A linear accelerator can be used to accelerate which of the following?
Question 3 options:

protons and electrons

protons and neutrons

protons only

protons, electrons, and neutrons

Answer:

True

Explanation:

This is because of the point where the forces are applied by our muscles and

the angle they have about the bones. Take for example the  diagram I uploaded.

If we do a free body diagram and a sum of torques, we would get that:

[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]

In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:

[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]

If we solve the equation for the force of the muscle we would get that:

[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]

Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.

Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:

[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]

which yields:

[tex]F_{muscle}=4.04 F_{hand}[/tex]

so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.

Answer:

the required solution is; x(t) = 0.675sin( 2.222t )

Explanation:

Given the data in the question;

Using both Newton's and Hooke's law;

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5

given that mass m = 9 kg

[tex]x[/tex] = 1.8 m

k is F / x

hence

k = F / x

given that, F = 80 N

we substitute

k = 80 / 1.8

k = 44.44

so

m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,

we input

9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,

[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0

so auxiliary equation is,

r² + 4.9377 = 0

r² = -4.9377

r = √-4.9377

r = ±2.222i

hence, the solution will  be;

x(t) = A×cos( 2.222t ) + B×sin( 2.222t )

⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )

using initial conditions

x(0) = 0

⇒ 0 = A

[tex]x^t[/tex](t) = 1.5

1.5 = 2.222B

so

B = 1.5 / 2.222 = 0.675

Hence, the required solution is; x(t) = 0.675sin( 2.222t )

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]

Force:

[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]

Answer:

kinetic and static

Explanation:

hope it helps! ^w^

Answer:

13.3 pounds.

Explanation:

For a spring of constant K, with an attached object of mass M, the period can be written as:

T = 2*π*√(M/K)

Where π = 3.14

First, we know that the period is 4 seconds, then we have:

4s = (2*π)*√(M/K)

We know that if the mass is reduced by 10lb, the period becomes 2s.

Then the new mass of the object will be: (M - 10lb)

Then the period equation becomes:

2s = (2*π)*√((M-10lb)/K)

So we have two equations:

4s = (2*π)*√(M/K)

2s = (2*π)*√((M-10lb)/K)

We want to solve this for M.

First, we need to isolate K in one of the equations.

Let's isolate K in the first one:

4s = (2*π)*√(M/K)

(4s/2*π) = √(M/K)

(2s/π)^2 = M/K

K = M/(2s/π)^2 = M*(π/2s)^2

Now we can replace it in the other equation.

2s = (2*π)*√((M-10lb)/K)

First, let's simplify the equation:

2s/(2*π) = √((M-10lb)/K)

1s/π =  √((M-10lb)/K)

(1s/π)^2 =  ((M-10lb)/K

K*(1s/π)^2 = M - 10lb

Now we can use the equation: K =  M*(π/2s)^2

then we get:

K*(1s/π)^2 = M - 10lb

(M*(π/2s)^2)*(1s/π)^2 = M - 10lb

M/4 = M - 10lb

10lb = M - M/4

10lb = (3/4)*M

10lb*(4/3) = M

13.3 lb = M

Answer:

196 megajoules

Explanation:

Since you are talking about the gravitational potential I am assuming 10km is the height of the object in free fall.

PEg = mgh    2000kg×9.8m/s²×10000m = 196 megajoules

Answer:

Explanation:

When a falcon is hovering, the force of up thrust is balanced by the weight.

When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.

Answer:

V = k Q / R       potential at distance R for a charge Q

R = k Q / V

R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m

Note: Our equation says that if R if infinite then V must be zero.

Answer:

Answer is B (Decomposition)

Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(

Answer:

[tex]T=3*10^-3 N/m[/tex]

Explanation:

From the question we are told that:

Radius :

[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]

Work [tex]W=1.5 * 10^{-4}[/tex]

Generally the equation for Work done  is mathematically given by

[tex]W=TdA[/tex]

Where

[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]

[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]

[tex]dA=0.050m^2[/tex]

Therefore

[tex]W=TdA[/tex]

[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]

[tex]T=3*10^-3 N/m[/tex]

Answer:

standard kilogram is the SI unit of mass

Answer:

The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram

Answer:

A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.

