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2 years ago

10

A ball with a mass of 5.0 g is moving at a speed of 2.0 m/s. Would doubling the mass or doubling the speed have a greater effect

on the kinetic energy of the ball? Explain.

2 answers:

otez555 [7]2 years ago

7 0

Answer:

Yes

Explanation:

doubling up would have a greater effect.

Arte-miy333 [17]2 years ago

5 0

Answer:

Yes

Explanation:

Kinetic energy is K.E1/2mv2 so that means it is directly proportional to mass and velocity.

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Which molecule will form a solution with a very low pH?

ser-zykov [4K]

The correct answer is A: H3PO4

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3 years ago

A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0

3 years ago

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

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3 years ago

A ray diagram shows an object placed between 2F and F of a convex lens.The image produced is

nekit [7.7K]

The image produced is magnified and real.

4 0

3 years ago

Read 2 more answers

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0

3 years ago