PLEASE HELP ASAP URGENT!!!!!

Help immediately! Answer I need help.

garri49 [273]

Answer:

0.80 m

Explanation:

elastic potential energy formula

elastic potential energy = 0.5 × spring constant × (extension) 2

4 0

2 years ago

Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri

Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

Q = 24 pC = 24 10⁻¹² C

E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

E₂ = 7.2 N / C

let's find the difference between these two fields

ΔE = E₂ -E₁

ΔE = 7.2 - 5.4

ΔE = 1.8 N / C

the minimum detection field is

E_minimum = 0.77 N / C

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0

3 years ago

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las

topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

$w=\frac{2.44 \lambda L}{D}$

where $\lambda$ is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

$L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m$

4 0

3 years ago

A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a ne

blagie [28]

Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

3 0

3 years ago

What accounts for the two precipitation peaks in mbandaka?

slava [35]

The two precipitation peaks in Mbandaka during March to April and September to November is due to the intertropical convergence zone.

Intertropical convergence zone is a narrow zone located near the equator. It is where the northern and southern air masses intersect which results to low atmospheric pressure. Due to the intertropical convergence zone’s meeting of air masses, often times the air pressure are lower which will results to colder air, or even rainfall during the period of March to April, and most especially September to November in Mbandaka.

<span>Since Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named as a Wetland of International importance, there is really a bigger chance that this place experience above 60mm precipitation in a year, temperatures averaging from 23 – 26 degrees Celsius.</span>

7 0

3 years ago