The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)?
A. 3/2000s
B. 3/5s
C. 1/15s
D. 1/4

Answer:

A13 mm is about 1/4 A5 mm

Explanation:

Find the attachment

Answer:

Ke- = Ke+ = 0.294MeV

Explanation:

To fins the kinetic energy of both electron and positron you use the following formula, for the case of annihilation of one electron an positron:

2[tex]E_p=2E_o+K_{e^-}+K_{e^+}[/tex]   (1)

Ep: photon energy = 0.804MeV

Eo: rest energy of one electron (and positron) = 0.51MeV

Ke-: kinetic energy of electron

Ke+: kinetic energy of positron

You replace the values of Ep and Eo in the equation (1):

[tex]K_{e^-}+K_{e^+}=2E_p-2E_o=2(0.804MeV-0.51MeV)=0.588MeV[/tex]

Iy you assume both positron and electron have the same speed, then, the kinetic energy of them are equal, and the kinetic energy of each one is:

[tex]K_{e^-}=K_{e^+}=\frac{0.588MeV}{2}=0.294MeV[/tex]

Answer:

2Ω

Explanation:

Ohm's law:

V = IR

10 V = (5 A) R

R = 2 Ω

Answer:

3985 s or 66.42 mins

Explanation:

Given:-

- The length of the rod, L = 83.0 cm

- The cross sectional diameter of rod , d = 2.4 cm

- The temperature of reservoir, Tr = 540°C

- The amount of ice in chamber, m = 1.43 kg

- The temperature of ice, Ti = 0°C

- Thermal conductivity of silver, k = 406 W / m.K

- The latent heat of fusion of water, Lf = 3.33 * 10^5 J / kg

Find:-

How much time is required for the ice to melt completely

Solution:-

- We will first determine the amount of heat ( Q ) required to melt 1.43 kg of ice.

- The heat required would be used as latent heat for which we require the latent heat of fusion of ice ( Lf ). We will employ the first law of thermodynamics assuming no heat is lost from the chamber ( perfectly insulated ):

                              [tex]Q = m*L_f\\\\Q = ( 1.43 ) * ( 3.33 * 10 ^5 )\\\\Q = 476190 J[/tex]

- The heat is supplied from the hot reservoir at the temperature of 540°C by conduction through the silver rod.

- We will assume that the heat transfer through the silver rod is one dimensional i.e along the length ( L ) of the rod.

- We will employ the ( heat equation ) to determine the rate of heat transfer through the rod as follows:

                             [tex]\frac{dQ}{dt} = \frac{k.A.dT}{dx}[/tex]

Where,

                           A: the cross sectional area of the rod

                           dT: The temperature difference at the two ends of the rod

                           dx: The differential element along the length of rod ( 1 - D )

                           t: Time ( s )

- The integrated form of the heat equation is expressed as:

                            [tex]Q = \frac{k*A*( T_r - T_i)}{L}*t[/tex]

- Plug in the respective parameters in the equation above and solve for time ( t ):

                           [tex]476190 = \frac{406*\pi*0.024^2 * ( 540 - 0 ) }{0.83*4}*t \\\\t = \frac{476190}{119.49619} \\\\t = 3985 s = 66.42 mins[/tex]

Answer: It would take 66.42 minutes to completely melt the ice

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

[tex]\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}[/tex]

(see the attached graphic)

We have

[tex]\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]

[tex]\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)[/tex]

[tex]\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)[/tex]

Then by setting components equal, we have

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0[/tex]

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]

We only care about the direction for this question, which we get from the first equation:

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}[/tex]

[tex]\cos\theta=-\dfrac57[/tex]

[tex]\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)[/tex]

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

Answer:

SO THAT

Answer:

The moment of inertia is [tex]I = 1.8 \ kg m^2[/tex]

Explanation:

From the question we are told that

   The force applied is  [tex]F = 10 \ N[/tex]

    The distance of the knob to the hinge is  [tex]d = 0.9 \ m[/tex]

     The angular acceleration is  [tex]a = 5 \ rad/s[/tex]

