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2 years ago

HELP ME PLZ !!!!!!!!!!!!!!!!!!!!

Physics

2 answers:

malfutka [58]2 years ago

7 0

Answer:

Explanation:

Its the 2nd one on the left (the bolded one)

stira [4]2 years ago

5 0

The bolded box at the bottom

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3 years ago

Read 2 more answers

A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?

Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, $F_w$ ↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, $F_{(helicopter)}$ ↑ = The weight of the crate = 2,452.5 N

The net force, $F_{(net)}$ = $F_{(helicopter)}$ ↑ - $F_w$ ↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0

3 years ago

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b

Lesechka [4]

Answer:

a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,

c) U = 54.7 nJ , d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

C = Q / V

Q = C V

can also be calculated using geometry consideration

C = e or A / d

we reduce to the SI system

A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

d = 1.53 cm = 1.53 10⁻² m

we substitute

Q = eo A / d V

Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

Q = k Q₀

Q = 80 397.57

Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

C = k C₀

C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

C = 1.157 10⁻¹⁰ F

V = Vo / k

V = 275/80

V = 3.4375 V

c) The stored energy is

U = ½ C V²

U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²

U = 5.47 10⁻⁸ J

let's reduce to nJ

109 nJ = 1 J

U = 54.7 nJ

d) energy after submerging

U = ½ (kCo) (Vo / k) 2

U = ½ Co Vo2 / k

U = U₀ / k

U = 54.7 / 80 nJ

U = 0.68375 nJ

the energy change is

ΔU = U₀ -U

ΔU = 54.7 - 0.687375

6 0

3 years ago