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4 years ago

Why do the group 1 elements tend to become +1 ions

Physics

1 answer:

Likurg_2 [28]4 years ago

7 0

because there small elemts like atoms but bigger than atoms we all know that

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A wheel of radius 0.23 m, which is moving initially at 25.0 m/s, rolls to a stop in 246.0 m. The wheel's rotational inertia is 0

Digiron [165]

Answer: -1.27 m/s^2

Explanation:

a = - V^2 / 2x

a = -(25^2) / 2 x (246) = 1.27 m/ s^2

Therefore the linear acceleration of the wheel is - 1.27 m/s^2

8 0

3 years ago

Need help with this question. Thirty points.

Anon25 [30]

$1)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12 mv^2 \\\\\implies m = \dfrac{2E_k}{v^2} = \dfrac{2\times 4500}{30^2}=10~ kg\\ \\2)\\\\\text{Kinetic energy,}\\\\E_k = \dfrac 12mv^2 \\\\\implies v^2 = \dfrac{2E_k}m\\\\\implies v = \sqrt{\dfrac{2E_k}{m}}=\sqrt{\dfrac{2 \times 320}{\dfrac{20}{9.8}}}= 17.709 ~ ms^{-1}\\\\3)\\\\\text{Kinetic energy,}\\\\E_k=\dfrac 12 mv^2=\dfrac 12 \times 50 \times 10^2=2500 ~J\\\\\\$

$4)\\\\\text{Potential energy,}\\\\E_p =mgh = 5\times 9.8\times 1.5 = 73.5 ~J$

5 0

2 years ago

Please help me on due today

Bad White [126]

Subject-Matter Achievement Goal

1. My subject-matter achievement goal is:

Answer: Read 1 full chapter book in 1 week

2. My action steps are:

Answer: Read an equal amount of chapters every day for a week

3. My time frame is:

Answer: 1 week

4. Assets and Strengths:

Answer: ability to focus,

5. Motivation:

Answer: Great interest in the book I chose

6. Obstacles:

Answer: Might lose page number, forget where I am at in the book, forget to read one day

7. Solutions:

Answer: Get or make a bookmark, keep a personal reading log, set alarm for a certain time each day

Study/Organizational Skills Goal

8. My study/organizational skills goal is:

Answer: manage my time better

9. My action steps are:

Answer: planning things out to be more organized

10. My time frame is:

Answer: from now to the end of the school year

11. Assets and Strengths:

Answer: a planner so I can pace myself

12. Motivation:

Answer: give breaks and take time to do things I enjoy

13. Obstacles:

Answer: leaving things last minute

14. Solutions:

Answer: planning to do certain things before a certain day

This can be for any of those goals

Personal Enrichment Goal

15. My personal enrichment goal is:

Answer: Stay healthy

16. My action steps are:

Answer: eat healthier foods

17. My time frame is:

Answer: now to the end of the school year

18. Assets and Strengths:

Answer: I don't make a big deal out of eating veggies and stuff

19. Motivation:

Answer: treat myself every so often

20. Obstacles:

Answer: Craving junk food

21. Solutions:

Answer: find a way to make healthier foods taste similar to junk food (EX. Put cheese on broccoli)

4 0

3 years ago

The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

8 0

4 years ago

A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha

slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity $\rho$ , the electric field $E$ , and the current density $J$ is given by

$\rho = \dfrac{E}{J}$

This equation can be solved for $J$ to get:

$J = \dfrac{E}{\rho}$

Since the current is $I = J\cdot A$

$I= J\cdot A = \dfrac{E}{\rho} \cdot A$

Now, for the tube of mercury $\rho = 9.84*10^{-7}\: \Omega \cdot m$ , $E = 23N/C$ , and the area is $A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2$ ; therefore,

$I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2$

$\boxed{I = 73.39A.}$

Hence, the current through the mercury tube is 73.39A.

5 0

4 years ago