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julia-pushkina [17]
2 years ago
11

Write the maximum number of electrons that can be accommodated in the s,p,d,f subshells​

Physics
1 answer:
ki77a [65]2 years ago
3 0

Answer:

S=2,P=6,D=10,F=14. The total number of electrons contained by s,p,d,f subshells is 32

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The answer is C I just took this quiz.
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Grrr i don't have much time, help please T^T
mart [117]

Answer:

2. ( b ) zero

3. ( c ) 10 s

4. Uniform then decreasing

Explanation:

2.

Since the motion is uniform, initial and final velocity will be 0, hence acceleration will be zero.

3.

Initial velocity ( u ) = 5 m/s

Final velocity ( v ) = 35 m/s

Acceleration ( a ) 3 m/s^2

To find : Time ( t )

Formula : -

t = v - u / a

 = 35 - 5 / 3

 = 30 / 3

t = 10 s

4 0
1 year ago
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An object is traveling with a constant velocity of 5 m/s. How far will it have gone after 7 s?
kozerog [31]
<h3>Answer:</h3>

35 meters

<h3>Explanation:</h3>

<u>Data given;</u>

  • Velocity of an object = 5 m/s
  • Time taken = 7 s

We are required to calculate how far the object traveled.

  • We need to know that;

Velocity = Displacement ÷ time

  • Therefore;

Displacement = Velocity × time

                       = 5 m/s × 7 s

                      = 35 m

Therefore; the object traveled 35 meters

6 0
3 years ago
A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a
scZoUnD [109]

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
T_h = \frac{T}{2}

Now, we can determine the velocity at which the can was launched at using the following equation:
v_f = v_i + at

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}

***vsinθ is the vertical component of the velocity.

Solve for 'v':
vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}

Now, recall that:
W = \Delta KE = \frac{1}{2}m(\Delta v)^2

Plug in the expression for velocity:
W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}

B.

We can use the same process as above, where T' = 2T and Th = T.

v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}

C.

The work done in part B is 4 times greater than the work done in part A.

\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}

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2 years ago
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Recruitment i think.
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