At A, coaster is only associated with potential energy.
At B, coaster is associated with kinetic as well as potential energy.
Since the track is frictionless, no energy will be lost when coaster reaches from point A to point B. Therefore, according to conservation of energy, total energy at A should be equal to total energy at B.
Total energy at A = mgh = mg(12)
Total energy at B = mgh+ mv²/2 = mg(2) + mv²/2
∴12mg = 2mg + mv²/2
∴(12g-2g)×2 = v²
∴v² = 20g
∴v = 14m/s.

Again conserving energy at points B and C.
Total energy at B = 2mg + m(14)²/2
Total energy at C = 4mg + mv²/2
∴2mg + m(14²)/2 = 4mg + mv²/2
Solving this you get,
v = 12.52 m/s.
Therefore, speed of roller coaster at point C is 12.52 m/s.

8 0

4 years ago

A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 10 nC is located at a distance r = 10

jok3333 [9.3K]

Answer:

q₁ = -2.92 nC

Explanation:

Given;

first point charge, q₁ = ?

second point charge, q₂ = 10 nC

net flux through the surface of the sphere, Φ = 800 N.m²/C

According to Gauss’s law, the flux through any closed surface (Gaussian surface), is equal to the net charge enclosed divided by the permittivity of free space.

$\phi = \frac{q_{enc.}}{\epsilon_o}$

where;

Φ is net flux

$q_{enc.}$ net charge enclosed

ε₀ is permittivity of free space.

$q_{enc.}$ = Φε₀

= 800 x 8.85 x 10⁻¹²

= 7.08 x 10⁻⁹ C

$q_{enc.}$ = 7.08 nC

q₁ + q₂ = $q_{enc.}$

q₁ = $q_{enc.}$ - q₂

q₁ = 7.08nC - 10 nC

q₁ = -2.92 nC

4 0

3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 55.0° above the horizontal. It explodes on a mountainside 42.

GuDViN [60]

Answer:

x=7227

y=1678

(7227,1678)

Explanation:

Ok, check the picture I attached you, because of the problem don't give us aditional information, let's asume that the projectile is fired from an initial position x=0 and y=0. Now let's use projectile motion equations, but firs let's find the initial velocity components in x-axis and y-axis:

$v_ox=v_o*cos(\theta_o)=300*cos(55)=172.0729309m/s$