Answer:

a)  v = 0.18 m / s, b)  K₀ₐ = 0.0378 J, c) K_{ob}= 0, d)  K = 0.02349 J,

Explanation:

a) For this exercise we must define a system formed by the two cars, so that the forces during the collision are internal and the moment is conserved

initial instant. Before the hole

         p₀ = ma v₀ₐ

final intnate. After the crash

         p_f = (mₐ + m_b) v

the moment is preserved

         p₀ = p_f

         mₐ v₀ₐ = (mₐ + m_b) v

         v = [tex]\frac{m_a}{m_a+m_b} \ v_{oa}[/tex]

let's calculate

         v = [tex]\frac{0.900}{0.900+0.550} \ 0.29[/tex]

         v = 0.18 m / s

in the same direction of the movement of carriage A

b) the initial kinetic energy car A

         K₀ₐ = ½ m  v₀ₐ²

         K₀ₐ = ½ 0.900 0.29²

         K₀ₐ = 0.0378 J

c) kinetic energy of carriage B

          k_{ob} = 0

because the car is stopped

d) the kinetic energy of the system

          K = ½ (mₐ + m_b) v²

           K = ½ (0.900 + 0.550) 0.18²

           K = 0.02349 J

E) we see that part of the kinetic energy is lost, therefore the scientific reeling is not conserved in inelastic collisions

F) and momentum is conserved since it is equal to the variation of the moment and this is conserved in all collisions

Answer:

x = 4 t^2 - 15 t + 25        displacement of particle

dx / dt = 8 t - 15        velocity of particle

d^2x / dt^2 = 8       acceleration of particle

If 8 t -15 = o     then t = 8 / 15

Since acceleration is a constant 8 then motion has uniform acceleratkon

The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.

An item or particle that is propelled in a gravitational influence, such as from the crust of the Ground, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.

A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m.

The initial velocity is zero. Then the time taken to reach the ground is given as,

h = ut + 1/2at²

- 24 = 0×t + 1/2 (-9.8)t²

24 = 4.9t²

t² = 4.8979

t = 2.213 seconds

Then the horizontal speed is given as,

v = 18 / 2.213

v = 8.133 meters per second

The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.

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Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

Answer:

First remember that the distance between two points (a, b) and (c, d) is given by the equation:

[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]

Now let's define the position of the radar as:

(0mi, 0mi)

Then we can write the position of the plane as:

(480mi/h*t, 1mi)

where t is time in hours.

Then we can write the distance equation as:

[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]

Now we want to get:

the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.

So first we want to find the value of t such that:

d(3) = 3mi

We will look at the positive value of t, because at this point the plane is increasing its distance to the station.

[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]

The rate of change when the plane is 3 mi away from the station is:

d'(0.0059h)

remember that:

d'(t) = dd(t)/dt

We can write:

d(t) = h( g(t) )

such that:

h(x) = √x

g(t) = (480mi/h*t)^2 + (1mi)^2

then:

d'(t) = h'(g(t))*g'(t)

This is:

[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]

The rate of change at t = 0.0059h is then:

[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]

Answer:

a) fem = 5.709 V,  b)  t = 0.196 s,  c)  t = 0.589 s, d)   T = 0.785 s

Explanation:

This is an exercise in Faraday's law

          fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]

          fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]

The magnetic field and the area are constant

          fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]

          fem = - N B A (-sin θ)  [tex]\frac{d \theta}{dt}[/tex]

          fem = N B (π d² / 4) sin θ   w

          fem= [tex]\frac{\pi }{4}[/tex]  N B d² w sin θ

with this expression we can correspond the questions

a) the peak of the electromotive force

this hen the sine of the angle is 1

         sin θ = 1

         fem = [tex]\frac{\pi }{4}[/tex]   77  1.18  0.10² 8.0

         fem = 5.709 V

b) as the system has a constant angular velocity, we can use the angular kinematics relations