The moment is mathematically represented as

        [tex]I = \frac{d Fsin(\theta)}{a}[/tex]

Here [tex]\theta = 90^o[/tex] This is because the force direction is perpendicular to the plane of the door

substituting values

          [tex]I = \frac{0.9 * 10 * sin (90)}{5}[/tex]

          [tex]I = 1.8 \ kg m^2[/tex]

Answer:

 v = 126 m / s

Explanation:

Let's analyze this exercise a little, they give us the thrust that is the applied force and the time that it lasts, and they ask us for the final speed, so we can use the Impulse ratio and the variation of the amount of movement

       I = F t = Dp

       F t = pf -p₀

      Now let's use Newton's second law to find the net thrust

        F = E - fr

the friction force has the formula

       fr = μ N

let's write Newton's second law on the y-axis

       N-W = 0

       N = W

we substitute

      fr = μ mg

we look for the net out

       F = 200 - μ mg

With the skater starting from rest, the initial speed is zero (vo = 0)

we substitute

      (200 - very m g) t = m v

       v = (200 µm - very g) t

let's calculate

     v = (200/75 - 0.10 9.8) 75

      v = 126 m / s

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g[tex]/m^{2}[/tex] = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = [tex]\frac{1}{2L}[/tex][tex]\sqrt{\frac{T}{M} }[/tex]

⇒      T = 4[tex]L^{2}[/tex][tex]f^{2}[/tex]M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × [tex]2^{2}[/tex] ×[tex]65^{2}[/tex] × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  [tex]Q =2.094 C[/tex]

Explanation:

From the question we are told that

    The diameter of the wire is  [tex]d = 0.205cm = 0.00205 \ m[/tex]

     The radius of  the wire is  [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]

     The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]

       The electric field change is mathematically defied as

         [tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]

Generally the charge is  mathematically represented as

       [tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]

Where A is the area which is mathematically represented as

       [tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]

 So

       [tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]

Therefore

      [tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]

substituting values

      [tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]

     [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]

From the question we are told that t =  5 sec

           [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]

          [tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]

         [tex]Q =2.094 C[/tex]

The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.

Given the following data:

Conversion:

Diameter of wire = 0.205 cm to m = 0.00205 meter.

Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]

To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:

First of all, we would find the area of the wire.

[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]

Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:

[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]

[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]

When t = 5 seconds:

[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]

Q = 2.094 Coulomb.

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Answer: Tech A is correct

Explanation:

Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.

Answer:

Its called PSY

Explanation: I so do not know why they named it this way but, hope i helped.

Answer:

1.been both -ve charged or both +be charged particles

2. 3.52mC

Explanation:

For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

F = Ke;

Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

F = 320 × 0.033 =10.56N

2.But according to columns law of charge;

F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

Hence Q=√F/K

Q= √10.56/9×10^9

=3.52×10^-3C

= 3.52mC

With smaller gravitational forces and therefor less vertical acceleration, the projectile launched on the moon ... with the same initial speed and direction ...

-- climbs faster,

-- spends more time climbing,

-- reaches a higher peak,

-- falls slower,

-- spends more time falling, and

-- covers more horizontal distance

than the projectile launched on the Earth.

This is not because of air resistance.  It would be true even if there were no air resistance on the Earth.  It's entirely a gravity thing.  

Answer:

x = 3.76 cm

y = 3.76 cm

Explanation:

This composite shape can be modeled as a square (7.2 cm × 7.2 cm) minus a quarter circle in the lower left corner (3.6 cm radius) and a right triangle in the upper right corner (3.6 cm × 3.6 cm).

The centroid of a square (or any rectangle) is at x = b/2 and y = h/2.

The centroid of a quarter circle is at x = y = 4r/(3π).

The centroid of a right triangle is at x = b/3 and y = h/3.

Build a table listing each shape, the coordinates of its centroid (x and y), and its area (A).  Use negative areas for the shapes that are being subtracted.

Next, multiply each coordinate by the area (Ax and Ay), sum the results (∑Ax and ∑Ay), then divide by the total area (∑Ax / ∑A and ∑Ay / ∑A).  The result will be the x and y coordinates of the center of mass.

