whether it's a person, a ball, or a particle) at a given moment in time.

Displacement is the difference in the object's position from one time to another.

Distance is the total amount the object has traveled in a certain period of time.

I hope this helps!

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2 years ago

At 3.00 m from a source that is emitting sound uniformly in all directions, the sound level (b) is 60.0 dB.

grigory [225]

Given that,

At 3.00 m from a source that is emitting sound uniformly in all directions, the sound level is 60.0 dB.

To find,

The distance from the source would the sound level be one-fourth the sound level at 3.00 m.

Solution,

The intensity from a source is inversely proportional to the distance.

Let I₁ = 60 dB, r₁ = 3 m, I₂ = 60/4 = 15 dB, r₂ =?

Using relation :

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\\r_2^2=\dfrac{I_1r_1^2}{I_2}\\\\r_2^2=\dfrac{60\times (3)^2}{15}\\\\r_2=6\ m$

So, at a distance of 6 m the sound level will be one fourth of the sound level at 3 m.

8 0

2 years ago

During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth

Marta_Voda [28]

(a) $2.68\cdot 10^{-6} W/m^2$

The intensity of an electromagnetic wave is given by

$I=\frac{P}{A}$

where

P is the power

A is the area of the surface considered

For the waves in the problem,

$P=182 kW = 1.82\cdot 10^5 W$ is the power

The area is a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

So, the intensity is

$I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2$

(b) $5.9\cdot 10^{-7} W$

In this case, the area of the reflection is

$A=0.22 m^2$

So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:

$P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W$

(c) $8.7\cdot 10^{-18} W/m^2$

We said that the power of the waves reflected by the aircraft is

$P=5.9\cdot 10^{-7} W$

If we assume that the reflected waves also propagate over a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

which has an area of

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

Then the intensity of the reflected waves at the radar site will be

$I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2$

(d) $8.1\cdot 10^{-8} V/m$

The intensity of a wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E_0^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

$E_0$ is the maximum value of the electric field vector

Solving the equation for $E_0$ ,

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m$

(e) $1.9\cdot 10^{-16} T$

The maximum value of the magnetic field vector is given by

$B_0 = \frac{E_0}{c}$

Substituting the values,

$B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T$

And the rms value of the magnetic field is given by

$B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T$

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2 years ago

Check out the picture of the periodic table of elements. Follow the arrow from left to right. What is an important trend you wou

soldi70 [24.7K]

C) the elements change from metal to non metal

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3 years ago

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