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Ugo [173]
3 years ago
12

What is the displacement of the particle in the time interval 5 seconds to 8 seconds?

Physics
2 answers:
Hoochie [10]3 years ago
5 0
The displacement of the particle in the time interval 5 seconds to 8 seconds will be 1.5 meters. 
Displacement is a vector quantity; meaning it has both magnitude and direction; magnitude being the distance between the interval at 5 seconds and position at 8 seconds, while the direction is from position at 5 second to position at 8 second. This is irrespective of the route followed during the time interval.
Gekata [30.6K]3 years ago
3 0

Answer:

1.5 meters plato users ;)

Explanation:

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If the distance between two asteroids is halved, the gravitational force they exert on each other will
mamaluj [8]

Answer:

e) Be four times greater

Explanation:

Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses (m_{1} & m_{2}) and the distance between them (r) in the next way:

F=G\frac{m_{1}m_{2}}{r^{2}} (2)

Now if distance between asteroids is halved:

F_{2}=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}}

F_{2}=G\frac{m_{1}m_{2}}{\frac{r^{2}}{4}}

F_{2}=G\frac{4m_{1}m_{2}}{r^{2}}=4G\frac{m_{1}m_{2}}{r^{2}}

Note that G\frac{m_{1}m_{2}}{r^{2}} because (1) is F so:

F_{2}=4F

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3 years ago
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Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Verno
iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is v_{A} =\frac{x_{A} }{t_{A} } (Equation A) and Average speed of the car B is v_{B} =\frac{x_{B} }{t_{B} } (Equation B), where x_{A} and x_{B} are the distances and t_{A} and t_{B} are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

96 minutes . \frac{1 hour}{60 minutes} = 1.6 h

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h} (Equation C)

v_{B} = \frac{208-x}{1.6h} (Equation D)

From Equation A and C, we have:

\frac{208-x}{1.6}+10 = \frac{x}{1.6}

208-x+16 = x

208 + 16 = 2x

x = \frac{224}{2}

x = 112 miles

Replacing x in Equation A:

v_{A}  = \frac{112miles}{1.6h}

v_{A} = 70 miles per hour

Replacing x in Equation B:

v_{B}  = \frac{208miles-112miles}{1.6h}

v_{B}  = \frac{96miles}{1.6h}

v_{B}  = 60 miles per hour

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Solve the given parallel circuit by computing the desired quantities.
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Hope this provides some answers    

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