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3 years ago

9

Whether an ionic bond or a covalent bond forms, bonds form in order to make newly created substances ________ than the substance

s they were created from.
A) less stable
B) more stable
C) less efficient
D) more efficient

1 answer:

Nezavi [6.7K]3 years ago

6 0

The answer would be less stable

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Explanation:

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Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and

djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

$x_f$ = Final length of the spring .

= $\sqrt{1.25^2+1.8^2} -0.60$

= 1.591 m

$x_i$ = The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

= 1.80 m.

m= The mass of the collar.

=3.55 kg

$v_c$ = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Or, $\frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2} = 3.55\times 9.81\times 1.80$

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

Learn more about friction:

brainly.com/question/24338873

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Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0

2 years ago

The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t

slamgirl [31]

Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length $L=5.0\times5.0\times10^{-4}\ m$

We need to calculate the linear charge density

Using formula of linear charge density

$E=\dfrac{2k\lambda}{r}$

$\lambda=\dfrac{Er}{2k}$

Put the value into the formula

$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

$\lambda=1.1\times10^{-10}\ C/m$

We need to calculate the charge

Using formula of charge

$Q=\lambda\timesL$

Put the value into the formula

$Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})$

$Q=2.75\times10^{-13}\ C$

Hence, The charge is $2.75\times10^{-13}\ C$

5 0

3 years ago