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ruslelena [56]
3 years ago
15

CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do

es the elastic potential energy of the spring change during this process?
Physics
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.    

Explanation:

Given that,

Spring constant of the spring, k=1.2\times 10^4\ N/m

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J

So, the elastic potential energy of the spring change during this process is 21.6 J.

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What is sex And new born baby why connected to women's veins?
san4es73 [151]

Sex is when a male and a female have intercourse, The reason new born babys are connected to a women's vein is because the baby is connected too the mother and as it grows the vein kinda like ''gets smaller''

3 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
3 years ago
Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es
mojhsa [17]

Answer:

 D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

       D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

        D = 2 3960 1100 + 1100²

        D = 8.712 10⁶ + 1.21 10⁶

        D = 9.92 10⁶ mi

         D = 9.9 10⁶ mi

8 0
3 years ago
Independent practice
coldgirl [10]

Explanation:

formula: <u>Mass</u>

Density x volume

2a) m=10kg v=0.3m³

10÷0.3=33.3 kg/m

2b) m = 160 kg V=0.1m³

160÷0.1=1600 kg/m

2c) m = 220 kg V = 0.02m³

220÷0.02=11000 kg/m

A wooden post has a volume of 0.025m³ and a mass of 20kg. Calculate its density in kg/m.

density = volume ÷ mass

20÷ 0.025=800 kg/m

Challenge: A rectangular concrete slab is 0.80m long, 0.60 m wide and 0.04m thick. Calculate its volume in m³.

Formula : Length x width x height = Volume

0.80 x 0.60 x 0.04 = 0.0192m³

B) The mass of the concrete slab is 180 kg. Calculate its density in kg/m.

density = volume ÷ mass

180 ÷ 0.0192 = 9375 kg/m

4 0
2 years ago
A baseball travels 200 metes in 6 seconds, what is the baseball’s velocity?
goldenfox [79]

Answer:

33.33 m/sec

Explanation:

A baseball travels 200 metes in 6 seconds,

what is the baseball’s velocity?

use the formula: velocity = distance over time

where (d) distance = 200 m

and (t) time = 6 sec.

plugin values into the formula:

v = d / t

  = 200 m / 6 sec

  = 33.33 m/sec.

therefore, the baseball's velocity is 33.33 m/sec

8 0
3 years ago
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