The phrase change in which a substance changes from a gas directly to a solid is

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi

True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0

3 years ago

What replaces a cold current that sinks to the ocean floor?

FrozenT [24]

Well simple the warm water then replaces the cold current that sinks to the ocean floor.

5 0

3 years ago

Which of the following is occurring while a satellite is in orbit around Earth? O It is continuously pulling away from Earth It

hoa [83]

AnswerIt is continuously falling towards the surface of the earth

Explanation:

since gravity from earth is the thing that keeps it constantly in orbit

4 0

3 years ago

When you rub your hands together, you cannot push harder on one hand than the other. can push harder on one hand than the other.

Sholpan [36]

I think it means that your putting in an equal amount of force on each hand because it's the same brain that controls what you do. I think that's what it means, not entirely sure but hope this helps

3 0

3 years ago

A hollow, uniformly charged sphere has an inner radius of r1 = 0.105 m and an outer radius of r2 = 0.31 m. The sphere has a net

Sliva [168]

Answer:

E = 77532.42N/C

Explanation:

In order to find the magnitude of the electric field for a point that is in between the inner radius and outer radius, you take into account the Gauss' law for the electric flux trough a spherical surface with radius r:

$\int E\cdot dS=\frac{Q}{\epsilon_o}$ (1)

Q: net charge of the hollow sphere = 1.9*10-6C

ε0: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Furthermore, you have that the net charge contained in a sphere of radius r is:

$Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}$ (2)

with the charge density is:

$\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}$ (3)

r2: outer radius = 0.31m

r1: inner radius = 0.105m

The electric field trough the Gaussian surface is parallel to the normal to the surface, the, you have in the integral of the equation (1):

$\int E\cdot dS=E(4\pi r^2)$ (4)

where you have used the expression for a surface of a sphere.

Next, you replace the expressions of equations (2), (3) and (4) in the equation (1) and solve for E:

$E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}$

you replace the values of all parameters, and with r = 0.17m

$E=\frac{(1.6*10^{-6}C)((0.17m)^3-(0.105m)^3)}{4\pi(8.85*10^{-12}C^2/Nm^2)(0.17m)^2((0.31m)^3-(0.105m)^3)}\\\\E=77532.42\frac{N}{C}$

The magnitude of the electric field at a distance r=0.17m to the center of the hollow sphere is 77532.42N/C

5 0

3 years ago