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3 years ago

12

How might "Roller Coaster Physics" have been different if the author's purpose had been to warn

1 answer:

BigorU [14]3 years ago

6 0

Answer: The answer is hupa loopa

Explanation:

You might be interested in

How does Newton's second law of motion can be used to calculate the acceleration of an object?

Alex777 [14]

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

4 0

4 years ago

A train accelerated from rest to a speed of 60m/s in 200 seconds it continues with the speed reached for 40 seconds before comin

Olenka [21]

I don't know to be honest

Explanation:

6 0

3 years ago

how do you Compute the approximate acceleration of gravity for an object above the earth's surface, assigning accel gravity with

solong [7]

The approximate acceleration of gravity for an object above the earth's surface is 9.8 m/s².

<h3>Acceleration due to gravity </h3>

The acceleration due to gravity of an object above the Earth's surface is calculated from Newton's second law of motion and Newton's law of universal gravitation.

<h3>Force between the object and Earth</h3>

$F_g = \frac{GmM}{R^2}$

<h3>Weight of the object</h3>

$F_g = mg\\$

Solve (1) and (2)

mg = GmM/R²

g = GM/R²

where;

M is mass of Earth
G is gravitational constant
R is radius of the Earth = 6,371 km

g = (6.673 x 10⁻¹¹ x 5.98 x 10²⁴)/(6,371 x 10³)²

g = 9.8 m/s²

Learn more about acceleration due to gravity here: brainly.com/question/88039

#SPJ1

7 0

2 years ago

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

3 years ago

If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, t

muminat

Answer:

75.5g

Explanation:

From the ionic equation, we can write

$CU^{2+}+SO^{2-}_{4}\\$

next we find the number of charge

Note Q=it

for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs

hence

$Q=8.5*13500\\Q=114750C$

Since one faraday represent one mole of electron which equal 96500C

Hence the number of mole produced by 114750C is

114750/96500=1.2mol

The mass of copper produced is

$mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g$

Hence the amount of copper produced is 75.5g

7 0

4 years ago

Read 2 more answers