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Dvinal [7]
4 years ago
6

What are the bonds that connect water molecules?

Physics
2 answers:
velikii [3]4 years ago
6 0

The bonds that hold water molecules together are called covalent bonds.

Burka [1]4 years ago
3 0
The type of bond that connects water molecules together is called covalent bond~
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What is the mass of an object that has a weight of 80.0 N?
Ira Lisetskai [31]

Explanation:

SUPONIENDO QUE LA ACELERACIÓN DE LA GRAVEDAD ES 9.80 m/s^{2}

USANDO LA SEGUNDA LEY DE NEWTON:

<em>m</em> = 80.0 N/9.80 m/s^{2} = 8.16 kg

4 0
3 years ago
Read 2 more answers
Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) 0.080, (b) 0.26, (c) 0.49, (d) 0.90, (e)
mote1985 [20]

Answer:

Explanation:

This is the case of forced oscillation . The pendulum having the same or matching time period  or angular frequency with that of angular frequency  of external periodic force , will be in resonance having largest amplitude.

Angular frequency of pendulum having length .9 m

= \sqrt{\frac{g}{l } }

l = .9

angular frequency

= \sqrt{\frac{10}{0.9 } }

= 3.33 rad / s

If we calculate angular frequencies of pendulum of all lengths given , we will find that other lengths do not give angular frequency falling between 2 and 4 radian . So only pendulum having length of .9 m will have vibration of maximum amplitude.

8 0
3 years ago
A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th
Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = .250\times 10^{- 3}\ m

Now,

Applying the equilibrium condition at the left end for torque:

T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}

T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}

T_{C} = 341.357\ Nm

The weight of the wire balances the tension in both the wires collectively:

W = T_{Al} + T_{C}

531 = T_{Al} + 341.357

T_{Al} = 189.643\ Nm

Now,

The fundamental frequency is given by:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

\mu = A\rho = \pi R^{2}\rho

(a) For the fundamental frequency of Aluminium:

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}

where

\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz

(b)  For the fundamental frequency of Copper:

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}

where

\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz

7 0
4 years ago
If an object accelerates from rest, with a constant acceleration of 4.4 m/s2, what will its velocity be after 28s?
Norma-Jean [14]

Answer:

123.2 m/s after 28s

Explanation:

Vi= 0 m/s

a= 4.4 m/s^2

t=28s

Vf after 28s

To find Vf use your kinematics formula Vf=Vi+at

Vi is Zero so it gets removed and the equation becomes

Vf=at  

Simply Plug and Solve

Vf= 4.4(28)

Vf=123.2 m/s after 28s

6 0
3 years ago
What forces act on human body in equilibrium.​​
shusha [124]

Explanation:

ear drum

muscular force

8 0
3 years ago
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