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AnnyKZ [126]
1 year ago
9

Which nucleus completes the following equation?

Physics
1 answer:
melomori [17]1 year ago
4 0

The nucleus that completes the equation given above is 218/85 At, which is option D.

<h3>What is a nuclear fission?</h3>

Nuclear fission is a nuclear reaction in which a large nucleus splits into smaller ones with the simultaneous release of energy.

According to this question, the following nuclear fission reaction is given: 222/87Fr = 4/2He+ ?

To find the missing nucleus, we subtract 4 from 222 to get 218, therefore, the nucleus that completes the equation given above is 218/85 At.

Learn more about nuclear fission at: brainly.com/question/913303

#SPJ1

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What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?
LenaWriter [7]

Answer:

Hello,~There!

What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

<h2><u>33% According to the Graph</u></h2>

Hope this helps!

6 0
3 years ago
A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
dimaraw [331]

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

7 0
2 years ago
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

  max height is at  t= - b/2a = 10/9.8 =1.02

     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will <u>also</u> still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

8 0
2 years ago
Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
A string vibrates at a frequency of 20Hz. what is its period?
Andreas93 [3]
We know, Period = 1/frequency = 1/20 sec.

So, your final answer is 0.05 sec

Hope this helps!
7 0
3 years ago
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