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-Dominant- [34]
4 years ago
3

Which describes how the spring constant affects the potential energy of an object for a given displacement from an equilibrium p

osition?
A the higher the spring constant, the greater the gravitational potential energy.
B The lower the spring constant, the greater the gravitational potential energy.
C The higher the spring constant, the greater the elastic potential energy.
D The lower the spring constant, the greater the elastic potential energy
Physics
2 answers:
sergij07 [2.7K]4 years ago
6 0

Answer:

A

Explanation:

A cause thats the anwser

brainliest pls

deff fn [24]4 years ago
4 0

Answer:

Based on the answer choices provided, the correct answer is:

<u>C, The higher the spring constant, the greater the elastic potential energy.</u>

Explanation:

The spring constant is best defined as the measure of a spring's resistance to force (in regards to factors such as stretching and compression).

Thus, the higher the spring constant, <u>the stiffer the string</u> - this equates to greater elastic potential energy (think of stretching a rubber band or using it as a slingshot).

Hope this helps!

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A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no
frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

4 0
4 years ago
What skills are used to move the body from one place to another? *
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Your muscles, bones, heart and other things in the body.
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3 years ago
Read 2 more answers
8. An airplane of mass 8500 kg dives has an altitude of 15,000 m. It then dives steeply to an
Ivan

Answer:

333.5 MJ

Explanation:

ΔV = m·g·Δh

     = 8500 · 9.81 · (15000-11000)

     = 8500 · 9.81 · 4000

     = 333 540 000

     ≈ 333.5 ·10⁶J = 333.5 MJ

8 0
3 years ago
Which event is an example of a contact force?
Scorpion4ik [409]

Answer:

D. a person pulling a sled.

Explanation:

contact force only occurs when something directly comes in contact with another object.

a. is wrong because that is called magnetic force.

b. gravitational force

c. is an electrical force

4 0
4 years ago
As a particle moves 10.0 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 × 10-16
Aleks04 [339]

Answer:

6.4×10⁻¹⁹ C

Explanation:

From the question,

V = E×d.............. Equation 1

Where V = electric potential, E = Electric Field, d = distance moved by the particle.

Given: E = 75 N/C, d = 10 m.

Substitute into equation 1

V = 75×10 = 750 V.

Also Using,

W = qV.................. Equation 2

W = Work done by the particle when it moves, q = Particles charge

make q the subject of the equation,

q = W/V............ Equation 3

Given: W = 4.8×10⁻¹⁶ J, V = 750 V

Substitute into equation 3

q = 4.8×10⁻¹⁶/750

q = 6.4×10⁻¹⁹ C

6 0
3 years ago
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