You will create a Scientific Argument in the CER (claims, evidence, reasoning) format to answer the following 2 questions:

In a butcher shop, a horizontal steel bar of mass 4.94 kg and length 1.46 m is supported by two vertical wires attached to its e

mojhsa [17]

Answer:

Tension in right wire = 25.9N

Explanation:

I have attached a free body diagram to depict this question.

From the diagram, i have labelled the tensions in the strings T1 and T2.

While i labelled the weight of the bar as Wb and weight of sausage as Ws.

Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)

Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.

Thus;

T1 + T2 = Wb + Ws

We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N

Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N

Thus,

T1 + T2 = 48.46N + 24.43

T1 + T2 = 72.89N - - - - - (eq 1)

Now, let's take moments about the left end of the bar.

The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m

So, moments about left end;

T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)

1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)

1.46T2 = 35.373 + 2.443

1.46T2 = 37.816

T2 = 37.816/1.46 = 25.9N

3 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = $\frac{k\ q_1\ q_2}{r^2}$ .....................1

and here k is 9 × $10^9$ N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = $\frac{k\ q^2}{d^2}$

and force between charges at A to C, at 2d distance

F1 = $\frac{k\ q^2}{(2d)^2}$ = $\frac{k\ q^2}{4d^2}$

force between charges at A to D, 3d distance

F1 = $\frac{k\ q^2}{(3d)^2}$ = $\frac{k\ q^2}{9d^2}$

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a) = -F1 - F2 + F3 ....................2

put here value

F(a) = $-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

solve it

F(a) = $\frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

F(a) = $-\frac{41}{36}\ F1$ = 1.13 F1

and

Charge b It receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2 ....................3

F(b) = $\frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}$

F(b) = $\frac{1}{4} \ F1$ = 0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1 ....................4

F(c) = $\frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}$

F(c) = $\frac{9}{4} \ F1$ = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1 ....................5

F(d) = $-\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}$

so

F(d) = $-\frac{49}{36} \ F1$ = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = $\frac{2.25}{0.25}$

ratio = 9 : 1

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aliya0001 [1]

Answer:

Marie Curie

Explanation:

I hope to see you helped :D?

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I just did this last semester... I know this is wave height!!

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