Which statement describes the distribution of charge in an atom?

A 45-gram object that has a volume of 9 mL has a density of

Elan Coil [88]

The answer : a. 5 g/mL

7 0

3 years ago

If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinit

sp2606 [1]

Answer:

W = 0×(kq2L)

Explanation:

We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r

If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them

WABC = kq2/(rAB + rAC + rBC)

So, to assemble four charges A, B, C, & D....

WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)

Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative

rAB = L

rAC = L√2

rAD = L (-)

rBC = L

rBD = L√2 (-)

rCD = L (-)

⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)

⇒ ∴ W = 0 × (kq2/L)

This way, working through each option...

(a)

The positive charges are equidistant from each other at a distance of L.

rAB = L

rAC = L

rAD = ½L⋅sin(60) (-)

rBC = L

rBD = ½L⋅sin(60) (-)

rCD = ½L⋅sin(60) (-)

Wa = kq2/(3L - (3/2)L⋅(0.866))

⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)

(b)

rAB = L

rAC = 2L

rAD = 3L (-)

rBC = L

rBD = 2L (-)

rCD = L (-)

Wb = kq2/(4L - 6L)

⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)

(c)

The factor doesn't matter, so Wc = 0 × (kq2/L)

In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.

This way, charge configuration (a) requires the least amount of work.

5 0

3 years ago

One way to force air into an unconscious person's lungs is to squeeze on a balloon appropriately connected to the subject. What

frozen [14]

Answer:

The force that you must exert on the balloon is 1.96 N

Explanation:

Given;

height of water, h = 4.00 cm = 4 x 10⁻² m

effective area, A = 50.0 cm² = 50 x 10⁻⁴ m²

density of water, ρ = 1 x 10³ kg/m³

Gauge pressure of the balloon is calculated as;

P = ρgh

where;

ρ is density of water

g is acceleration due to gravity

h is height of water

P = 1 x 10³ x 9.8 x 4 x 10⁻²

P = 392 N/m²

The force exerted on the balloon is calculated as;

F = PA

where;

P is pressure of the balloon

A is the effective area

F = 392 x 50 x 10⁻⁴

F = 1.96 N

Therefore, the force that you must exert on the balloon is 1.96 N

3 0

4 years ago

An antelope moving with constant acceleration covers the distance 68.0 m between two points in time 7.50 s. Its speed as it pass

Darya [45]

Answer:

A)The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope is 1.77 m/s²

Explanation:

The equations of the position and velocity of the antelope is given by the following expressions:

x = x0 + v0 · t + 1/2 · a ·t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) Let´s place the center of the frame of reference at the first point. The equation of position at t = 7.50 s will be:

x = x0 + v0 · t + 1/2 · a ·t²

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

We also know that at the second point the velocity is 15.7 m/s. Then at t = 7.50 the velocity will be 15.7 m/s.

v = v0 + a · t

15.7 m/s = v0 + a · 7.50 s

We can solve this equation for "a" and replace it in the equation of height to obtain "v0". Then:

a = (15.7 m/s - v0) / 7.50 s

Replacing it in the equation for position:

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

68.0 m = v0 · 7.50 s + 1/2 · (15.7 m/s - v0) / 7.50 s · (7.50 s)²

68.0 m = v0 · 7.50 s + 7.85 m/s · 7.50 s - 3.75 s · v0

68.0 m - 7.85 m/s · 7.50 s = 3.75 s · v0

(68.0 m - 7.85 m/s · 7.50 s) / 3.75 s = v0

v0 = 2.43 m/s

The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope will be:

a = (15.7 m/s - v0) / 7.50 s

a = (15.7 m/s - 2.43 m/s) / 7.50 s

a = 1.77 m/s²

The acceleration of the antelope is 1.77 m/s²

5 0

3 years ago

Helium gas at 20 °C is confined within a rigid vessel. The gas is then heated until its pressure is doubled. What is the final t

victus00 [196]

Answer:

586 K

Explanation:

Let P is the initial pressure.

Initial temperature, T₁ = 20°C = 293 K

Final pressure, P₂ = 2P

We need to find the final temperature of the gas.

The relation between the pressure and the temperature is as follows

$P\propto T\\\\or\\\\\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}$

Put all the values,

$\dfrac{P}{2P}=\dfrac{293}{T_2}\\\\\dfrac{1}{2}=\dfrac{293}{T_2}\\\\T_2=2\times 293\\\\T_2=586\ K$

So, the final temperature of the gas is 586 K.

4 0

3 years ago