Answer:
let M be the mass of brick
let m be the mass of the pebble
.5M(3)^2 =KE
.5m(5)^2= KE
.5M9 = .5m25
9M = 25m
(9/25)M = m
Answer:
A
Explanation:
e=1/2M[tex]V^{2}[/tex]
1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25
[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]
Why dose a hot air balloon float
In seismology, the P wave is a longitudinal wave. As a P wave travels through the Earth, the relative motion between the P wave and the particles is
Answer:
A longitudinal wave is a wave where the displacement of the medium is in the same direction than the propagation of the wave.
This means that as a P wave travels through the Earth, the relative motion between the p wave and the particles is near zero, as the motion of the particles is parallel to the motion of the wave.
An example of this would be the waves generated when you throw a rock in water, you can see how the water particles move along the waves in the water.
The relative motion between the p wave and the particles is almost negligible.
In case of a longitudinal wave, the displacement of the medium is in the same direction as the direction of the propagation of the wave. This property suggests that as a P wave travels through the Earth, the particles also move along in the same direction. The relative motion between the p wave and the particles is almost negligible. In case of transverse waves, the displacement of particles of the medium and the direction of the propagation of the wave is perpendicular to each otherLearn more:
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A 3 Kg exercise ball is held 2m above the ground. What is the gravitational potential energy?
Answer:
58.8
Explanation:
we should apply formula
m*g*h
3*9.81*2
A 15 N force is applied to a 12 kg box for 6 s. The box is initially at rest. What is the speed of the box at the end of the 6 s interval
Answer:
The speed of the box at the end of the 6 s interval is 7.5 m/s.
Explanation:
Given;
magnitude of the force, f = 15 N
mass of the box, m = 12 kg
initial velocity of the box, u = 0
time of the box motion, t = 6s
The applied force on the box is given by Newton's second law of motion;
f = ma
where;
a is the acceleration of the box
a = f / m
a = (15) / (12)
a = 1.25 m/s²
The final velocity of the box is calculated as;
v = u + at
v = 0 + (1.25 x 6)
v = 7.5 m/s.
Therefore, the speed of the box at the end of the 6 s interval is 7.5 m/s.
3. Ohm's law is represented by the equation I = V/R. Explain how the current would change if the amount of resistance decreased and the voltage stayed the same.
Answer:
Current will increase
Explanation:
In Ohm's law the equation for current is current = voltage / resistance.
In order to explain how current is effected when resistance decreases while voltage stays the same, lets represents the situations with some possible inputs.
Let's compare 3 different closed circuits all with a voltage of 10V.
In circuit 1, the resistance is 5 ohm.
In circuit 2, the resistance is 2 ohms.
In circuit 3, the resistance is 1 ohms.
The current of:
Circuit 1 = Voltage / Resistance = 10 V / 5 ohms = 2 Amps
Circuit 2 = Voltage / Resistance = 10V / 2 ohms = 5 Amps
Circuit 3 = Voltage / Resistance = 10V / 1ohm = 10 Amps
As you can see in this representation, as the resistance in a circuit increases while the voltage is constant, the total current is increased.
As the resistance decreases, the amount of current increases.
What is Ohm's law?Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Ohm's law is represented by the equation
I = V/R
Ohm's Law tells us that the electrical current in a circuit can be calculated by dividing the voltage by the resistance.
As the current change if the resistance is decreased,
In this case, there is a inverse relationship between the two variables. As the resistance increases, the current decreases, provided all other factors are kept constant.
In other words, the current is directly proportional to the voltage and inversely proportional to the resistance.
As the current is directly proportional to the voltage and inversely proportional to the resistance.
So,
an increase in the voltage will increase the current as long as the resistance is held constant.
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Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as much as Jack's power output. four times Jack's power output. twice Jack's power output. one-half as much as Jack's power output. the same as Jack's power output.
Answer:
Second Choice.
