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bixtya [17]
3 years ago
6

Based on the information in the table, which two elements are most likely in the same group, and why? bismuth and thallium, beca

use their atomic masses are very similar nitrogen and oxygen, because they were both first isolated in the same year sodium and thallium, because their names both end in the same suffix: -ium bismuth and nitrogen, because they have the same number of valence electrons
Physics
2 answers:
Oksanka [162]3 years ago
8 0

D. bismuth and nitrogen, because they have the same number of valence electrons

JUST TOOK THE TEST

kolezko [41]3 years ago
7 0
The answer would be Bismuth and Nitrogen because they have the same number of valence electrons.

Groups in the periodic table are arranged in columns. When it comes to neutral atoms, the number of valence electrons it has is equal to the main group of that element. For example, Bismuth is under group 5A of the periodic table. It has 5 valence electrons. If you look for Nitrogen in the periodic table, you will find it in Group 5A as well and this means that it also has 5 valence electrons. 
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Which of the following are results of the force of gravity?
docker41 [41]
Hello,

Here is your answer:

The proper answer for this question is option B "When released,a book falls to the ground". That's because of gravity the book will hit the ground!

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps.
7 0
3 years ago
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Help on this pls. !!!!!
Luba_88 [7]

Answer:

C

Explanation:

4 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim
never [62]

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

\text{Distance}=20\times 0.5=10\ m

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

\text{Distance}=u\times 0.5=0.5u\ m

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0

u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s

Maximum speed of the car by which it will not hit the deer is 25.1 m/s

5 0
3 years ago
Read 2 more answers
When might i travel at a low velocity but high acceleration?
SCORPION-xisa [38]
Maybe when you ice-skate 10 times per second around a 1-foot circle.
7 0
3 years ago
At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until
Sav [38]

Answer:

a) \Delta \theta = 617.604\,rad, b) \Delta t = 10.392\,s, c) \alpha = -14.128\,\frac{rad}{s^{2}}

Explanation:

a) The final angular speed at the end of the acceleration stage is:

\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)

\omega = 111.5\,\frac{rad}{s}

The angular deceleration is:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}

\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}

\alpha = -14.128\,\frac{rad}{s^{2}}

The change in angular position during the acceleration stage is:

\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}

\theta = 177.604\,rad

Finally, the total change in angular position is:

\Delta \theta = 440\,rad + 177.604\,rad

\Delta \theta = 617.604\,rad

b) The time interval of the deceleration interval is:

\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }

\Delta t = 7.892\,s

The time required for the grinding wheel to stop is:

\Delta t = 2.50\,s + 7.892\,s

\Delta t = 10.392\,s

c) The angular deceleration of the grinding wheel is:

\alpha = -14.128\,\frac{rad}{s^{2}}

4 0
3 years ago
Read 2 more answers
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