Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 50 centimeters to 80 centimeters

Answer:

R = 1138.9 Ω

Explanation:

Hello,

In this case, for the given power (P) and current (I), we can compute the resistance (R) via:

R = P / I²

Thus, we obtain:

R = 1.64x10⁵ W / (12.0 A)²

R = 1138.9 Ω

Best regards.

Answer:

v = 8.5 m/s

Explanation:

In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,

Weight = Centripetal Force

but,

Weight = mg

Centripetal Force = mv²/r

Therefore,

mg = mv²/r

g = v²/r

v² = gr

v = √gr

where,

v = minimum speed required = ?

g = 9.8 m/s²

r = radius = 7.4 m

Therefore,

v = √(9.8 m/s²)(7.4 m)

v = 8.5 m/s

Minimum speed for a roller coaster while travelling upside down  so that the person will not fall out = 8.5 m/s

For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.

In the given condition the weight of the person must be balanced by the centrifugal force.

and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person

According to the Newton's Second Law of motion we can write force balance

[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]

Given Radius of loop = r = 7.4 m

Putting the value  of r = 7.4 m  in equation (1) we get

[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]

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In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm

(a) 7.392cm

(b) -15.32 cm/s

(c) -184cm/s²

(d) 0.4πs and 8.00cm

The general equation of a simple harmonic motion (SHM) is given by;

x(t) = A cos (wt + Φ)        --------------(i)

Where;

x(t) = position of the body at a given time t

A =  amplitude or maximum displacement during oscillation

w = angular velocity

t = time

Φ = phase constant.

Given from question:

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

(a) At time t = 0;

The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;

x(0) = 8.00 cos (5(0) + π / 8)

x(0) = 8.00 cos (π /8)

x(0) = 8.00 x 0.924

x(0) = 7.392 cm

Therefore, the position of the piston at time t = 0 is 7.392cm

(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;

v(t) = [tex]\frac{dx(t)}{dt}[/tex]

v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]

v(t) = 8 (-5 sin (5t + π / 8))

v(t) = -40sin(5t + π / 8)     --------------------(iii)

Now, substitute t=0 into the equation as follows;

v(0) = -40 sin(5(0) + π / 8)

v(0) = -40 sin(π / 8)

v(0) = -40 x 0.383

v(0) = -15.32 cm/s

Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s

(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;

a(t) = [tex]\frac{dv(t)}{dt}[/tex]

a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]

a(t) = -200 cos (5t + π / 8)

Now, substitute t = 0 into the equation as follows;

a(0) = -200 cos (5(0) + π / 8)

a(0) = -200 cos (π / 8)

a(0) = -200 x 0.924

a(0) = -184.8 cm/s²

Therefore, the acceleration of the piston at time t = 0 is -184cm/s²

(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;

x(t) = A cos (wt + Φ)                   --------------(i)

x(t) = 8.00 cos (5t + π / 8)         ---------------(ii)

From these equations it can be deduced that;

Amplitude, A = 8.00cm

Angular velocity, w = 5 rads/s

But;

w = [tex]\frac{2\pi }{T}[/tex]           [Where T = period of oscillation]

=> T = [tex]\frac{2\pi }{w}[/tex]

=> T = [tex]\frac{2\pi }{5}[/tex]

=> T = 0.4π s

Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm

Answer:

344.8 x10^-8J/m³

Explanation:

Using=> energy intensity/ speed oflight

= 1000/2.9x10^8

= 344.8 x10^-8J/m³

The electromagnetic energy is 344.8 x10⁻⁸J/m³

We have to use the formula which says

Electromagnetic energy = energy intensity/ speed of light

We are given intensity as 1000 W/m²

Electromagnetic energy    = 1000/2.9 x 10⁸

                                             = 344.8 x 10⁻⁸J/m³

Therefore the electromagnetic energy is contained in each cubic meter will be  344.8 x 10⁻⁸J/m³

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Answer:

The moment of inertia of the merry-go round is 453.125 kg.m²

Explanation:

Given;'

angular velocity of the merry-go round, ω = 1.6 rad/s

rotational kinetic energy, K =  580 J

Rotational kinetic energy is given as;

K = ¹/₂Iω²

Where;

I is the moment of inertia of the merry-go round

[tex]I = \frac{2K}{\omega^2} \\\\I = \frac{2*580}{1.6^2} \\\\I = 453.125 \ kg.m^2[/tex]

Therefore, the moment of inertia of the merry-go round is 453.125 kg.m²

Since the small merry-go round is spinning about its center in a clockwise direction, its moment of inertia is equal to 453.13 [tex]Kgm^2[/tex]

Given the following data:

To calculate the moment of inertia of the small merry-go round:

Mathematically, the rotational kinetic energy of an object is giving by the formula:

[tex]E_{rotational} = \frac{1}{2} Iw^2[/tex]

Where:

Making moment of inertia (I) the subject of formula, we have:

[tex]I = \frac{2E_{rotational}}{w^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]I = \frac{2(580)}{1.6^2}\\\\I = \frac{1160}{2.56}[/tex]

Moment of inertia (I) = 453.13 [tex]Kgm^2[/tex]

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Answer:

2404 MPa

Explanation:

See attachment for solution

The maximum stress that exists at the tip of the internal crack is 3,400 Mpa.

