Mode of physical activity <br> A.frequency <br> B.intensity <br> C.time <br> D.type

The graph of the relationship between the volume of a gas at constant temperature and its pressure is a(n)?

timurjin [86]

<span>When the difference between two results is larger than the estimates error, the result is</span>

5 0

2 years ago

How does the configuration of the electric field occur between a "parallel plate" setup in a lab?What is the effect of conductor

sleet_krkn [62]

Explanation:

The magnitude of the electric field between the plates is given by

E = -ΔV/d

minus sign indicates Potential decreases in the direction of electric field

where

ΔV is the potential difference between the plates

D is the distance between the plates.

The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.

8 0

2 years ago

Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-

n200080 [17]

Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

$I_0$ = Intensity of unpolarized light = $1.1\times 10^{-10}\ W/m^2$

$\theta$ = Angle of the filter = $30^{\circ}$

Intensity of light is given by

$I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2$

The intensity of light detected by the camera is $4.125\times 10^{-11}\ W/m^2$

7 0

3 years ago

A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T

NeX [460]

Answer

i dont know

Explanation:

4 0

2 years ago

Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

3 years ago

Mode of physical activity A.frequency B.intensity C.time D.type