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Ivan
3 years ago
5

In the lab, you performed a trial where the fan was turned

Physics
2 answers:
beks73 [17]3 years ago
7 0

Answer:

Simple Response The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

Ksju [112]3 years ago
6 0

Answer:

Sample Response: The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

on edge :)

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A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,
Rus_ich [418]

Answer:

\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\

Explanation:

Given that

Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\

As both charges are negative so there exist force of repulsion in direction as shown in figure.

F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N

Angle at which force F12 is acting is

\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o

F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\

\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

F_{21}=-5.63\times 10^{-11}N

\vec{F}_{21}=

7 0
3 years ago
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Answer:

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Explanation:

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Answer:

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