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3 years ago

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What are the units for the spring constant, k? A. newton meters B. newton seconds C. newtons/meter D. newtons/second E. newtons/

seconds2

2 answers:

mr Goodwill [35]3 years ago

5 0

Answer:

C. N/m (newtons/meter)

Explanation:

Since the equation for potential energy for a spring is PE = 1/2kx², after you know <em>k </em>and <em>x</em>, you will get an answer in Joules.

Please let me know if you want me to explain further!

Thanks!

lapo4ka [179]3 years ago

4 0

Answer:

Option (c)

Explanation:

The spring constant is defined as the force per unit length of compression or expansion in spring. So it's unit is newton per metre.

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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa

Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

$\frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

a ) when charge remains constant

energy = $\frac{q^2}{2C}$

q is charge and C is capacity

energy stored initially E₁= $\frac{q^2}{2C_1}$

energy stored finally E₂ = $\frac{q^2}{2C_2}$

$\frac{E_1}{E_2} = \frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

$E_2$ = $\frac{1.15}{2.3 } \times E_1$

= $\frac{1.15}{2.3 } \times 8.38$

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

$\frac{E_2}{E_1} = \frac{C_2}{C_1}$

$\frac{E_2}{8.38} = \frac{2.3}{1.15}$

$E_2$ = 16.76 .

8 0

3 years ago

You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is

tangare [24]

$1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }$

<u>Explanation</u>:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

$\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}$

Where, F = force, $G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right)$ , $mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}$ , $mass \left M_{2}= 0.31 \mathrm{kg}$ and Distance(r) = 0.42 m.

Substitute the values in the formula.

$\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}$

$\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}$

$\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}$

$\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }$

3 0

4 years ago

HELP HELP HELP ME!!!

zzz [600]

I believe the answer is C. Hope this helps!!

7 0

3 years ago

Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41

alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 = 0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

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3 years ago

g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug t

mihalych1998 [28]

Answer:

a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

a = v² / r

angular and linear variables are related

v = w r

we substitute

a = w² r

where r is the radius of the wheel

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3 years ago