Answer:
A) the frequency and amplitude of the output voltag
Explanation:
Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.
Changing the speed regulator will change the engine throttle setting to maintain the speed.
While the power, torque, current, fuel flow rate and torque angle will have decreased.
The bar magnet is pushed toward the center of a wire loop. Looking down from the top view (would appear the magnet is coming up toward the observer); Which is true? A. There is no induced current in the loop B. There is a counterclockwise induced current in the loop C. There is not enough information to correctly answer the question D. There is a clockwisee induced current in the loop
Answer:
Explanation:
B. There is a counterclockwise induced current in the loop
Explanation:
This in line with the right hand grip rule,
The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.
The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at time t = -0.0s, with a velocity of 30m/s. The acceleration due to gravity at the surface of the moon is 1.62m/s2 the time when the rock is ascending at a height of 180m is closest to:______.
a. 8s .
b. 12s.
c. 17s.
d. 23s.
e. 30s
For the previous situation, the height of the rock when it is descending with a velocity of 20m/s is closest to:_____.
A. 115m.
B. 125m.
C. 135m.
D. 145m
E. 155m.
Explanation:
Given that,
Initial speed of the rock, u = 30 m/s
The acceleration due to gravity at the surface of the moon is 1.62 m/s².
We need to find the time when the rock is ascending at a height of 180 m.
The rock is projected from the surface of the moon. The equation of motion in this case is given by :
[tex]h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2[/tex]
It is a quadratic equation, after solving whose solution is given by:
t = 7.53 s
or
t = 8 seconds
(e)If it is decending, v = -20 m/s
Now t' is the time of descending. So,
[tex]v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s[/tex]
Let h' is the height of the rock at this time. So,
[tex]h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m[/tex]
or
h' = 155 m
An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 82.0 V. The circuit has a resistance of 71.0 Ω and an impedance at this frequency of 107 Ω and an impedance at this frequency of 105Ω. What average power is delivered to the circuit by the source?
Explanation:
Given that,
Frequency of LCR circuit is 120 Hz
RMS voltage, [tex]V_{rms}=82\ V[/tex]
Resistance of circuit, R = 71 Ω
Impedance, Z = 107 Ω
We need to find the average power is delivered to the circuit by the source. Firstly, finding the rms value of current,
[tex]I_{rms}=\dfrac{V_{rms}}{Z}\\\\I_{rms}=\dfrac{82}{105}\\\\I_{rms}=0.78\ A[/tex]
Power is given by :
[tex]P=I_{rms}V_{rms}\cos\phi[/tex]
[tex]\cos\phi = \dfrac{R}{Z}\\\\\cos\phi = \dfrac{71}{105}\\\\\cos\phi =0.676[/tex]
Now, power,
[tex]P=0.78\times 82\times 0.676\\\\P=43.23\ W[/tex]
So, the average power of 43.23 watts is delivered to the circuit by the source.
What is the answer?
Answer: i think it is d. none of them.
Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.
Suppose there are only two charged particles in a particular region. Particle 1 carries a charge of +q and is located on the positive x-axis a distance d from the origin. Particle 2 carries a charge of +2q and is located on the negative x-axis a distance d from the origin.
Required:
Where is it possible to have the net field caused by these two charges equal to zero?
1. At the origin.
2. Somewhere on the x-axis between the two charges, but not at the origin.
3. Somewhere on the x-axis to the right of q2.
4. Somewhere on the y-axis.
5. Somewhere on the x-axis to the left of q1.
Answer:
x₂ = 0.1715 d
1) false
2) True
3) True
4) false
5) True
Explanation:
The field electrifies a vector quantity, so we can add the creative field by these two charges
E₂-E₁ = 0
k q₂ / r₂² - k q₁ / r1₁²= 0
q₂ / r₂² = q₁ / r₁²
suppose the sum of the fields is zero at a place x to the right of zero
r₂ = d + x
r₁ = d -x
we substitute
q₂ / (d + x)² = q₁ / (d-x)²
we solve the equation
q₂ / q₁ (d-x)² = (d + x) ²
let's replace the value of the charges
q₂ / q₁ = + 2q / + q = 2
2 (d²- 2xd + x²) = d² + 2xd + x²
x² -6xd + d² = 0
we solve the quadratic equation
x = [6d ± √ (36d² - 4 d²)] / 2
x = [6d ± 5,657 d] / 2
x₁ = 5.8285 d
x₂ = 0.1715 d
with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d
this value remains on the positive part of the x axis, that is, near charge 1
now let's examine the different proposed outcomes
1) false
2) True
3) True
4) false
5) True
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge.
