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solong [7]
3 years ago
13

Using Hooke's law F spring = k dela x, find elastic constant of a spring that stretches 2cm when a 4 newton force is applied to

it
Answers in the picture

Physics
1 answer:
photoshop1234 [79]3 years ago
6 0

Answer: <em>2 N/cm  (B)</em>

Given that:

Spring stretches = 2 cm ,

Force  = 4 N,

<em>determine </em>

            the spring constant/elastic constant ( k)= ?

            We know that from hook's law, <em>F = k.x</em>  N

                                                                 4 = k × 2

                                                                => k  =4/2

                                                                         = 2 N/cm

The elastic constant is 2 N/cm

           

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Answer:

Reflection

Explanation:

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3 years ago
What makes these image a represent of what research <br>​
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The economic down fall
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3 years ago
a baseball player is dashing toward home plate with a speed of 200 feet per hour. When he decides to hit the dirt. he slides for
Serggg [28]

To solve this problem we will use the kinematic formula for the final velocity.


V_f_x = V_0_x + a_xt


The final speed is 0 at the moment the player stops.


The time until it stops is 1.3 s


The initial speed is 200 feet / s  Note (check the speed units in the problem statement, 200ft / s is very much and 200ft / h is very small)

Then, we clear the formula.


a_x = \frac{(Vfx-V0x)}{t}\\ a_x = \frac{(0-200)}{1.3}\\ a_x = -153.5 ft / s ^ 2

Because the player is slowing down, the acceleration goes in the opposite direction to the player's movement, and that is why it is negative.



To answer part b) we use the following formula.


Vf ^ 2 = Vo ^ 2 + 2a * (x_2 - x_1)\\\\ (x_2 - x_1)= \frac{(V_f ^ 2-V_0 ^ 2)}{2a}\\\\ (x_2 - x_1)= \frac{(0-200 ^ 2)}{- 2 * 153.5}\\\\ (x_2 - x_1)= 130.29 feet

8 0
3 years ago
A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from
s2008m [1.1K]

E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0
2 years ago
I don’t get this question and i need help!!!
galina1969 [7]

i would say number 2

3 0
3 years ago
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