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irina [24]
3 years ago
14

What could serve as ballast for a balloon? Help pls...thx

Physics
1 answer:
I am Lyosha [343]3 years ago
6 0
With the advent of the plastic balloon and the beginning of the unmanned ... That would lead to a more sophisticated ballast system that uses fine steel or iron
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The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A
Bogdan [553]

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

5 0
3 years ago
An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi
Amanda [17]

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

8 0
3 years ago
2. Can the frictional force in this experiment be ignored? Explain.
neonofarm [45]

Answer:

i dont know

Explanation:

i dont know

5 0
2 years ago
Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of substances on opposi
Mrac [35]

When two sides of a membrane are in contact with each other, the distribution of ions will alter as a result of the binding of a signal molecule to a ligand-gated ion channel.

<h3>What is a ligand-gated ion channel?</h3>

Ligand-gated ion channels (LGICs) are membrane proteins that are structurally integral and feature a pore that permits the controlled passage of particular ions across the plasma membrane. The electrochemical gradient for the permeant ions drives the passive ion flux.

When a chemical ligand, such as a neurotransmitter, attaches to the protein, ligand-gated ion channels open. Changes in membrane potential cause voltage channels to open and close. When a receptor physically deforms, as in the case of pressure and touch receptors, mechanically-gated channels open.

Learn more about ligand-gated ion channel here:

brainly.com/question/15215628

#SPJ4

3 0
2 years ago
How does this relate to Newton's second law?
taurus [48]

Answer:

In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.

Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.

Explanation:

In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.

Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.

4 0
3 years ago
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