How far are the goggles from the edge of the pool?
Answer:
Explanation:
Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/
α = 26.56°
Laser angle with Normal = 90 - 26.56 = 63.44 °
Assuming a red laser, refractive index in water is 1.331.
Angle of refraction in water is given by:
Ref Ind = Sin i / Sin r
1.331 = Sin 63.44 / Sin r
Sin r = 0.8945 / 1.331 = 0.6721
Angle r = 42.22°
For the path in water:
Tan 42.22 = x / 3.2
x = 2.9m where x is the lateral displacement of the laser ince it hits the water
So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool
Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. SOLUTION
Answer:
Some parts of your question is missing attached below is the missing parts and the answer provided is pertaining to your question alone
answer : -6661.59 volts
Explanation:
The total electric potential can be calculated using this relation
V = k [tex](\frac{q1}{r1} + \frac{q2}{r2})[/tex]
q 1 = 1.62 uc
r1 = 4.00 m
q2 = -5.73 uc
r2 = 5.00 m
k = 8.99 * 10^9 N.m^2/c^2
insert the given values into the above equation
V = ( 8.99 * 10^9 ) * [tex](\frac{1.62*10^{-6} }{4} + \frac{-5.73*10^{-6} }{5})[/tex] = -6661.59 volts
A rollercoaster is not moving and has 50,000 J of GPE at the top of a hill. How much kinetic energy will it have halfway down the hill, assuming there is no friction
Answer:
The kinetic energy is 25000 J
Explanation:
At the top of the hill, the potential energy = 50000 J
the potential energy = mgh
where m is the mass
g is the acceleration due to gravity
h is the vertical height at the top of the hill
Note the mass of the roller coaster and acceleration due to gravity will always remain constant, so that halfway down the hill, only the height changes by half its initial value.
This means that at halfway down the hill, the potential energy of the roller coaster is
PE = [tex]mg\frac{h}{2}[/tex] = 50000/2 = 25,000 J
We also know that the total mechanical energy of a system is given as
ME = KE + PE = constant
where
ME is the mechanical energy of the system
PE is the potential energy of the system
KE is the kinetic energy of the system
Let us now analyse.
At the top of the hill, all the mechanical energy of the roller coaster is equal to its potential energy due to the height on the hill above ground, since the roller coaster is not moving (kinetic energy is energy due to motion). Halfway down, the mechanical energy of the roller coaster is due to both the kinetic energy and the potential energy, since the roller coaster is moving down, and is still at a given height above the ground. Having all these in mind, we can proceed and say that at halfway down the hill, ignoring friction,
ME = KE + PE = constant
50000 = KE + 25000
therefore
KE = 25000 J
If a ray of light traveling in the liquid has an angle of incidence at the interface of 33.0 ∘, what angle does the refracted ray in the air make with the normal?
Answer:
29°
Explanation:
because the refracted ray angle is small than angle of incidence
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days? Also find its Standard Deviation
Answer:
so the probability will be = 0.062
Standard deviation = 0.8925
Explanation:
The probability of rain = 15% = 15/100= 0.15
and the probability of no rain=q = 1-p= 1-0.15= 0.85
The number of trials = 7
so the probability will be
7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062
Taking this as binomial as the p and q are constant and also the trials are independent .
For a binomial distribution
Standard deviation = npq= 0.15 *0.85 *7= 0.8925
a reagent is added to the blue solution to identify the copper 2 ions name the blue solution
Answer:
buriret i belive
Explanation:
.
Answer:
The blue solution is named copper sulfate
a person Travels along a straight road for half the distance with velocity V1 and the remaining half the distance with velocity V2 the average velocity is given by
Answer: (V1+V2)/2
Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2
If the mass of an object is 10 kg and the
velocity is -4 m/s, what is the momentum?
A. 4 kgm/s
B. -40 kgm/s
C.-4 kgm/s
D. 40 kgm/s
Answer:
B. -40 kgm/s is the answer
The valid digits in a measurement are called _____ digits. Question 10 options: insignificant significant uncertain non-zero
Answer:
Significant
Explanation:
Valid digits in measurements are called significant digits, or also called significant figures.
