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WINSTONCH [101]

3 years ago

10

Two identical point charges in outer space are held apart at a distance D. As soon as the charges are released, each begins movi

ng at an acceleration a. If instead they were released at a distance D/2, the acceleration of one charge would be ____.
A. 2a
B. 4a
C. a/2
D. a/4

2 answers:

AVprozaik [17]3 years ago

6 0

Answer:

B. 4a

Explanation:

Force between the charges is inversely proportional to the square of the distance

=> Force will be 4 times and acceleration will be 4a

=> Answer b).

yanalaym [24]3 years ago

6 0

Answer: B. 4a

Explanation: The force experienced by the two identical charge is related to thier acceleration a by

F = m*a

m is mass of charge

a is thier acceleration

Also,

Force F is inversely proportional to D². From Coulomb's law

So we have

F ~1/D²

Since the distance D was reduced to D/2

We have that

F¹ ~ 1/(D/2)²

F¹~ 1/D²/4

From the law of reciprocal we have

F¹~ 4*1/D²

Which is the new force due to the decrease in distance to D/2

But 1/D² is original force F.

F¹ ~ 4*F

But F= m*a

F¹ ~ 4*m*a

But m is constant so the only thing changing is the acceleration. So we have, the new acceleration increased by a factor of 4 to be 4*a = 4a.

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Flowchart symbols

MArishka [77]

Answer:

Option (A)

Explanation:

A flowchart can be described as a representation of a process or an algorithm in a sequential manner (stepwise) in the form of diagrams. These steps are made of symbols that are connected to one another with the help of arrows. These arrows are used to show the direction of the process.

There are mainly four symbols that are used to describe a flow chart, which includes-

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Thus, the correct answer is option (A).

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3 years ago

What important milestone occurred as a result of the Soviets' use of the R-7

AleksAgata [21]

Answer:

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Explanation:

4 0

2 years ago

Read 2 more answers

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

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Nana76 [90]

Not sure.can you give me a clue?

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