Currents exist in the ocean without the need of electric fields. True False

A small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground. Until it reaches terminal speed

goldenfox [79]

Answer:

Explanation:

If a small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground and reaches a terminal speed after falling 0.400m, the Change in distance will be 1.90m - 0.400 = 1.50m

If it takes 5.4secs fo r the Styrofoam to reach the ground, the terminal velocity will be expressed as;

Vt = change in distance/time

Vt = 1.5m/5.4s

Vt = 0.28m/s

Note that the Styrofoam reaches its final velocity when the acceleration is zero.

To get the constant value B from the equation a = g-Bv

a = 0m/s²

g = 9.81m/s²

v = 0.28m/s

Substituting the parameters into the formula.

0 = 9.81-0.28B

-9.81 = -0.28B

Divide both sides by -0.28

B = -9.81/-0.28

B = 35.04

b) at t = 0sec, the initial terminal velocity is also zero.

Substituting v = 0 into the equation to get the acceleration.

a = g-Bv

a = g-B(0)

a = g

Hence the acceleration at t =0s is equal to the acceleration due to gravity which is 9.81m/s²

c) Given speed v = 0.150m/s

g = 9.81m/s²

B = 35.04

Substituting the given data into the equation a = g-Bv

a = 9.81-35.04(0.15)

a = 9.81 - 5.26

a = 4.55m/s²

7 0

3 years ago

Read 2 more answers

LAST ONE! ASAP PLEASE

Doss [256]

Answer:

There are so many questions which one you don't know

4 0

3 years ago

Read 2 more answers

a 20 g bead is attached to a light 120 cm long string. This bead moves in a horizontal circle with a constant speed of 1.5 m/s.

earnstyle [38]

Answer:

0.22 N

Explanation:

The tension in the spring is expressed as

$Tcos\theta=mg$ and making T the subject then

$T=\frac {mg}{cos\theta}$ where T is the tension in the string, m is the mass of the bead, g is acceleration due to gravity and $\theta$ is the angle which here is given as 25 degrees. Taking the value of g as 9.81 then

$T=\frac {0.02\times 9.81}{cos 25}=0.216482748 N\approx 0.22 N$

7 0

3 years ago

Is this equation balanced H2O → H2 + O2 –

sasho [114]

Yess
Hopes this helps

4 0

3 years ago

According to Coulomb’s Law, what happens to the force when the distance increase between 2 particles?

ohaa [14]

Answer:

The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.

Explanation:

Coulomb’s law, mathematical description of the electric force between charged objects. Formulated by the 18th-century French physicist Charles-Augustin de Coulomb, it is analogous to Isaac Newton’s law of gravity.

Both gravitational and electric forces decrease with the square of the distance between the objects, and both forces act along a line between them. In Coulomb’s law, however, the magnitude and sign of the electric force are determined by the electric charge, rather than the mass, of an object. Thus, charge determines how electromagnetism influences the motion of charged objects. Charge is a basic property of matter. Every constituent of matter has an electric charge with a value that can be positive, negative, or zero.

Coulomb's Law says that the force between 2 charges is proportional to the product of the quantities of charge on each and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is $F=k\frac{q_{1}q_{2} }{r^{2} }$ .

$F$ is the force.

$k$ is the Coulomb's constant ( $8.987*10^{9} \frac{Nm^{2} }{C^{2} }$ ).

$q_{1}$ is the electric charge of object 1.

$q_{2}$ is the electric charge of object 2.

$r$ is the distance between the two charges.

Electric force is inversely proportional to ( $r^{2}$ ) instead of ( $r$ ). As the distance between charges increases, the electric force decreases by a factor of $\frac{1}{r^{2} }$ .

8 0

3 years ago