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exis [7]
3 years ago
6

Explain how its possible for a compound to have both iconic and covalent bonds.Thanks!

Physics
1 answer:
Evgesh-ka [11]3 years ago
3 0
The atoms which make up the ion are covalently bonded to one another. 19) It is possible for a compound to possess both ionic and covalent bonding. a. If one of the ions is polyatomic then there will be covalent bonding within it.
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If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat
Maksim231197 [3]

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

    v_{max} = Aw

  Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to w\\ of the oscillation .

  Since  w = 2 \pi f

      v_{max}^{|}/v_{max} = 2f/f = 2

  Where v_{max}^{|} is the maximum speed when frequency is doubled .

6 0
3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0
3 years ago
Read 2 more answers
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
2 years ago
During the storm a tree fell over into a river what might happen to the tree
mart [117]
The tree might get swept away by the current and it will disappear when it catches on something
4 0
3 years ago
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