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3 years ago

An electron in a(n) __________ subshell experiences the greatest effective nuclear charge in a many-electron atom.

Physics

1 answer:

bixtya [17]3 years ago

6 0

The electrons float around in an outer sub shell

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¿Qué cantidad de calor absorbió una masa de 4 g de cinc al pasar de 20 °C a<br> 180 °C?

Alborosie

(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,

Ce = 0.093 cal/g. °c

Q = m C ΔT

Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )

Q= 4×0.093 × 160

Q = 59.52 cal

I hope I helped you^_^

7 0

3 years ago

Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth

Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

4 0

3 years ago

Read 2 more answers

A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is

valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

$\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2$

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

$\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}$