The given parameters;

The maximum stress that exists at the tip of the internal crack is calculated as follows;

[tex]\sigma _{max} = 2\sigma \times \sqrt{(\frac{l}{r} )} \\\\\sigma _{max} = 2 \times 170 \times 10^6 \times \sqrt{(\frac{2.5\times 10^{-2}}{2.5 \times 10^{-4}})} \\\\\sigma _{max} = 3.4 \times 10^{9} \ Pa\\\\\sigma _{max} = 3,400\ Mpa[/tex]

Thus, the maximum stress that exists at the tip of the internal crack is 3,400 Mpa.

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Answer:

The hydrostatic force on one end of the trough is 54994.464 N

Explanation:

Given;

liquid density, ρ = 810 kg/m³

side of the equilateral triangle, L = 8m

acceleration due to gravity, g =  9.8 m/s²

Hydrostatic force is given as;

H = ρgh

where;

h is the vertical height of the equilateral triangle

Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.

let the half side of the triangle = x

x = ⁸/₂ = 4m

The half section of the triangle forms a right angled triangle

h² = 8² - 4²

h² = 48

h = √48

h = 6.928m

F = ρgh

F = 810 x 9.8 x 6.928

F = 54994.464 N

Therefore, the hydrostatic force on one end of the trough is 54994.464 N

Answer:

Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.

Answer:

Order of 10^(-35) m.

Explanation:

The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.

Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.

This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.

Given that,

Outer diameter = 14 mm

Inner diameter = 11 mm

Length = 21 cm

Young's modulus = 68 GPa

Tensile strength = 7 Mpa

(a). We need to calculate the effective spring constant of the straw with respect to elongation

Using formula of effective spring constant

[tex]\dfrac{Y}{\Delta l}=\dfrac{YA}{l}[/tex]

[tex]k=\dfrac{YA}{l}[/tex]

Where, k = effective spring constant

Y= Young's modulus

A = area

l = length

Put the value into the formula

[tex]k=\dfrac{68\times10^{9}\times\dfrac{\pi}{4}(0.014^2-0.011^2)}{21\times10^{-2}}[/tex]

[tex]k=1.90\times10^{7}\ N/m[/tex]

(b). Force = 90 N

We need to calculate the stretch the straw

Using formula of stretch

[tex]\Delta l=\dfrac{F}{k}[/tex]

Where, F = force

k = effective spring constant

Put the value into the formula

[tex]\Delta l=\dfrac{90}{1.90\times10^{7}}[/tex]

[tex]\Delta l=0.0000047\ m[/tex]

[tex]\Delta l=4.7\times10^{-6}\ m[/tex]

(c). We need to calculate the extend the length of the straw before it breaks

Using formula of extend length

[tex]E_{max}=\dfrac{Y\Delta l}{l}[/tex]

[tex]\Delta l=\dfrac{E_{max}l}{Y}[/tex]

Put the value into the formula

[tex]\Delta l=\dfrac{7\times10^6\times21\times10^{-2}}{68\times10^{9}}[/tex]

[tex]\Delta l=0.0000216\ m[/tex]

[tex]\Delta l=0.0216\times10^{-3}\ m[/tex]

[tex]\Delta l=0.0216\ mm[/tex]

Hence, (a). The effective spring constant of the straw with respect to elongation is [tex]1.90\times10^{7}\ N/m[/tex]

(b). The stretch the straw is [tex]4.7\times10^{-6}\ m[/tex]

(c). The extend length of the straw before it breaks is 0.0216 mm.

Answer:

b

Explanation:

determine the rotational kinetic energy about it's center of mass

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is [tex]v = 0.0002808 \ m/s[/tex]

Explanation:

From the question we are told that

    The current on the copper is  [tex]I = 20 \ A[/tex]

     The cross-sectional area is  [tex]A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2[/tex]

The number of copper atom in the wire is  mathematically evaluated

      [tex]n = \frac{\rho * N_a}Z}[/tex]

Where [tex]\rho[/tex] is the density of copper with a value [tex]\rho = 8.93 \ g/m^3[/tex]

          [tex]N_a[/tex] is the Avogadro's number with a value [tex]N_a = 6.02 *10^{23}\ atom/mol[/tex]

         Z  is the molar mass of copper with a value  [tex]Z = 63.55 \ g/mol[/tex]

