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ch4aika [34]
3 years ago
5

A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br

eak when the tension reached how many newtons
Physics
1 answer:
serg [7]3 years ago
7 0

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

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2) A car travels 5 miles north, and then 10 miles south. What is the person's DISTANCE and
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A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

f_{k}=2450N*0.12

so

f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

so

750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

5 0
3 years ago
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