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4 years ago

What direction is the net force on a falling skydiver before she reaches terminal velocity?

Physics

1 answer:

GarryVolchara [31]4 years ago

8 0

Answer:

The net force and the acceleration on the falling skydiver is upward

Explanation:

An upward net force on a downward falling object would cause that object to slow down. the skydiver this slows down. As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.

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describe what happens in terms of energy when you blow up a ballon and release it without trying its neck

const2013 [10]

Answer:

The answer is below

Explanation:

When a gallon is blown the air energy is saved as potential energy which is compressed inside the balloon. However, when the balloon is released without trying its neck, the air energy quickly turned to kinetic energy, in which the air energy keeps moving as the balloon float around the space.

Hence, in short, what happens in terms of energy when you blow up a balloon and release it without trying its neck is that the air energy moves from potential energy to kinetic energy.

3 0

3 years ago

What wattage is a toaster that uses 5 amps from a 120 volt outlet

nevsk [136]

Power = (voltage) x (current)

Power = (120 volts) x (5 amperes) = 600 watts

8 0

3 years ago

Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to

Zigmanuir [339]

Answer:

the same

Explanation:

3 0

3 years ago

A thin stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it h

marta [7]

Answer:

$v = \sqrt{3gL}$

Explanation:

As we know that length of the rod is L and mass is M

so here it one end of the rod is stationary and other end of the rod is rotating about one end

then we will have energy conservation to find the total rotational kinetic energy of rod about its one end is given as

$mg\frac{L}{2} = \frac{1}{2}I\omega^2$

$mg\frac{L}{2} = \frac{1}{2}(\frac{mL^2}{3})\omega^2$

$g = \frac{L}{3} \omega^2$

$\omega = \sqrt{\frac{3g}{L}}$

so the linear speed of the other end of the rod just before it hit the ground is given as

$v = L\omega$

$v = \sqrt{3gL}$

8 0

3 years ago

Three cylindrical wires, 1, 2, and 3 are made of the same materialand have resistances R1, R2, and R3, respectively. Wires 1 and

bonufazy [111]

Answer:

1) R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2

Explanation:

1) The resistance of a wire is given by the expression

R = $\rho \ \frac{l}{A}$

where ρ is the resistivity of the material, l the length of the wire and A the area of the wire

The area is given by

A = π r² = π d² / 4

we substitute

R = (ρ 4 /π) $\frac{l}{d^2}$

the amount in parentheses is constant for this case

let's analyze the situation presented, to find the resistance of each wire

* indicate l₁ = l₂ and d₂ = 2 d₁

the resistance of wire 1 is

R₁ = (ρ 4 /π) $\frac{l_1}{d_1^2}$

the resistance of wire 2 is

R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}

R₂ = (ρ 4 /π) $\frac{l_1}{ (2 d_1)^2}$

R₂ = (ρ 4 /π ) $\frac{l_1}{d_1^2}$ ¼

R₂ = ¼ R₁

* indicate that d₂ = d₃ and l₃ = 2 l₂

R2 = (ρ 4 /π) $\frac{l_2}{d_2^2}$

the resistance of wire 3 is substituting the indicated condition

R3 = (ρ 4 /π 2) \frac{l_3}{d_3^2}

R3 = (ρ 4 /π) $\frac{2 \ l_2}{d_2}$

R3 = 2 R₂

let's write the relations obtained

R₁ = (ρ 4 /π) $\frac{I_1}{ d_1^2}$

R₂ = ¼ R₁

R₃ = 2 R₂

let's write everything as a function of R1

R₁ =(ρ 4 /π) $\frac{l_1}{d_1^2}$

R₂ = ¼ R₁

R₃ = ½ R₁

the resistance of the wire in decreasing order is

R₁ > R₃ > R₂

2) The power dissipated by a wire is

P = V I

the voltage is

V = I R

I = V / R

substituting

P = V² / R

therefore the power dissipated by each wire is

wire 1

P₁ = V² / R₁

wire 2

P₂ = V² / R₂

P₂ = $\frac{V^2}{ \frac{1}{4} R_1}$

P₂ = 4 P₁

wire 3

P₃ = V² / R₃

P₃ = $\frac{V^2}{ \frac{1}{2} R_1}$

P₃ = 2 P₁

Therefore, the wire that dissipates the most is wire 2

6 0

3 years ago