These significant digits allow data and measurements to be more accurate and exact.
Answer:
Significant digits
Explanation
Took the test got it right
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current
Answer:
Explanation:
From the question we are told that
The radius is [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]
The current is [tex]I = 4.5 \ A[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{J}{\sigma }[/tex]
Where [tex]\sigma[/tex] is the conductivity of aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]
J is the current density which mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Here A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (1.4*10^{-3})^2[/tex]
[tex]A = 6.158*10^{-6} \ m^2[/tex]
So
[tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]
[tex]J = 730757 A/m^2[/tex]
So
[tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]
[tex]E = 0.021 \ N/C[/tex]
Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?
Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]
So, the force has a magnitude of 4212 N
You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.
Answer:
It will take. the same distance up as before, but take a longer time
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 34.1 m/s2 with a beam of length 5.55 m , what rotation frequency is required?
Answer:
f = 0.4 Hz
Explanation:
The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:
f = (1/2π)√(ac/r)
where,
f = frequency of rotation = ?
ac = radial acceleration = 34.1 m/s²
r = radius = length of beam = 5.55 m
Therefore,
f = (1/2π)√[(34.1 m/s²)/(5.55 m)]
f = 0.4 Hz
White light is spread out into spectral hues by a diffraction grating. If the grating has 1000 lines per cm, at what angle will red light (λ = 640 nm) appear in first order?
Answer:
3.67°
Explanation: Given that λ=640nm , m = 1
Considering the slit separation
d = 1cm/1000
= 1.000×10^-3cm
= 1.000×10-5m
We then have
Sinθ = mλ/d
Sinθ= (1×640×10^-9)/1.000×10-5m
Sinθ = 0.064
θ= sin-1 0.064
θ= 3.669°
= 3.67°
An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.
Answer:
(A). 1620 watt.
(B).0.8885.
Explanation:
So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;
Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.
(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).
(B). the rms values to find the power efficiency Pout/Pin of the inverter.
P(in) = 195 × 9.35 = 1823.3 watt.
Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.
A jumbo jet has a mass of 100,000 kg. The thrust of each of its four engines is 50,000 N. What is the jet's acceleration in meters per second squared right before taking off? Neglect air resistance and friction.
Answer:
The acceleration is [tex]a =2\ m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the jumbo jet is [tex]m_j = 100000\ kg[/tex]
The thrust is [tex]F_k = 50000 \ N[/tex]
Generally given that the jet has four engines the total thrust is
[tex]F_t = 4 * F_k[/tex]
substituting values
[tex]F_t = 4 * 50000[/tex]
[tex]F_t = 200000 \ N[/tex]
Generally the acceleration of the is mathematically represented as
[tex]a = \frac{F_t}{m}[/tex]
substituting values
[tex]a =2 \frac{N}{kg}[/tex]
Now
[tex]N = kg \cdot m/s^2[/tex]
Hence
[tex]a =2 \frac{kg * \cdot m/s^2}{kg}[/tex]
[tex]a =2\ m/s^2[/tex]
A 10-cm-long thin glass rod uniformly charged to 6.00 nC and a 10-cm-long thin plastic rod uniformly charged to - 6.00 nC are placed side by side, 4.4 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3
Answer:
A) E(r) = 1.3957 × 10^(5) N/C
B) E(r) = 9.8864 × 10⁴ N/C
C) E(r) = 1.13 × 10^(5) N/C
Explanation:
We are given;
q = 6 nc = 6 × 10^(-9) C
L = 10 cm = 0.1 m
d = 4.4 cm = 0.044 m
r1 = 1 cm = 0.01 m
r2 = 2 cm = 0.02 m
r3 = 3 cm = 0.03 m
Formula for the electric field strength in this question is given as;
E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)
When factorized, we have;
E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]
Plugging in the relevant values for q/(2π(ε_o)L)
We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m
Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53
Thus;
E(r) = 1078.52 [1/r + 1/(d - r)]
A) E1 is at r = 1 cm = 0.01m
Thus;
E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))
E(r) = 1.3957 × 10^(5) N/C
B) E2 is at r = 2 cm = 0.02 m
Thus;
E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))
E(r) = 9.8864 × 10⁴ N/C
C) E2 is at r = 3 cm = 0.03 m
Thus;
E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))
E(r) = 1.13 × 10^(5) N/C
Consult Interactive Solution 27.18 to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 653 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference
Answer:
Explanation:
In the given case for destructive interference , the condition is,
path difference = (2n+1)λ /2 where n is an integer and λ is wavelength
2 μ d = (2n+1)λ /2
Putting λ = 653 nm
for minimum thickness n = 0
2 μ d = 653 / 2 nm
= 326.5 nm
For constructive interference the condition is
2 μ d = n λ₁
326.5 nm = n λ₁
λ₁ = 326.5 / n
For n = 1
λ₁ = 326.5 nm ,
or , 326.5nm .
