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3 years ago

Please help! Thanks so much! :)

Physics

1 answer:

notsponge [240]3 years ago

7 0

Answer: B

Explanation: I'm not 100% sure tho sorry if i'm wrong

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A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.

s344n2d4d5 [400]

Answer:the answer is wave.

Explanation:

8 0

3 years ago

Describe at least two changes that occurred to the atomic model over time, along with the experimental evidence that made these

e-lub [12.9K]

I think you could use Bohr model and the Plum Pudding Model

Bohr Model : the one that states atoms consist of positively charged nucleus and orbited by negatively charged electrons

Plum Pudding model : Atom consist of objects made to pieces with positive and negative charges

hope this helps

6 0

4 years ago

Read 2 more answers

"The density of material at the center of a neutron star is approximately 1.00 × 1018 kg/m3. What is the mass of a cube of this

andriy [413]

Q: "The density of material at the center of a neutron star is approximately 1.00 × 10¹⁸ kg/m3. What is the mass of a cube of this material that is 1.76 microns on each side. (One micron is equal to 1.00 × 10-6 m.)"

Answer:

5.452 kg

Explanation:

Density: This can be defined as the ratio of the mass of a substance to its volume. The S.I unit of density is kg/m³.

Mathematically, Density can be represented as

D = M/V

M = D×V ............................ Equation 1

Where D = density of the material, M = mass of the material, V = volume of the cube of the material,

But

V = L³

Where L = Length of the cube.

If 1 micron = 1.00×10⁻⁶ m,

Then, 1.76 microns = 1.76×10⁻⁶ m

Therefore L = 1.76×10⁻⁶ m

V = (1.76×10⁻⁶ )³

V = 5.452×10⁻¹⁸ m³

Given: D = 1.0×10¹⁸ kg/m³, and V = 5.452×10⁻¹⁸ m³

Substitute into equation 1

M = 1.0×10¹⁸×5.452×10⁻¹⁸

M =5.452 kg.

Hence the mass of the cube material = 5.452 kg

3 0

4 years ago

Two stationary positive point charges, charge 1 of magnitude 3.00 nC and charge 2 of magnitude 1.80 nC , are separated by a dist

MA_775_DIABLO [31]

Answer: U = -4.97*10^-17 J

Explanation:

Potential Energy of point charges,

U = kqq• / r, where

U = Potential Energy

q, q• = value of electric charges

k = 8.99*10^9 N.m²/C² constant of proportionality

r = distance between two charges

a) first electric potential due to electric field of first charge

q = 3*10^-9 C

q• = q(electron) = -1.602*10^-19 C

r = 0.5 * 31 cm = 15.5 cm = 0.155 m

U1 = kqq•/r

U1 = (8.99*10^9 * 3*10^-9 * -1.602*10^-19) / 0.155

U1 = -4.32*10^-18 / 0.155

U1 = -2.79*10^-17 J

Second electric potential due to electric field of second charge

U2 = kqq•/r

U2 = (8.99*10^9 * 1.8*10^-9 * -1.602*10^-19) / 0.155

U2 = -2.59*10^-18 / 0.155

U2 = -1.67*10^-17 J

U = U1 + U2

U = -2.79*10^-17 + -1.67*10^-17

U = -4.46*10^-17 J

b) first electric potential due to electric field of the first charge

q = 3*10^-9 C

q• = q(electron) = -1.602*10^-19 C

r = 10 cm = 0.1 m

U1 = kqq•/r

U1 = (8.99*10^9 * 3*10^-9 * -1.602*10^-19) / 0.1

U1 = -4.32*10^-18 / 0.1

U1 = -4.32*10^-17 J

Second electric potential to the electric field of second charge

q = 1.8*10^-9

q• = -1.602*10^-19 C

r = 50 - 10 = 40 cm = 0.4m

U2 = (8.99*10^9 * 1.8*10^-9 * -1.602*10^-19) / 0.4

U2 = -2.59*10^-18 / 0.4

U2 = -6.48*10^-18 J

U = U1 + U2

U = -4.32*10^-17 + -6.48*10^-18

U = -4.97*10^-17 J

5 0

3 years ago

Read 2 more answers

PLEASE HELP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Vikentia [17]

C it’s C

good luck

Dkdkkdisoskjska

4 0

3 years ago