          θ = w₀ t

          t = θ/w₀

Recall that the angles are in radians, so the angle for the maximum of the sine is

           θ= π/2

           t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]

           t = 0.196 s

c) for the electromotive force to be negative, the sine function of being

            sin θ= -1

whereby

          θ = 3π/ 2

          t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]  

          t = 0.589 s

d) This electromotive force has values ​​that change sinusoidally with an angular velocity of

          w = 8 rad / s

angular velocity and period are related

          w = 2π / T

          T = 2π / w

          T = 2π / 8

          T = 0.785 s

Answer:

the person's heart rate

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

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Explanation:

The sum of the voltages of the components connected in a series circuit is equal to the voltage across the battery.

[tex]V_T = V_1 + V_2 +V_3[/tex]

From Ohm's law ([tex]V=IR[/tex]) and in a series circuit, the amount of current flowing through the components is the same for all. So we can write [tex]V_T[/tex] as

[tex]V_T= 75\:\text{V} = I(170)+I(190) + 21\:\text{V}[/tex]

[tex]I(170+190)=54\:\text{V}[/tex]

[tex]I= \dfrac{54\:\text{V}}{360\:\text{ohms}}=0.15\:\text{A}[/tex]

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]

ωf = 8.8 rad/s

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

v = 2.2 m/s

[tex]4.4×10^{-7}\:\text{m}[/tex]

Explanation:

The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:

[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]

or

[tex]\dfrac{hc}{\lambda} = \phi[/tex]

Solving for the wavelength [tex]\lambda[/tex],

[tex]\lambda = \dfrac{hc}{\phi}[/tex]

[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]

[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]

Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.

The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.

To find the wavelength, the given values are,

Energy (E) = 4.50×10¯¹⁹ J

The distance between two consecutive crests and troughs is called the wavelength of a wave.

Here, for the wavelength,

Energy (E) = 4.50×10¯¹⁹ J

Planck's constant (h) = 6.626×10¯³⁴ Js

Speed of light (v) = 3×10⁸ m/s

The wavelength of the light can be obtained as illustrated below:

E = hv / λ

Cross multiply λ,

E × λ = hv

Divide both sides by E,

λ = hv / E

Substituting all the values,

λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹

λ = 0.000000441733 m

λ = 441.73nm

λ - The maximum wavelength of light.

Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm

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Answer:

thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.

Answer:

log T = 3.72 to 3.77

Explanation:

Temperature range is

T = 5200 to 5900

Take the log

So,

log T = log 5200 to log 5900

log T = 3.72 to 3.77

Answer:

49 N

Explanation:

Given,

Mass ( m ) = 50 kg

To find : Weight ( W ) = ?

Take the value of acceleration due to gravity as 9.8 m/s^2

Formula : -

W = mg

W = 50 x 9.8

W = 49 N

Answer:

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Explanation:

a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:

[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)

Where:

[tex]W[/tex] - Useful work, in joules.

[tex]E[/tex] - Food energy, in joules.

If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:

[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]

[tex]\eta = 9.082\,\%[/tex]

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:

[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]

[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]

[tex]E = 7.60\times 10^{5}\,J[/tex]

[tex]E = 181.644\,kcal[/tex]

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Answer:

19m

Explanation:

we have proper length L = 100m

the speed of the train v = 0.95

the speed of light is given as = 3x10⁸

length of the tunnel is given as = 50 meters

we can solve for the lenght contraction as

LX√1-v²/c²

= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )

= 31.22 metres

the train would be well seen at

50 - 31.22

= 18.78

= this is approximately 19 metres

we conclude tht the trains ends clears the ends of the tunnel by 19 meters.

thank you!

Answer:

hgff

Explanation:

Answer:

The charge moves to equilibrium.

E.e = B.e.V

E is electric field force.

e is the charge.

B is magnetic field force.

V is acceleration voltage.