A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.

Answer:

a)   # _photon = 2.5 10¹⁸ photons / s,   b) E = 10⁻² N / C,  c)     B = 3 10⁻¹¹ T

d)  r=  2 10⁹ m

Explanation:

a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation

          E₀ = h f

          c = λ f

          E₀ = h c /λ

          E₀ = 6.63 10⁻³³⁴   3 10⁸/500 10⁻⁹

          E₀ = 3.978 10⁻⁻¹⁹ J

Let's use a direct ratio rule to find the number of photons

         #_foton = E / Eo

         #_fototn = 1 / 3.978 10⁻¹⁹

         # _photon = 2.5 10¹⁸ photons / s

b) The intensity received by the detector is related to the electric field

          I = E²

Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space

          I = P / A

          P = I A

Let's use index 1 for the point on the bulb and index 2 for the point on the detector.

The area of ​​a sphere is

          A = 4π r²

         P = I₁ A₁ = I₂ A₂

         I₁ r₁² = I₂ r₂²

         I₂ = I₁  r₁²/r₂²

         I₂ = I₁    1 / 100²

         I₂ = I₁ 10⁻⁴

we must know the intensity at the output of the bulb suppose that I₁ = 1 J

          I₂ = 10⁻⁴ J

let's look for the electric field

         E =√I

         E = √10⁻⁴

         E = 10⁻² N / C

c) for the calculation of the magnetic field we use that the field is in phase

               E / B = c

               B = E / c

               B = 10⁻² / 3 10⁸

               B = 3 10⁻¹¹ T

d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area  A = 4π R² where R = 100 m how many photons are there in the area of ​​the detector r = 1 cm,   A’= 10⁻⁴ m²

             #_photons = 2.5 10¹⁸ A_detector / A_sphere

             #_photons = 2.5 1018 10-4 / 4π 10⁴

             #_photons = 2 10⁹ photons in the detector area

for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  

So, option B is the answer

The question is incomplete, the complete question is;

Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect.

Part A If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be

K/2.

K.

2K.

greater than 2K.

Answer:

2K

Explanation:

Given that the kinetic energy of photo electrons is given by;

K= E -Wo

Where;

K = kinetic energy

E= energy of incident photon

Wo = work function

But;

E= hf

Wo = fo

h= Plank's constant

f= frequency of incident photon

fo= Threshold frequency

So:

K= hf - hfo

Where the frequency of incident light is doubled;

K= 2hf - hfo

Hence, maximum kinetic energy of the emitted electrons in this case will be 2K

Answer:

the density of the salt water is 1030 kg/m³

Explanation:

Given;

radius of the cylindrical pool, r = 2 m

depth of the pool, h = 1.3 m

specific gravity of the salt water, γ = 1.03

The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa

Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³

The density of the salt water is calculated as;

[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]

Therefore, the density of the salt water is 1030 kg/m³

Answer:

  Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

        a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

   a = 2 lines 1/600

   a = 2/600

    a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

      tan θ = y / L

as the measured angles are small

      tan θ = sin θ / cos θ sin θ

      sin θ = y / L

we substitute

     a  y/L = λ

     y = λ L / a

for λ = 400 10-9 m

      I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

      i = 346.89 10⁻⁶ m

f

or λ = 700 nm

        y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

        y_f = 609.609 10⁻⁶ m

the separation of this spectrum

        Δr = v_f - i

        Dr = (609.609 - 346)  10 ⁻⁶

        Dr = 263 10⁻⁶ m

Answer:

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

Explanation:

Concepts and reason

The concept required to solve this problem is Newton’s second law of motion.

Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.

Fundamentals

According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:

F = maF=ma

Here, m is mass and a is the acceleration.

(a)

Rearrange the equation F = maF=ma for a.

a = \frac{F}{m}a=  

m

F

​

Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=  

m

F

​

 .

\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}  

a=  

(2300kg+2400kg)

18,000N

​

=  

(4700kg)

18,000N

​

=3.83m/s  

2

​

(b)

Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 for a and 2400 kg for m in the equation F = maF=ma .

\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}  

F=(2400kg)(3.83m/s  

2

)

=9120N

​

Ans: Part a

The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 .

Part b

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².

The force of the SUV's bumper on the truck's bumper is 10000N

Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.

According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.

