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2 years ago

What family of elements has 7 valence electrons?

2 answers:

6 0

The elements in this family are fluorine, chlorine, bromine, iodine, and astatine. Halogens have 7 valence electrons, which explains why they are the most active non-metals. They are never found free in nature.

ivann1987 [24]2 years ago

4 0

Group 17 (VII) (halogens)

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find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²

Ierofanga [76]

Answer:

$T_{m }$ = 4.86 s

$T_{e}$ = 1.98 s

Explanation:

Given:

Length = l = 1 m

Acceleration due to gravity of moon = $g_{m}$ = 1.67 m/s²

Acceleration due to gravity of Earth = $g_{e}$ = 10 m/s²

Required:

Time period = T = ?

Formula:

T = 2π $\sqrt{\frac{l}{g} }$

Solution:

For moon

Putting the givens,

T = 2(3.14) $\sqrt{\frac{1}{1.67} }$

T = 6.3 $\sqrt{0.6}$

T = 6.3 × 0.77

T = 4.86 sec

For Earth,

Putting the givens

T = 2π $\sqrt{\frac{1}{10} }$

T = 2(3.14) $\sqrt{0.1}$

T = 6.3 × 0.32

T = 1.98 sec

3 0

2 years ago

Find the poing of center of gravity plz show the steps...

olasank [31]

Answer:

Explanation:

For a uniformly distributed mass, the center of gravity is also the geometric center. For this shape, the center is at point C.

7 0

2 years ago

Water flows steadily through a pipe of length L and radius R=75mm. The velocity distribution across the outlet is given by u=uma

ruslelena [56]

Answer:

4/3

Explanation:

Shown in the picture attached

8 0

3 years ago

Please can some one explaine fore me What is work done

olga nikolaevna [1]

In Physics, 'work' has a very clear definition:

It's (strength of a force) times (distance through which the force acts).

'Work' has the units of Energy.

If you push against a shopping cart with 30 newtons of force, and
you keep pushing while the cart moves 4 meters, then you have
done (30 x 4) = 120 newton-meters of work = 120 "Joules".

5 0

2 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$