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Degger [83]
3 years ago
9

A systematic error in data is called

Physics
1 answer:
d1i1m1o1n [39]3 years ago
7 0
<span>an error having a nonzero mean, so that its effect is not reduced when observations are averaged.</span>
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What is the magnitude of the electric field at the point if the electric potential in the region is given by V 2.00xyz2, where V
Hatshy [7]

Answer:

Electric field at a point ( x , y , z) is E=-2yz^2-2xz^2-4xyz .

Explanation:

Given :

Electric potential in the region is , V = 2xyz^2\ .

We need to find the electric field .

We know , electric field , E=-\dfrac{dV}{dr}  { Here r is distance }

In coordinate system ,

E=-\dfrac{dV}{\delta x }-\dfrac{dV}{\delta y }-\dfrac{dV}{\delta z }  { \delta is partial derivative }

Putting all values we get ,

E=-\dfrac{2xyz^2}{\delta x }-\dfrac{2xyz^2}{\delta y }-\dfrac{2xyz^2}{\delta z }\\\\E=-2yz^2-2xz^2-4xyz

Hence , this is the required solution.

3 0
2 years ago
Which goes faster? *<br> light in space<br> sound in air<br> sound in space<br> light in water
andrezito [222]

Answer:

Hello! Your answer is, sound in the air is faster

Explanation:

The speed of sound through air is about 340 meters per second. It's faster through water and it's even faster through steel. Light will travel through a vacuum at 300 million meters per second. So they're totally different scales.

Hope I helped! Ask me anything if you have any questions! Brainiest plz. Hope you make an 100% and have a nice day! -Amelia♥

8 0
3 years ago
A body is moving vertically upwards. It’s velocity changes at a constant rate from 50m/s to 20m/s in 3 sec. What is it’s acceler
sertanlavr [38]

Answer:

-10 m/s²

Explanation:

a = Δv / Δt

a = (20 m/s − 50 m/s) / 3 s

a = -10 m/s²

3 0
3 years ago
Read 2 more answers
Determine the magnitude and direction of the force on an electron traveling 3.58 E 6 m/s horizontally to the west in a verticall
olya-2409 [2.1K]

Answer:banana

Explanation:

Because the amount of potassium is that

4 0
2 years ago
A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t
Murljashka [212]

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

8 0
3 years ago
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