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Leokris [45]
3 years ago
8

If a force of 3000 N is applied to a large rock, but the rock does not move, how much work is done on the rock?

Physics
1 answer:
kkurt [141]3 years ago
8 0

Answer:

No work was done.

W = 0

Explanation:

Work is said to be done whenever a force of one newton moves a body of one kilogram through a distance of one meter. Meaning the applied force has to move the body from a point of rest through certain distance.

Work = force × distance

So, in the case of this question, we only have the force been applied, but no distance was covered. Hence, no work was done.

W = 3000× 0 meter

W = 0

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What is the spring coefficient of a spring that stores 4000 joules of energy
Verizon [17]

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

6 0
3 years ago
What type of change occurs when ice absorbs energy and melts?
anygoal [31]
The answer is C.endothermic change.

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3 0
3 years ago
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Two resistors are connected in parallel to a 12 V battery. The potential difference across one of the resistors is 12 v . Calcul
JulsSmile [24]

Answer:

See the explanation below.

Explanation:

We have to take into account that the potential difference is equal to the voltage, and this is measured between two points as the resistors are connected in parallel to the voltage source, the resistors will have the same voltage.

For ease, we will take the attached image of resistors connected in parallel.

As both resistors at their ends share the A & B connection points, these are at a voltage of 12V

5 0
3 years ago
Consider helium at 350 K and 0.75 m3/kg. Using Eq. 12-3, determine the change in pressure corresponding to an increase of (a) 1
Alex777 [14]

Answer:

a) (dP)_{v} = 9.692 kPa

b) (dP)_{T} = -9.692 kPa

c) dP = 0 Pa

Explanation:

The specifies equation is :

dz = (\frac{\delta z}{\delta x}) _{y} dx + (\frac{\delta z}{\delta y}) _{x} dy

Note that:

dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV

1% increase in temperature at specific volume:

dT = \frac{0.01}{1} *350\\dT = 3.5 K

a) Change in pressure of helium at constant volume:

(dP)_{v} = \frac{R}{v} dT

R = 2.0769 kJ/kg-K

dT = 3.5 K

v = 0.75 m³/kg

(dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa

b)

dv = (1%/100%) *0.75

dv = 0.0075 m³/kg

Change in pressure of helium at constant temperature:

(dP)_{T} = \frac{-RT}{v^{2} } dv

R = 2.0769 kJ/kg-K

T = 350 K

v = 0.75 m³/kg

dv = 0.0075 m³/kg

(dP)_{T} = \frac{-(2.0769*350)}{0.75^{2} } *0.0075\\(dP)_{T} = -9.692 kPa

c) The change in pressure of helium :

dP = (dP)_{v} + (dP)_{T}

dP = 9.692 - 9.692

dP = 0

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4 years ago
What is a community of organisms at a major regional or global level
Jet001 [13]
That my good sir would be an ecosystem 
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4 years ago
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