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meriva
2 years ago
10

It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring? (b) When a 3.0 kg block is pushed a

gainst a massless spring of force constant 4.5. 10' N/m, the spring is compressed 8.0 cm. The block is released and it slides 2.0 m from the point at which it is released across a horizontal surface before friction stops it. What is the coefficient of kinetic friction between the block and the surface?
Physics
1 answer:
NARA [144]2 years ago
8 0

Answer:

Explanation:

a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²

500 = 5 x 10⁻³ k ,

k = (500/5) x 10³ = 10⁵ N/m

b )

k = 4.5 x 10¹ = 45 N/m

Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J

This energy gets dissipated by friction .

work done by friction = μ mg d

d is the distance traveled under friction

so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2

μ = 245 x 10⁻⁴  or 0.00245 which appears to be very small. .

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jeyben [28]

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

= 4.03 ×10-15m2/s

4 0
2 years ago
At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc
lys-0071 [83]

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

5 0
2 years ago
A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav
fenix001 [56]

Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

B = \frac{\mu I}{2\pi r}

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

7 0
2 years ago
What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a
Umnica [9.8K]

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec              

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity g=9.8m/sec^2

From first equation of motion we know that v=u+gt

v=0+9.8\times 9=88.2m/sec

So instantaneous velocity after 9 sec will be 88.2 m/sec  

5 0
2 years ago
The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1
Andrei [34K]

Answer:

The expression of gravitational field due to mass m_! at a distance l_a

Explanation:

We have given mass is m_1

Distance of the point where we have to find the gravitational field is l_a

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by g=\frac{Gm_1}{l_a^2}

This will be the expression of gravitational field due to mass m_! at a distance l_a

4 0
2 years ago
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