Where would a car traveling on a roller coaster have the most kinetic energy ? and why?

Physics

2 answers:

goldenfox [79]3 years ago

4 0

Answer:

As the car travels up the coaster it is gaining potential energy.

Explanation:

Because It has the greatest in amount of potential energy at the top of the coaster. when the car travels down the roller coaster it obtains speed and kinetic energy.

Soloha48 [4]3 years ago

3 0

Most potential energy at the top of a hill/loop and the most kinetic energy at the bottom of a hill/loop

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Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the bea

mylen [45]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

$I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2$

The maximum magnetic field is calculated using the following intensity formula;

$I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T$

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

5 0

3 years ago

Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci

EleoNora [17]

We subtract the velocity of car 1 from the velocity of car 2:
$v=(30\ mph\ North)-(20\ mph\ South)$
$=(30\ mph\ North)+(20\ mph\ North)$
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