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3 years ago

Do metals gain or lose electrons in order to acquire a full octet

1 answer:

8 0

Metals are just over a full octet, and non-metals just under. It just occurs as metals have 1-3 valence electrons so it is cooler for metals to lose electron than to gain electron for the reason that they would have to get more electrons as likened to losing electrons. So the answer is lose.

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A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

$P = mv$

$P = 1.25 \times 84.3$

$P = 105.4 kg m/s$

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3 years ago

What is normal for a spring that obeys hook's law ?

Bezzdna [24]

A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm

7 0

3 years ago

Read 2 more answers

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

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3 years ago

Students with disabilities who require assistance in class are generally mainstreamed into regular classes rather than putting t

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7 0

3 years ago

Read 2 more answers

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

7.5

×m

...........................(

)

in Above equation in x and t. Separating the variables and integrating,

∫

/7.5

×=

∫

−

4.7619

Here C =constant of integration.

0 at t

, we get: C

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

−

4.7619

(

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

−

4.7619

= −

3.7619

or x

7.15

(b) To find the jogger's acceleration in m

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

7.5×

Now differentiating above equation w.r.t time we get:

−

0.675

the joggers acceleration is :

−

0.675

−

4.34

6 in equation (2). We get:

−

4.7619

(

−

(

0.04

6 )

10=

−

4.7619

0.832

6 0

3 years ago