So

     [tex]n = \frac{8.93 * 6.02 *10^{23}}{63.55}[/tex]

     [tex]n = 8.46 * 10^{28} \ atoms /m^3[/tex]

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         [tex]N = 8.46 * 10^{28} \ electrons[/tex]

The current through the wire is mathematically represented as

         [tex]I = N * e * v * A[/tex]

substituting values

        [tex]20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}[/tex]

=>     [tex]v = 0.0002808 \ m/s[/tex]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

[tex]a = \frac{v^2}{r}[/tex]

At constant speed, we will have;

[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

Answer:

the ray of light should approach normal

Explanation:

When light passes through two means of different refractive index, it fulfills the equation

              n₁ sin  θ₁ = n₂ sin θ₂

where index 1 and 2 refer to each medium

In this problem, they tell us that light passes to a medium with a higher index, which is why

               n₁ <n₂

let's look for the angle in the second half

            sinθ₂ = n₁ /n₂  sin θ₁

            θ₂ = sin⁻¹ (n₁ /n₂  sin θ₁)

let's examine the angle argument the quantity n₁ /n₂ <1   therefore the argument decreases, therefore the sine and the angle decreases

Consequently the ray of light should approach normal

Answer:

ML²/6

Explanation:

Pls see attached file

The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,

The length of the rod is L. It has a non-uniform distribution of mass given by:

dm/dx = Cx

where C has units kg/m²

dm = Cxdx

the total mass M of the rod can be calculated by integrating the above relation over the length:

[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]

Thus,

C = 2M/L²

Now, the moment of inertia of the small element dx of the rod is given by:

dI = dm.x²

dI = Cx.x²dx

[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]

I = ML²/2

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Explanation:

velocity of disc [tex]=\sqrt((gh)/0.75)[/tex]

lets call (h) 1 m to make it simple.

= 3.614 m/s

[tex]\sqrt((4/3) x 1 x 9.8) = 3.614[/tex] m/s pointing towards this:

[tex]4×V_d=\sqrt(4/3hg)[/tex]

[tex]V_h=\sqrt(hg)[/tex]

velocity of hoop=[tex]\sqrt(gh)[/tex]

lets call (h) 1m to make it simple again.

[tex]\sqrt(9.8 x 1) = 3.13[/tex] m/s

[tex]\sqrt(gh) = sqrt(hg)

so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)[/tex]

The disc is the fastest.

While i'm on this subject i'll show you this:

Solid ball [tex]=0.7v^2= gh[/tex]

solid disc [tex]= 0.75v^2 = gh[/tex]

hoop [tex]=v^2=gh[/tex]

The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.

The solid ball will be the faster of the 3, like above i'll show you.

solid ball: velocity [tex]=\sqrt((gh)/0.7)[/tex]

let (h) be 1m again to compare.

[tex]\sqrt((9.8 x 1)/0.7) = 3.741[/tex] m/s

solid disk speed [tex]=\sqrt((gh)/0.75)[/tex]

uniform hoop speed [tex]=\sqrt(gh)[/tex]

solid sphere speed [tex]=\sqrt((gh)/0.7)[/tex]

Answer:

The period is [tex]T = 0.700 \ s[/tex]

Explanation:

From the question we are told that  

    The mass is [tex]m = 0.350 \ kg[/tex]

     The extension of the spring is  [tex]x = 12.0 \ cm = 0.12 \ m[/tex]

The spring constant for this is mathematically represented as

       [tex]k = \frac{F}{x}[/tex]

Where F is the force on the spring which is mathematically evaluated as

       [tex]F = mg = 0.350 * 9.8[/tex]

       [tex]F =3.43 \ N[/tex]

So  

    [tex]k = \frac{3.43 }{ 0.12}[/tex]

    [tex]k = 28.583 \ N/m[/tex]

The period of oscillation is mathematically evaluated as

      [tex]T = 2 \pi \sqrt{\frac{m}{k} }[/tex]

substituting values

     [tex]T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }[/tex]

     [tex]T = 0.700 \ s[/tex]

Answer:

straight

Explanation:

because it is good in that direction

Answer:

The total work done by the person is given as = m g h

= 4.5kg x 9.8m/s²x1.2m

= 52.92J

This is the work done in moving the block in a vertical distance

However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.

Explanation:

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

Answer:

Such limitations are given below.

Explanation:

Answer:

The correct options are:

B) They have the same wave speed as visible light

D) They have a lower frequency than gamma rays.

Explanation:

B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s

D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).

We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.

Answer: c. expansion coefficients.

Explanation: Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients.

I found the answer on Quizlet. :)

Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients. The correct option is c.

The coefficient of thermal expansion (CTE) is the rate at which a material expands as its temperature rises. This coefficient is determined at constant pressure and without a phase change, i.e. the material is expected to remain solid or fluid.