F = ma

And, a =  F/m

Given, the mass of the ruck , m = 2000 Kg

The mass of the SUV, M = 2500 Kg

The total mass of the both = 2000 + 2500 = 4500 Kg

The maximum force on the trick , F = 18000 N

The maximum acceleration of the truck can give the SUV:

[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]

a = 18000/4500

a = 4 m/s²

The force of the SUV's bumper on the truck's bumper will be:

[tex]F_{max} -f= ma_{max}[/tex]

[tex]f= 18000-2000\times 4[/tex]

[tex]f =10000N[/tex]

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Answer:

where are the options

it's not full question

Answer:

    T₂ > T₁ > T₃ >T₄

Explanation:

In a simple harmonic motion the angular velocity is

         w = [tex]\sqrt{\frac{k}{m} }[/tex]

angular velocity and period are related

         w = 2π / T

we substitute

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

let's find the period for each case

a) mass m

   spring constant k

          T₁ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) mass 2m

   spring constant k

          T₂ = 2π [tex]\sqrt{\frac{2m}{k} }[/tex]

          T₂ = T₁ √2

          T₂ = T₁ 1.41

c) masses 3m

   spring constant 6k

          T₃ = 2π [tex]\sqrt{\frac{3m}{6k} }[/tex]

          T₃ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.5}[/tex]

          T₃ = T₁ 0.707

d) mass m

    spring constant 4k

          T₄ = 2π [tex]\sqrt{ \frac{m}{4k} }[/tex]

          T₄ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.25}[/tex]

          T₄ = T₁ 0.5

now let's order the periods in decreasing order

           T₂ > T₁ > T₃ >T₄

Answer:

2164 Hz

Explanation:

Since point P is 4.70 m away from one speaker and 3.60 m away from the other speaker, the path length difference ΔL = 4.70 m - 3.60 m = 1.1 m.

The path length difference ΔL = nλ for a constructive interference where n is an integer and λ = wavelength of sound from oscillator = v/f where v = speed of sound in air = 340 m/s and f = frequency of sound from oscillator.

So, ΔL = nλ = nv/f

So, the frequency from the oscillator is f = nv/ΔL

Substituting the values of v and ΔL into the equation, we have

f = nv/ΔL

f = n340 m/s/1.1 m

f = n309.09 /s

f = 309.09n Hz

We now insert values of n that will gives us a frequency in the range 1908 Hz to 2471 Hz.

The value of n that will give us a frequency in the range is n = 7

So, when n = 7,

f = 309.09n Hz

f = 309.091 × 7 Hz

f = 2163.64 Hz

f ≅ 2164 Hz

So, the frequency of the oscillator that will produce a constructive interference at P is 2164 Hz.

Answer:

Friction always opposes the motion

HAVE A GREAT DAY :)

Answer:

Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.

Answer:

The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.

Answer:

approaches infinity

Explanation:

There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.

The relativistic momentum can be written as:

[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

where

u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.

m = mass of the object

c = speed of light.

So, as u tends to c, we will have:

[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.

Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.

Answer:

[tex]D_l=d[/tex]

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

[tex]V_{e1}=V_{e2}[/tex]

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

[tex]D=\frac{at^2}{2}[/tex]

Where

Acceleration is given as

[tex]a=\frac{V_o}{2d}[/tex]

And

Time

[tex]T=\frac{d}{v_0}[/tex]

Therefore horizontal displacement towards the left is

[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]

[tex]D_l=d[/tex]

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex]  Rounding to the correct number of sig fig's to simplify:

[tex]v=\sqrt{400+2.0*10^2}[/tex] to get

v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

Answer:

Work Done= 3150J

Power= 1.75W

Explanation:

Work Done= Force x the distance travelled in the direction of the force (W= f x d)

Weight is a force, i think the qn. stated it wrongly, it should be 70N not 70kg.

Work Done= 70 x 45

=3150J

Power= Work Done/Time

=3150/(30x60)

*convert minutes to seconds since the S.I. unit of Power is joules/seconds(J/s) or watts(W)

=1.75W

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]

F = ma

∴

[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]

[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]

[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]

At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]

The velocity of the charge after the time t(sec) is expressed by using the formula:

[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]

Answer:

ΔT = 0.017 °C

Explanation:

According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:

Change in Internal Energy = (0.85)(Kinetic Energy)

[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]

where,

ΔT = increase in temperature = ?

v = speed of block = 4 m/s

C = specific heat capacity of copper = 389 J/kg.°C

Therefore,

[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]

ΔT = 0.017 °C

Answer:

a) Fnet = mg - Fb - Fr

b) 8.67 secs

Explanation:

mass of object = 80 kg

Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696

Proportionality constant = 10 N-sec/m

a) Calculate  equation of motion of the object

Force of resistance on object  due to water = Fr ∝ V

                                                                         = Fr = Kv = 10 V

Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object  while mg ( weight ) acts in the negative y - direction on the object.

Fnet = mg - Fb - Fr

∴ Equation of motion of the object ( Ma = mg - Fb - Fr )

b) Calculate how long before velocity of the object hits 40 m/s

Ma = mg - Fb - Fr

a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V

V = u + at ---- ( 1 )

u = 0

V = 40 m/s

a = 9.6138 - 0.125 V

back to equation 1

40 = 0 + ( 9.6138 - 0.125 (40) ) t

40 = 4.6138 t

∴ t = 40 / 4.6138 = 8.67 secs

Answer:

D) Q = 80 C

Explanation:

Given;

current flowing in the light bulb, I = 2A

time of current flow, t = 40,000 ms = 40,000 x 10⁻³ s = 40 s

The quantity of the charge is calculated as;

Q = It

where;

Q is the quantity of the charge (Coulombs)

Q = (2 ) x (40)

Q = 80 C

Therefore, the quantity of charge flowing in the circuit is 80 C

D) Q = 80 C

Answer:

to prevent friction on the surfaces

Answer:

Explanation:

Oiling reduces friction between parts with relative motion between them.

Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.

Oil also becomes oxidized which reduces the oil's ability to decrease friction.

Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

Answer:

[tex]t_2=10[/tex]

Explanation:

From the question we are told that:

Velocity of both spaceships [tex]V=0.8c[/tex]

Time [tex]t_1=6[/tex]

Generally the equation for time of spaceship 2 through earth is mathematically given by

[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]

[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]

[tex]t_2=10[/tex]

Answer:

a) Dielectric constant ( λ ) = 750 * 10^-9 m

   minimum thickness of coating ( d )  = 187.5 nm

b) 3.6%

Explanation:

Given data:

wavelength of red light in air = 750 nm

εr = 4

μr = 1,  σ = 0

a) Determine the required dielectric constant and min thickness of coating used

Refractive index of coating ( n ) = √εr * μr =  √4*1 = 2

the refractive index of glass( ng)  = 1.5 which is < 2

λ = 750 * 10^-9 m

Dielectric constant ( λ ) = 750 * 10^-9 m

To determine the minimum thickness we will apply the formula below

d = m λ/2n ;  where  m = 1

∴ d = 750 nm / 2 ( 2 )

      = 187.5 nm

minimum thickness of coating ( d )  = 187.5 nm

b) Determine the percentage of the incident power that will be reflected

R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2

   = [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2

   = 0.03628 =  3.6%

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

Answer:

thanks for da 5points hoi

Explanation: thanks dawg

There can be uncertainty in calculating the wavelength of a laser light  due to experimental errors

All measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

Uncertainty  means the range of possible values within which the true value of the measurement lies.

The deviation  in the value of the measured quantity from the actual quantity or true value is called an error

(a) For the calculation of wavelength of laser light , the sources which can lead to uncertainty are

1. least count of measuring instruments like spectrometer or interferometer

2. Parallax error in the measurement

3. Error in identifying the order of fringes

4.. unable to identify the accurate  reading of Vernier or circular scales present in the measuring instruments.

5. Propagating errors

The least count of a measuring instrument is the smallest and accurate value in the measured quantity that can be measured by instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is that  all the errors add up. which increases the uncertainty

b. The uncertainty in measurement due to  least count depends on the instrument used for measurement f wavelength. A  Michelson's

interferometer has the least count of .0001mm. whereas spectrometer has a least count of 0.5⁰. Hence uncertainty in the measurement by Michelson's interferometer is very less as compared to any other instrument.

C. The maximum uncertainty arises due to the least count , as all other errors can be minimized by taking an average value of many observations but the least count of an instrument do not change so uncertainty within the least count arises.

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Answer:

a)[tex]L=0.00142H[/tex]

b) [tex]C=2.65*10^{-12}[/tex]

Explanation:

From the question we are told that:

Frequency[tex]F=13kHz[/tex]

Current [tex]I=0.14A[/tex]

Capacitor[tex]C_e=1.4*10^{-5}J[/tex]

Generally the equation for Energy in the inductor is mathematically given by

Where L is now subject

[tex]L=\frac{2C_e}{I^2}[/tex]

[tex]L=\frac{2*1.4*10^{-5}}{(0.14)^2}[/tex]

[tex]L=0.00142H[/tex]

Generally the equation for Value of Capacitor is mathematically given by

[tex]C=\frac{1}{(2 \pi f)^2} L[/tex]

[tex]C=\frac{1}{(2 3.142 13*10^3Hz)^2} *0.00142[/tex]

[tex]C=2.65*10^{-12}[/tex]

Complete Question

The Question diagram is attached below

Answer:

a)  [tex]W_{Fg}= 12500 Nm[/tex]

b)  [tex]W_{T_1}= - 6339.3Nm[/tex]

c)  [tex]W_{T_2}= - 3662.8Nm[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=255kg[/tex]

Distance [tex]d=4m[/tex]

Generally the equation for Work done is mathematically given by

[tex]W=F*d[/tex]

For [tex]F_g[/tex]

[tex]W_{Fg}=2500 x 5.3[/tex]

[tex]W_{Fg}= 12500 Nm[/tex]

For [tex]T_1[/tex]

[tex]W_{T_1}= - {1830 sin(60) x 4}[/tex]

[tex]W_{T_1}= - 6339.3Nm[/tex]

For [tex]T_2[/tex]

[tex]W_{T_2}= - {1295 sin(45) x 4}[/tex]

[tex]W_{T_2}= - 3662.8Nm[/tex]