See attached image.

Answer:

5.06*10^6 m/s

Explanation:

Given that

Initial velocity, u = 4*10^6 m/s

Acceleration, a = 6*10^12 m/s²

Distance traveled, s = 80 cm

Final velocity, v = ?

We can find the final velocity by using one of the equations of motion.

v² = u² + 2as

On substituting the values, we have

v² = (4*10^6)² + 2 * 6*10^12 * 0.8

v² = 2.56*10^13

v = √2.56*10^13

v = 5.06*10^6 m/s

Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s

The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].

The given parameters;

The final velocity of the proton over the given distance is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]

Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]

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Answer:

a) Fs = (μs*mp*g)/2  |  b) τ = Fs*lp  |  c) Wτ,constant = τΘ

Explanation:

a) Fs = (μs*mp*g)/2

b) τ = Fs*lp

c) Wτ,constant = τΘ

Answer:

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat

Explanation:

Fourier's Law of heat conduction, gives the following formula:

Q = - KAΔT/t

where,

Q = Rate of Heat Conduction out of window = ?

K = Thermal Conductivity of Glass = 0.84 W/m.°C

A =Surface Area of window = 2.82 m²

ΔT = Difference in Temperature of both sides of surface

ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)

ΔT = 15°C

t = thickness of window = 0.675 cm = 0.00675 m

Therefore,

Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat.

Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N 

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

Answer:

a. TRUTH

b. FALSE

c. TRUTH

d. FALSE

Explanation:

The emf (electromagnetic force) is generated in a loop or solenoid by the change in the magnetic flux in a closed conductor path (for example, a wire).

This can be noted in the following formula, which is known as the Lenz's law:

[tex]emf=-N\frac{d\Phi_B}{dt}=-N\frac{d(AB)}{dt}[/tex]   (1)

Then, the change, in time, of the area of the conductor, or the change in the magnitude of the magnetic field, the induced emf acquires different values. Furthermore, the loops have a resistance, then, a current can be measured when an emf is induced.

Based on this information you have:

a. an induced emf is caused by a changing magnetic flux. TRUTH

b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field. FALSE

c. an induced emf can be observed by measuring the current that is created. TRUTH

d. an induced emf and conventional induced current are in opposite directions. TRUTH (the minus sing in the equation (1) )

Answer:

A , D , E

Explanation:

Solution:-

- Consider the two identical objects with mass ( m ).

- The stiffness of the springs are ( k1 and k2 ).

- Both the spring store 55.0 J of potential energy.

- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.

- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.

- Apply Energy conservation for spring with stiffness ( k1 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

- Apply Energy conservation for spring with stiffness ( k2 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

Answer: Both objects will have the same maximum speed ( A )

- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:

                         k1 > k2

Where,

                 k1: The stiff spring

                 k2: The flexible spring

Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )

- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,

                      U1 = U2

                      0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2

                      [ k1 / k2 ] = [ x2 / x1 ]^2

Since,

                     k1 > k2 , then [ k1 / k2 ] > 1    

Then,

                     [ x2 / x1 ]^2 > 1

                     [ x2 / x1 ] > 1

                     x2 > x1                  

Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )

Answer:

  a = - e E / m

a = - 1,758 10¹¹ E

Explanation:

For this exercise we can use Newton's second law

        F = m a

where the force is electric

 the forces given by the product of the charge by the electric field

         F = q E

in this case tell us that the charge is the charge of the electron

         q = -e = - 1.6 10⁻¹⁹ C

we substitute

        - e E = m a

          a = - e E / m

we calculate

           a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E

           a = - 1,758 10¹¹ E

The negative sign indicates that the acceleration is in the opposite direction to the electric field

Answer:

a

The free body diagram is shown on the first uploaded image

b

The tension on the rope is  [tex]T=39.16 \ N[/tex]  

Explanation:

From the  question we are told that

    The mass of the box is  m

    The tension on the box is  T

     The angle at which it is pulled is  [tex]\theta = 40^o[/tex]

