Which of the following best describes the behavior of the above pair of substance.

A hooligan throws a stone vertically down with an initial speed of 17 m/s from the roof of a building 51 metres above the

lisov135 [29]

Answer:

Explanation:

v² = u² + 2as

v² = 17² + 2(-9.8)(-51)

v² = 1,288.6

v = ±35.89707...

v = 36 m/s

7 0

3 years ago

What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about

nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude n

• the car's weight with magnitude w

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ F (⟂) = N + W (⟂) = m a (⟂)

and

∑ F (//) = W (//) = m a (//)

Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to

N - m g cos(θ) = m a sin(θ)

and

m g sin(θ) = m a cos(θ)

The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get

tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→ θ ≈ 3.52°

8 0

3 years ago

Bats are extremely adept at catching insects in midair. If a 81.5-g bat flying in one direction at 7.21 m/s catches a 8.11-g ins

aivan3 [116]

You can look that up on google

7 0

3 years ago

550 nm light passes through a diffraction grating with 3000 lines per centimeter. The screen is 115 cm away from the grating. Wh

Sloan [31]

Answer:

69.7 cm

Explanation:

What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m

Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ

So, d/mλ = x/D for a second order bright fringe, m = 2.

So, d/2λ = x/D

x = dD/2λ

So, x =

For a dark fringe, we have

d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

d/(2 + 1/2)λ = x'/D

d/(5/2)λ = x'/D

2d/5λ = x'/D

x' = 2dD/5λ

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ

x" = dD/10λ

Substituting the values of the variables into the equation, we have

x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)

x" = 1.15/165 × 10² m

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

6 0

3 years ago

What's is the origin of Neptune's moon Triton?

Elden [556K]

Triton was discovered by British astronomer William Lassell on October 10, 1846, just 17 days after the discovery of Neptune.

Triton is named after the Greek sea god Triton (Τρίτων), the son of Poseidon (the Greek god corresponding to the Roman Neptune).

4 0

3 years ago