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2 years ago

You hang a 350 gram mass on a spring and its displaced 0.5 cm. Its spring constant is

Physics

1 answer:

Keith_Richards [23]2 years ago

4 0

Answer:

Explanation:

kx = mg

k = mg/x

k = 0.350(9.81)/0.005

k = 686.7

k = 687 N/m

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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

3 years ago

A long solenoid that has 1,130 turns uniformly distributed over a length of 0.430 m produces a magnetic field of magnitude 1.00

sweet-ann [11.9K]

Given Information:

Number of turns = N = 1130 turns

Length of solenoid = L = 0.430 m

Magnetic field = B = 1.0x10⁻⁴ T

Required Information:

Current = I = ?

Answer:

I = 0.0302 A

Explanation:

The current flowing in the solenoid winding can be found using

I = BL/μ₀N

Where μ₀ is the permeability of free space, N is the number of turns, B is the magnetic field and L is the length of solenoid

I = 1.0x10⁻⁴*0.430/4πx10⁻⁷ *1130

I = 0.0302 A

I = 30.28 mA

6 0

2 years ago

If an piano and an apple are dropped in a tunnel without air.which would hit the ground first

liq [111]

Answer:

the piano.

Explanation:

4 0

2 years ago

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

2 years ago

Please help! Will give brainly, 50 points!! I'm stuck with this question and I don't get it!!

IrinaVladis [17]

Answer:

3.52176 x 10^-10 N

Explanation:

Fg = 3.52176 x 10^-10 Newton

8 0

3 years ago