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3 years ago

Plsss help I will mark brainlist

Physics

2 answers:

gulaghasi [49]3 years ago

6 0

Metal improves conductivity since Metals usually have small amounts of valance electrons which allow smoother movement through them. Metalloids can either improve or weaken conductivity by adding or removing an electron. Non-metals have poor conductivity because their valance shell haves 4 or more valance electrons.

AKA It is B

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G/O/O/G/L/E

sp2606 [1]3 years ago

4 0

Answer:the 2nd one

Explanation:

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A 2.00 kg cat is in a 97.00 kg elevator. What force on the elevator cable would be needed to lower the cat/elevator pair with an

umka21 [38]

The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.

<u>Given the following data:</u>

Mass of cat = 2 kg

Mass of elevator = 97 kg

Acceleration = 2 $m/s^2$

To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:

First of all, we would calculate the total mass of the cat/elevator pair.

$Total \;mass=2 + 97$

Total mass = 99 kilograms

Mathematically, Newton's Second Law of Motion is given by this formula;

$Force = mass \times acceleration$

Substituting the given parameters into the formula, we have;

$Force = 99 \times 2$

Net force = 198 Newton

Read more here: brainly.com/question/24029674

3 0

3 years ago

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 .