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2 years ago

When forklift moves a crate, it performs _______.

Physics

2 answers:

Helga [31]2 years ago

4 0

Motion of the forklift

Katarina [22]2 years ago

4 0

Answer:

GPE

Explanation:

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You buy a 75 watt light bulb. the label means that

alukav5142 [94]

The label means that the bulb will consume 75 joules of electrical energy in 1 sec

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3 years ago

HELLPPPP. The metalloid that has three valence electrons is ___________?

Alja [10]

Answer:

Boron

Explanation:

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2 years ago

A spring is 17 cm long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force th

RideAnS [48]

Answer:

im pretty sure it B but I recommend waiting for another person. I used the workdone formula (Force*Dictance*cos(theta) and got 55 Joules

Explanation:

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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

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

7 0

2 years ago

Can someone help me with this problem?

emmasim [6.3K]

Answer:

Explanation:

u flipped it i cant see

5 0

2 years ago