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3 years ago

A 500.-kg roller coaster car starts from rest at the top of a 60.0-meter hill.

1 answer:

8 0

1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.

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How many discovered planets are there

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We have discovered 786 planets. Most of which were only recently discovered.

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2 years ago

1) A car is moving to the right with the constant speed of 1.2 m/s. If the car starts

Marina CMI [18]

(100, 108)
Due to
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100, 108

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2 years ago

If the velocity of an object is zero, then that object cannot be accelerating. | True or False

valkas [14]

Answer: False

Explanation:

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3 years ago

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti

allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q = −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

3 years ago