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lana [24]
3 years ago
9

What is the spring coefficient of a spring that stores 4000 joules of energy

Physics
1 answer:
Verizon [17]3 years ago
6 0

Answer:

222.2N/m

Explanation:

Given parameters:

Elastic potential energy = 4000J

Extension  = 6m

Unknown:

Spring coefficient  = ?

Solution:

The elastic potential energy is the energy stored within a string.

It is expressed as;

           EPE = \frac{1}{2} k e²  

k is the spring constant

e is the extension

         4000  =  \frac{1}{2} x k x 6²  

        8000  =  36k

          k  = 222.2N/m

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How many neutrons does element X have if its atomic number is 48 and its mass number is 167?
lys-0071 [83]
If its atomic number is 48, then it has 48 protons in the nucleus
of each atom.  Any more mass than that is supplied by the neutrons
that are mixed in there with the protons.

If the mass is 167, and 48 of those are protons, then there are

             (167 - 48)  =  119 neutrons

in each nucleus.

6 0
4 years ago
Name the Organs and functions of digestive system
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6 0
3 years ago
A trip is taken that passes through the following points in order
Inessa05 [86]

Answer:

B) - 5.0 m

Explanation:

B is located on a positive location, 15m from the starting point A. Hence, since E is located a positive distance 10m from A, the difference becomes 10 - 15 = - 5.0 m

5 0
3 years ago
two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

v_0_x=6380 m/s

v_0_y=6770 m/s

time period of engine running=763 s

Displacement in x=4.50\times 10^6

y=7.27\times 10^6

Using s=ut+\frac{at^2}{2} in x and y direction

x=v_0_x\times t+\frac{at^2}{2}

4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}

4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}

a=-1.23 m/s^2

In y direction

y=v_0_y\times t+\frac{a't^2}{2}

7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}

7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}

a=7.24 m/s^2

x component=-1.23 m/s^2

y component=7.24 m/s^2

3 0
3 years ago
Read 2 more answers
A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un
Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0
3 years ago
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