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Alenkasestr [34]
2 years ago
10

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False

Physics
2 answers:
Studentka2010 [4]2 years ago
4 0
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is? 
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
ohaa [14]2 years ago
3 0

Answer:

1.True

2.6

3.All of the above

Explanation:

1.In a single atom,  no more than 2 electrons can occupy a single orbital.

By pauli exclusion principle, an orbital can contain a maximum of only 2 electrons, the two electrons must have opposite spins.

Therefore, it is true.

2.We know that maximum number of electrons in p sublevel is 6.

Hence, the maximum number of electrons allowed in a p sublevel of the 3rd principle level is 6.

3.A neutral atom has a ground state electronic configuration of 1s^2 2s^2.

Therefore, Total number of electrons=4

Atomic number=4

Number of protons=Atomic number=4

The atom has  total number of orbital is 2.

The element has no unpaired electrons.

Hence, all of the above is correct option.

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A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when
choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

F_t - mg = ma

here since speed is constant or it is at rest

so we will have

a = 0

F_t = mg

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

F_t - mg = ma

F_t = mg + ma

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

F_t - mg = -ma

F_t = mg - ma

so tension force is less than the weight of crate if it is accelerating downwards

4 0
2 years ago
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor
nalin [4]

Answer:

u_x=38.13\ m/s

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

S=u_yt+\dfrac{1}{2}gt^2

Here u_y=0

S= 57 cm

0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2

t=0.34 s

Now in the horizontal direction

x=u_xt

Here x=13 m

t= 0.34 s

So

13=u_x\times 0.34

u_x=38.13\ m/s

So the initial speed of ball is 38.13 m/s.

7 0
1 year ago
If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?
Ghella [55]

<u>Given that</u>

mass (m) = 1300 Kg ,

height (h) = 1500 m

Determine the potential energy ?

     P.E = m × g × h

           = 1300 × 9.81 × 1500

           = 19129500  Joules

           = 19129.5 KJ

3 0
2 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
2 years ago
Two hockey players skating on essentially frictionless ice collide head-on. Madeleine, of mass 65.0 kg, is moving at 6.00 m/s to
xeze [42]

Explanation:

It is given that,

Mass of Madeleine, m_1=65\ kg

Initial speed of Madeleine, u_1=6\ m/s (due east)

Final speed of Madeleine, v_1=-3\ m/s (due west)

Mass of Buffy, m_2=55\ kg

Final speed of Buffy, v_2=3.5\ m/s (due east)

Let u_1 is the Buffy's velocity just before the collision. Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

65\times 6+55\times u_2=65\times (-3)+55\times 3.5

u_2=-7.13\ m/s

So, the initial speed of the Buffy just before the collision is 7.13 m/s and it is moving due west. Hence, this is the required solution.

5 0
2 years ago
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