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zepelin [54]
3 years ago
5

Why do you think scientists put their results in data tables and graphs?

Physics
1 answer:
Nikolay [14]3 years ago
8 0

Answer:Since most of the data scientist collect is quantitative, data tables and charts are usually used to organize the information • Graphs are created from data tables • They allow the investigator to get a visual image of the observations, which simplifies interpretation and drawing

Have a nice day (^-^) ;)

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During a 0.001 s interval while it is between the plates, the change of the momentum of the electron Δp with arrow is < 0, -8
Delvig [45]

Answer:

5.5 x 10^5 N/C

Explanation:

t = 0.001 s

Δp = - 8.8 x 10^-17 kg m /s

Force is equal to the rate of change of momentum.

F = Δp / Δt

F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N

q = 1.6 x 10^-19 C

Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)

E = 5.5 x 10^5 N/C

8 0
3 years ago
What process occurs when all of the energy from light waves is transferred to a medium?
Wittaler [7]
Absorption happens when <span>all of the energy from light waves is transferred to a medium.</span>
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3 years ago
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Which substance might lead to potentially addiction if abuse
uranmaximum [27]
Heroin is the number 1 most addictive substance.
4 0
3 years ago
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A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri
frutty [35]

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

8 0
3 years ago
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The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p
beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0
3 years ago
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