     The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]

At equilibrium the net force acting on the block along the horizontal axis is zero i.e

     [tex]Tcos \theta -F_w = 0[/tex]

substituting values

     [tex]Tcos (40) -30 = 0[/tex]

     [tex]Tcos (40) = 30[/tex]

     [tex]T(0.76604)) = 30[/tex]

     [tex]T=39.16 \ N[/tex]      

Answer:

Once at a steady cruising speed of about 16,150mph (26,000kph

Explanation:

Answer:

Explanation:

Ex(z,t) = Eocos(kz - ω t + φ)

k = 2π/λ  , ω = 2π f

φ = +30° , E₀ = 10³ V .

z/λ = 0.25 , ft = 0.125

Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)

Putting the values given above

Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )

= 1000cos (90⁰ - 45+30)

= 1000 cos 75

=258.8  V .

Answer:

f = 1.96 revolutions per minute

Explanation:

The formula for the the frequency of revolution of a satellite, to develop an artificial gravity, with the help of centripetal acceleration is given as follows:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = centripetal acceleration= apparent gravity or artificial gravity = 2.2 m/s²

r = radius of station or satellite = diameter/2 = 104 m/2 = 52 m

Therefore,

f = (1/2π)√[(2.2 m/s²)/(52 m)]

f = (0.032 rev/s)(60 s/min)

f = 1.96 revolutions per minute

Answer:

Fx = - (1/h)( wL2 + wL3 - wLend )

Explanation:

Assuming The twins Gilles and Jean has a weight ( w ) each

The torque that would balance the equation would be = wL2 + wL3 -------- 1

THEREFORE the ccw torques are = wLend + Fh ----------- 2

hence equation 2 equals equation 1

= wLend + Fh = wL2 + wL3 --------- 3

equation 3 can as well be represented as

F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4

From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative

therefore equation 4 is represented as

 Fx = - (1/h)( wL2 + wL3 - wLend )

Answer:

a. about 20 years younger

b. Malcolm sits around for 49.94 years

c. 2.268x[tex]10^{17}[/tex] m

Explanation:

light travels 3x[tex]10^{8}[/tex] m in one seconds

in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m

for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex]  = 3.78x[tex]10^{17}[/tex] m

Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex]  = 2.4x[tex]10^{8}[/tex]  m/s

time that will pass for Malcolm will be  distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]  

= 1575000000 s = 49.94 years

the relativistic time t' will be

t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]

t' = 49.94 x 0.6 = 29.96 years       this is the time that has passed for Flo

this means that Flo will be about 20 years younger than Malcolm when she returns

relativistic distance is

d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x 0.6

d' = 2.268x[tex]10^{17}[/tex] m     this is how far it is to Flo

Answer:

a = 5.34 m/s²

T = 37.86 N

Explanation:

This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:

a = g(m₂ - m₁)/(m₂ + m₁)

where,

a = acceleration of both masses = ?

g = 9.8 m/s²

m₂ = heavier mass = 8.5 kg

m₁ = lighter mass = 2.5 kg

Therefore,

a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)

a = (9.8 m/s²)(6 kg)/(11 kg)

a = 5.34 m/s²

The formula for tension in cable is derived to be:

T = 2m₁m₂g/(m₁ + m₂)

T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)

T = 37.86 N

Answer:

The force is  [tex]F = 2400 \ N[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

 From the question we are told that

   The mass of the dancer is  [tex]m_d = 60 \ kg[/tex]

From the diagram the

      The first distance is [tex]l_1 = 20 \ cm[/tex]

      The second distance is  [tex]l_2 = 5 \ cm[/tex]

At equilibrium the moment about the center of the dancers  feet  is mathematically represented as

      [tex]F * l_2 - (mg* l_1)[/tex]

   Where [tex]g= 10 \ m/s^2[/tex]

substituting values

      [tex]F * 5 - (60* 9.8 * 20)[/tex]

=>    [tex]F = \frac{60 * 10 * 30}{5}[/tex]

=>      [tex]F = 2400 \ N[/tex]