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kvasek [131]
3 years ago
11

Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of

Physics
1 answer:
vesna_86 [32]3 years ago
8 0

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

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How much charge passes through a wire in 4.0 s if the current is 3.0 A?
NemiM [27]

Answer:

12 coloumbs

Explanation:

Q = IT

I= 3A, T=4s

Q = 3×4 =12coloumbs

6 0
2 years ago
A car accelerates from zero to a speed of 110
Verizon [17]

The car's rate of  acceleration : a = 2.04 m/s²

<h3>Further explanation</h3>

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

3 0
3 years ago
A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC , is
ICE Princess25 [194]

A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC, is on the x axis at x=0.500 m.

No, there is no point at a finite distance where the electric potential is zero.

Hence, Option D) is correct.

What is electric potential?

Electric potential is the capacity for doing work. In the electrical case, a charge will exert a force on some other charge and the potential energy arises. For example, if a positive charge Q is fixed at some point in space, any other positive charge when brought close to it will experience a repulsive force and will therefore have potential energy.

It is also defined as the amount of work required to move a unit charge from a reference point to a specific point against an electric field.

To learn more about electric potential, refer to:

brainly.com/question/15764612

#SPJ4

4 0
1 year ago
When the temperature of the air is 50°C, the velocity of a sound wave traveling through the air is approximately?
noname [10]

The answer is:

C. 361 m/s

The explanation:

To calculate  the speed of sound at a given temperature (50°C) we are going to use this formula:

v = 331 + 0.6T

when V is the velocity

and T is the temperature = 50°C

by substitution:

v = 331 + 0.6(50)

v = 361 m/s

So, The correct answer is C.

because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the  sound in the air is change with temperature.

5 0
3 years ago
Read 2 more answers
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=\frac{N2}{mg}*l * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= \frac{241.75}{69.1*9.8} *l * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0
3 years ago
Read 2 more answers
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