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Valentin [98]
2 years ago
8

Which best describes deena’s error? the line for constant velocity is not horizontal; it is diagonal and slopes downward. the ar

ea of the rectangle and the triangle under the line should be added together for constant acceleration find acceleration by finding the area of the rectangle above the line. find velocity by calculating the slope of the line.
Physics
1 answer:
denis-greek [22]2 years ago
3 0

The statement best describes deena's error is: Find acceleration by finding the area of the rectangle above the line.

<h3>What is acceleration?</h3>

Acceleration is given by the ratio of Resultant or total force acting on any object and the its mass.

Deena made a chart to summarize features of a velocity vs. time graph for objects with constant acceleration and objects with constant velocity.

In the  velocity time graph, the area under the curve shows the displacement of the object and the slope represents the acceleration.

Thus, the deena's error is  Find acceleration by finding the area of the rectangle above the line.

Learn more about acceleration.

brainly.com/question/12550364

#SPJ1

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(a) 1.72\cdot 10^{-5} \Omega m

The resistance of the rod is given by:

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The radius of the rod is half the diameter: r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m, so the cross-sectional area is

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So, the resistance is

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(b) 8.57\cdot 10^{-4} /{\circ}C

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega

The equation that gives the change in resistance as a function of the temperature is

R(T)=R_0 (1+\alpha(T-T_0))

where

R(T)=0.86 \Omega is the resistance at the new temperature (92.0°C)

R_0=0.81 \Omega is the resistance at the original temperature (20.0°C)

\alpha is the temperature coefficient of resistivity

T=92^{\circ}C

T_0 = 20^{\circ}

Solving the formula for \alpha, we find

\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C

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