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2 years ago

A spring stores 10. joules of elastic potential

2 answers:

6 0

The answer would be . Since we are looking for the spring constant you would need to use the formula
$PEs = \frac{1}{2} k {x}^{2}$
. Then you'd substitute, for PEs and x.
$10j=1/2k(.20m^2).$
Then solve. k=500n/m

serg [7]2 years ago

6 0

<h2>Answer:</h2>

The spring constant is 50 N/m.

<h3>Explanation:</h3>

Spring constant is the specific property of a spring which tells about the stiffness of a spring.

It is the amount of force which is used to displace a spring one meter from its equilibrium position.

Its unit is N/m.

The formula of potential energy in a spring is :

P.E = 1/2Kx2

Here,

P.E = 10 joules

x = 0.2m

K = P.E × 2/x2

K = 10 × 2/0.4

k = 50 N/m

Hence the spring constant is 50 N/m.

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A 2 kg stone moves with a speed of 1 m/s. A second 2 kg stone is moving twice as fast. Compare their kinetic energies.

alekssr [168]

D
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5 0

2 years ago

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A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh

Jet001 [13]

Answer:

heat used to rise temperature pan = 30.1%

heat used to rise temperature water = 69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature = 80.0ºC

to find out

heat used to rise temperature of pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature ...........1

specific heat of water is 4186 J/kgºC

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

total heat = 62.8 + 27 = 89.8 kJ

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0

2 years ago

If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?

Gnom [1K]

The average current density in the wire is given by:

$J=\frac{I}{A}$

where I is the current intensity and A is the cross-sectional area of the wire.

The cross-sectional area of the wire is given by:

$A=\pi r^2$

where r is the radius of the wire. In this problem, $r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m$ , so the cross-sectional area is

$A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2$

and the average current density is

$J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2$

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2 years ago

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If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i

slavikrds [6]

Answer:

$3t_{1/2}$

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

$N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}$ (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em> $t_{1/2}$ <em>: is the half-lifes.</em>