Bimetallic strips, which are utilized as adjustable switches in electric appliances, are made up of metallic strips with differing expansion coefficients. The coefficient of thermal expansion indicates how the size of an object varies as temperature changes.

Therefore, the correct option is c. expansion coefficients.

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#SPJ2

Answer:

10.4 m/s

Explanation:

Given that

mass of the first snowball, m1 = 0.3 kg

mass of the second snowball, m2 = 0.7 kg

Magnitude of initial velocity for both masses, u = 10.4 m/s

To start with, we use the formula of conservation of linear momentum which states that

magnitude of initial momentum is equal to magnitude of final momentum.

m1u1 + m2u2 = V(m1 + m2)

0.3 * 10.4 + 0.7 * 10.4 = V(0.3 + 0.7)

3.12 + 7.28 = V(1)

10.4 = V

The 1 kg mass is an addition Of the 0.3 mass & 0.7 kg mass.

Thus, the speed of the 1 kg mass is 10.4 m/s

Answer:

a) 40 V

b) 69.23 V

c) 69.23 V

Explanation:

See attachment for solution

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is [tex]m_r = 2 \ kg[/tex]

      The  initial  speed of the rubber ball is  [tex]u = 3 \ m/s[/tex]

      The final speed at which it bounces bank [tex]v - 3 \ m/s[/tex]

      The mass of the clay ball  is  [tex]m_c = 2 \ kg[/tex]

       The  initial  speed of the clay  ball is [tex]u = 3 \ m/s[/tex]

       The final speed of the clay ball is  [tex]v = 0 \ m/s[/tex]

Generally Impulse is mathematically represented as

       [tex]I = \Delta p[/tex]

where [tex]\Delta p[/tex] is the change in the linear momentum so  

       [tex]I = m(v-u)[/tex]

For the rubber  is  

        [tex]I_r = 2(-3 -3)[/tex]

       [tex]I_r = -12\ kg \cdot m/s[/tex]

=>     [tex]|I_r| = 12\ kg \cdot m/s[/tex]

For the clay ball

       [tex]I_c = 2(0-3)[/tex]

        [tex]I_c = -6 \ kg\cdot \ m/s[/tex]

=>    [tex]| I_c| = 6 \ kg\cdot \ m/s[/tex]

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  [tex]P_{avg} = F * v_{avg}[/tex]

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

            [tex]P_{avg} = \frac{W }{\Delta t }[/tex]

Where W is  is the Workdone which is  mathematically represented as

         [tex]W = F * s[/tex]

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                [tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]

Now this  displacement can be represented mathematically as  

            [tex]s = v_{avg} * \Delta t[/tex]

Where [tex]v_{avg }[/tex] is the average  velocity and [tex]\Delta t[/tex] is the time  taken  

So  

            [tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]

=>         [tex]P_{avg} = F * v_{avg}[/tex]

Answer:

Pavg =  Fvavg

Explanation:

Since the P (power) done by the F (force) is:

P = Fs/t

and we are looking for the velocity, so then it would be:

P = Fv

with the average velocity the answer is:

Pavg =  Favg

If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v  is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).

Answer:

Higher voltages pose huge dangers to humans with changes in resistance however, the use of hand lotion and rubber gloves help to reduce the amount of current flowing through the human. Notwithstanding higher currents can lead to death by stopping blood flow to the heart causing a heart attack. Best way to survive voltages above 120 volts is to use a rubber glove and avoid touching bare wires.

Explanation:

Other classes of gloves that can be used include:  Class 1 gloves which can be used for current up to 7,500 volts of AC, Class 2 up to 17,000 volts AC, Class 3 up to 26,500 volts AC, and Class 4 up to 36,000 volts AC. However, cotton gloves can be used inside to absorb perspiration and to improve the comfort of the user.

Given Information:  

Radius of wire = r = 0.405 mm = 0.405×10⁻³ m

Length of wire = L = 15 m

Voltage = V = 12 V

Resistivity =  ρ = 100×10⁻⁸ Ωm

Required Information:  

Current = I = ?

Answer:  

Current = I = 0.412 A

Explanation:  

The current flowing through the wire can be found using Ohm's law that is

V = IR

I = V/R

Where V is the voltage across the wire and R is the resistance of the wire.

The resistance of the wire is given by

R = ρL/A

Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(0.405×10⁻³)²

A = 0.515×10⁻⁶ m²

So the resistivity of the wire is

R = ρL/A

R = (100×10⁻⁸×15)/0.515×10⁻⁶

R = 29.126 Ω

Finally, the current flowing through the wire is

I = V/R

I = 12/29.126

I = 0.412 A

Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

Substitute these values into equation (i) as follows;

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² +  [tex]\frac{1}{2}[/tex]m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) +  ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J