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3 years ago

When are the gravitational attractions are greater when large or small?

2 answers:

5 0

The gravitational attractions are greatest be when the objects have large masses and they're closer together. Their sizes don't matter.

castortr0y [4]3 years ago

5 0

Large
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A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball initially moving in th

inn [45]

Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

<h3>What is conservation of momentum?</h3>

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

$m_1u_1 + m_2u_2 = (m_1+m_2)v$

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

$v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}$

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

$v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}$

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

$v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s$

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Learn more about the conservation of momentum here;

brainly.com/question/7538238

#SPJ4

4 0

2 years ago

Fe₂O3<br> + co<br> →<br> Fe3O4 + CO₂

Goryan [66]

Explanation:

Fe₂O₃ + CO → Fe₃O₄ + CO₂

Balancing the equation above, we can derive simple mathematical equations that are very easy to solve.

aFe₂O₃ + bCO → cFe₃O₄ + dCO₂

a,b,c and d are the coefficients needed to balance the equation above;

Conserving Fe; 2a = 3c

O: 3a + b = 4c + 2d

C: b = d

let a = 1;

c = $\frac{2}{3}$

Since b = d

3a + d = 4c + 2d

3a = 4c + 2d - d

3a = 4c + d

a = 1, c = $\frac{2}{3}$

3 = 4 x $\frac{2}{3}$ + d

d = $\frac{1}{3}$

b = $\frac{1}{3}$

multiplying a, b, c and d by 3:

a = 3 b = 1 c = 2 and d = 1

3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂

Learn more:

Balanced equation brainly.com/question/2612756

#learnwithBrainly

6 0

3 years ago

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0

3 years ago

When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the _

fgiga [73]

When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the <span>number of successful effective collisions is higher. More of the reactants collide and are able to form products. Hope this answers the question. have a nice day.</span>

6 0

3 years ago

It takes you 9.8 min to walk with an average velocity of 1.9 m/s to the north from the bus stop to the museum entrance.

Dahasolnce [82]

60 x 9.8 = 588
588 x 1.9 = 1117.2

5 0

3 years ago