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3 years ago

When is the magnitude of the acceleration of a mass on a spring at its maximum value?

Physics

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

Explanation:

Now we know that Force is the rate of change of momentum meaning

F= mv/t

But

mv/t = Ke

v/t = ke/m

a= ke/m

Where a is acceleration

K is constant of proportionality of tension on a spring

e is the extension substended by a string.

From the formula of acceleration we can see that as mass decreases acceleration increases so we can see that m = 1

We would have a maximum value of acceleration.

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I'm very confused. Thanks for whoever helps me :)

sergij07 [2.7K]

(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**

6 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

3 0

3 years ago

When you stand with the wind blowing on your back, the low pressure center is

yan [13]

✷ Question: When you stand with the wind blowing on your back, the low pressure center is _____.

Answer: Letter A

✐ Explanation: When wind hits your back, all the molecules are in front of you.

Hope that helps! ★ If you have further questions or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia

5 0

3 years ago

Read 2 more answers

Why cant people with aids foght out the infection?

sveta [45]

They can fight the infection but not the disease

3 0

3 years ago

Read 2 more answers

Determine the average velocity and average speed for the following situation. A dog runs west for 35 meters then turns east for

mart [117]

Answer:

0.882 m/s average velocity and 1.71 m/s average speed

Explanation:

The dog travels a total of 35 m west and 110 m east.

110-35 = 75 m east of the starting position. Since velocity is a vector you must consider its first and final position and not the total distance traveled.

75 m / 85 s = 0.882 m/s average velocity

Speed is not concerned with direction so we instead add the total distance traveled which is 35+110 = 145 m. We then perform the same operation as before and divide by the time it took to run this distance.

145 m / 85 s = 1.71 m/s average speed

7 0

3 years ago