In the water circuit model which part represent the wire 1)pump 2)pipes 3)tap

The number of particles in a gas system inversely affects _____ and directly affects _____.

iragen [17]

The answer is A.
When a substance is a gas, it fills its container no matter how many particles it's made of, however the number of particles present will directly affect the pressure and temperature of the system.

5 0

3 years ago

You are making a telephone out of two aluminum cans and some string. You can choose between two types of string: a 2-m length of

Nataliya [291]

Answer:

C)You should use the thin cooking twine.

Explanation:

A)You can choose either because they are the same length and will produce the same wave speed.

B)You should use the heavy rope.

C)You should use the thin cooking twine.

The speed of wave in a string is given by the following formula:

| $v$ | = $\sqrt{\frac{F_T}{u} }$

Where | $v$ | = speed of wave, $F_T$ = tension in the string, and μ = mass per length of the string.

Even though the two strings have the same length, the μ (mass/length) for the heavy rope will be more than the that of a thin rope. Consequently, the $F_T$ :μ for the thin rope will be higher than that of the heavy rope and as such, gives a bigger | $v$ |.

Therefore, the thin rope should be used in order to get a faster wave speed in the telephone.

The correct option is C.

3 0

3 years ago

A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. a. centimeters per hour (cm/h) b. c

guajiro [1.7K]

It would be 300 cm/h
and 5 cm/m

4 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk

Westkost [7]

Answer:

1.843 x 10^-5 C

Explanation:

Givens:

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation

E = (Q/A)/∈o (1)

Where,

Q: total charge on the disk.

A: the area of the disk.

Calculations:

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold

Q = EA∈o

= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)

= 1.843 x 10^-5 C

note:

calculations maybe wrong but method is correct

8 0

3 years ago