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kondaur [170]
3 years ago
15

If calcium lost two electrons it would have the same number of electrons as

Physics
1 answer:
Anna11 [10]3 years ago
3 0
If Calcium lost two electrons, it would have the same number of electrons as Argon which has 18 electrons.
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You travel 23 meters north in 16 seconds, 5 meters south in 4 seconds, and 16 meters north in 18 seconds. Calculate your total d
dolphi86 [110]

Answer:

d: 44m

Δd:34  

Explanation:

d: 23+5+16= 44m

Δd: 23-5+16= 34m

4 0
2 years ago
A mechanic uses a jack to lift up a car. He exerts a force of 11,000 N at a distance of 3m from the axis of rotation. How much t
pshichka [43]

Answer:

<h2>The amount of torque put on the car is 33,000Nm</h2>

Explanation:

Formula for calculating torque is expressed as T = rFsin\theta\\ where;

r is the radius of the  of the arm of the jack = 3m

F is the force exerted = 11000

\theta\\ is the angle of rotation = 90°

On substituting;

T = 3*11000sin90^{o} \\T = 3*11000 (sin90^{o} =1)\\T = 33000Nm

6 0
3 years ago
If your
Oxana [17]
What are you asking?
4 0
3 years ago
Read 2 more answers
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
A child looks at his reflection in a spherical Christmas tree ornament 8.0 cm in diameter in season that the image of his face i
malfutka [58]

From the information given,

diameter of ornament = 8

radius = diameter/2 = 8/2

radius of curvature, r = 4

Recall,

focal length, f = radius of curvature/2 = 4/2

f = 2

Recall,

magnification = image d

8 0
1 year ago
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