Longest wavelength possible is 326.5
Nuclear energy is over ___ times stronger than the chemical bonds between the atoms
Answer:
1 millions times stronger
For a transverse wave, what is a wavefront?
A a line joining all points on the same crest of a wave
B a line showing the displacement of a wave
C the energy content of a wave
D the first part of a wave to reach a point
wavefront is the long edge that moves, for example, the crest or the trough
g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube
Answer:
The length of the tube is 85 cm
Explanation:
Given;
speed of sound, v = 340 m/s
first harmonic of open-closed tube is given by;
N----->A , L= λ/₄
λ₁ = 4L
v = Fλ
F = v / λ
F₁ = v/4L
Second harmonic of open-closed tube is given by;
L = N-----N + N-----A, L = (³/₄)λ
[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]
Third harmonic of open-closed tube is given by;
L = N------N + N-----N + N-----A, L = (⁵/₄)λ
[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]
The difference between second harmonic and first harmonic;
[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]
The difference between third harmonic and second harmonic;
[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]
Thus, the difference between successive harmonic of open-closed tube is
v / 2L.
[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]
Therefore, the length of the tube is 85 cm
When the adjustable mirror on the Michelson interferometer is moved 20 wavelengths, how many fringe pattern shifts would be counted
Answer:
The number of fringe pattern shift is m = 40
Explanation:
From the question we are told that
The Michelson interferometer is moved 20 wavelengths i.e [tex]20 \lambda[/tex]
Generally the distance which the Michelson interferometer is moved is mathematically represented as
[tex]d = \frac{m * \lambda}{2}[/tex]
Here [tex]m[/tex] is the number of fringe pattern shift
So
[tex]20 \lambda = \frac{m * \lambda}{2}[/tex]
[tex]40 \lambda = m * \lambda[/tex]
m = 40
A force of pounds makes an angle of with a second force. The resultant of the two forces makes an angle of to the first force. Find the magnitudes of the second force and of the resultant.
This question is incomplete, the complete question is;
A force of 193 pounds makes an angle of 79°14' with a second force. The resultant of the two forces makes an angle of 27°0' to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
magnitudes of the second force (vector) is 110.84 lb
the resultant force force has a magnitude of 239.85 lb
Explanation:
Given that;
Magnitude of resultant vector R = ?
Direction of resultant vector α = 27°0'
Magnitude of vector p = 193
Magnitude of vector Q = ?
Angle between two vectors ∅ = 79°14'
Using the formula
tan∝ = [ Qsin∅ / P + Qcos∅]
tan27°0' = [ Qsin79°14' / 193 + Qcos79°14' )
we cross multiply
(193tan27°0') + (Qcos79°14'tan27°0' ) = Qsin79°14'
Q = 193tan27°0' / (sin79°14' - cos79°14'tan27°0')
Q = 110.84 lb
Therefore magnitudes of the second force (vector) is 110.84 lb
Now
R = √( p² + Q² + 2PQcos∅ )
R = √( 193² + 110.84² + ( 2 × 193 × 110.84 × cos79°14'))
R = 239.85 lb
Therefore the resultant force force has a magnitude of 239.85 lb
A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?
Answer:
The work done on the system is -616 kJ
Explanation:
Given;
Quantity of heat absorbed by the system, Q = 767 kJ
change in the internal energy of the system, ΔU = +151 kJ
Apply the first law of thermodynamics;
ΔU = W + Q
Where;
ΔU is the change in internal energy
W is the work done
Q is the heat gained
W = ΔU - Q
W = 151 - 767
W = -616 kJ (The negative sign indicates that the work is done on the system)
Therefore, the work done on the system is -616 kJ