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exis [7]
3 years ago
5

1.Which statement is true when two objects are 20 miles apart but one is much bigger?

Physics
2 answers:
ryzh [129]3 years ago
8 0
The first one is c? and 2nd is d i think
Paha777 [63]3 years ago
7 0

1. C.  Gravitational attraction exists between the two objects.

Explanation:

Gravitational attraction is always exerted between two objects which have mass, and its magnitude is given by:

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 the masses of the two objects, and r the separation between them. Since the two objects have for sure non-zero masses m1 and m2, even if they are 20 miles apart, the value of the gravitational attraction F is non-zero, so the correct answer is C.


2. D.  Two atoms come together to form a molecule.

Explanation:

this outcome is actually caused by the electrostatic forces between the two atoms, not by gravitational force. In fact, gravitational force becomes relevant only when the masses of the two objects involved are large enough: this is the case for planets, stars, galaxies, and objects in the universe. However, two atoms have very small masses, so the gravitational force between them is really negligible. On this smaller scales, the electrostatic force is much stronger than the gravitational force, so the electrostatic force is the real responsible for the formation of bonds between atoms.

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2. Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 x 10-9 C and the oth
Maru [420]

Answer:

A. -2.16 * 10^(-5) N

B. 9 * 10^(-7) N

Explanation:

Parameters given:

Distance between their centres, r = 0.3 m

Charge in first sphere, Q1 = 12 * 10^(-9) C

Charge in second sphere, Q2 = -18 * 10^(-9) C

A. Electrostatic force exerted on one sphere by the other is:

F = (k * Q1 * Q2) / r²

F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²

F = -2.16 * 10^(-5) N

B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:

Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))

= - 6 * 10^(-9) C

Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C

Hence the electrostatic force between them is:

F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²

F = 9 * 10^(-7) N

7 0
3 years ago
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A


Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0
3 years ago
Make a
defon

Answer:

Yes, a mixture can be made up of just elements and no compounds. The elements never experiance a chemical interaction and stay only in physical contact with each other.

BRAILIEST PLS I NEED LEVEL UP!

3 0
2 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
What happens when an atom gives up loosely held valence electrons to another atom​
vazorg [7]

Answer:

The number of valence electrons increases to 8 or, the atom gives up loosely held valence electrons.

Explanation:

Please mark brainliest and have a great day!

3 0
3 years ago
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