Explanation:
Jack's Power = W/t
Jill's Power = 2W/(0.5)*t
2/0.5 = 4
Jill's Power = 4*W/t
Jill's Power is 4 times greater than Jack's
Second Choice
The energy released in the fission of one 235U92 is 206.6 MeV. A Nuclear reactor that uses this element has an output of 28.7 Mega Watts. What is the mass of 235U92 that is consumed in one day in this reactor
Answer: the required mass is 1.7628 × 10²⁵ μ
Explanation:
Given that;
energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV
power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s
now fission need per second will be;
⇒ power / energy released i fission
= 17.9375 × 10¹⁹ MeV / 206.6 MeV = 8.68 × 10¹⁷ per second
now fission need per day will be;
⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day
hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;
⇒ (235 × 7.5 × 10²²)μ
= 1.7628 × 10²⁵ μ
Therefore the required mass is 1.7628 × 10²⁵ μ
Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (568 nm) on a double slit with a separation of 0.107 mm. The diffraction pattern shines on the classroom wall 3.5 m away. Calculate the fringe separation between the third order and central fringe.
Answer:
18.58 mm
Explanation:
Formula for the fringe width is;
β = λL/d
We are given;
λ = 568 nm = 568 × 10^(-9) m
L = 3.5 m
d = 0.107 mm = 0.107 × 10^(-3) mm
β = (568 × 10^(-9) × 3.5)/(0.107 × 10^(-3))
β = 0.01858 m = 18.58 mm
A horse pulls a wagon with 2000N of at constant velocity. What is the friction acting on the wagon?
0N
1000N
1500N
2000N
Answer:
f = 0 N
Explanation:
It is given that,
A horse pulls a wagon with 2000N of at constant velocity.
We need to find the friction acting on the wagon.
As the wagon is pulled with a constant velocity. If its velocity is constant, it means its acceleration is 0. As a result friction force is 0.
Hence, the correct option is (a).
9. A 15kg mass is lifted upward at a constant speed to a height of 22 m. Calculate the work done by the lifting
force.
Work done by lifting is 3,300 Newton.
Given that;
Mass of thing = 15 kg
Height lifted = 22 m
Find:
Work done by lifting
Computation:
Work done = mgh
Work done by lifting = (15)(10)(22)
Work done by lifting = 3,300 Newton
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A 0.3 g mosquito is flying toward a girl with a speed of 4.5 mph. Just before landing on the girl, the fly is swatted straight back at a speed of 12 mph. If the fly swatter and the fly were in contact for 0.2 s, what is the force that was exerted on the fly
Answer:
1.1x10^-2N
Explanation:
We have the change in momentum as
P = 0.3(4.5+12)g.mph
= 0.3x0.447x(4.5+12)x10^-3
Then the force that is exerted will be
F = p/∆t
∆t = 0.2
= 0.3x0.447x(4.5+12)x10^-3/0.2
= 0.1341x16.5x10^-3/0.2
= 1.1x10^-2
Therefore the force that was exerted is equal to 1.1x10^-2
The required magnitude of the force exerted on the fly is of [tex]5.025 \times 10^{-3} \;\rm N[/tex].
Given data:
The mass of mosquito is, [tex]m =0.3 \;\rm g =3 \times 10^{-4} \;\rm kg[/tex]
The speed of flying is, u = 4.5 mph = 4.5 ( 0.447) = 2.01 m/s.
The swatting speed of mosquito is, v = 12 mph = 12 (0.447 ) = 5.36 m/s.
The time of contact is, t = 0.2 s.
In this problem, we will first calculate the change in momentum, and the change in momentum is given as,
p = m ( v - u)
Solving as,
[tex]p = 3 \times 10^{-4} (5.36 - 2.01)\\\\p = 1.005 \times 10^{-3} \;\rm kg.m/s[/tex]
Now as per the Newton's second law,
[tex]F = p/t\\\\F = 1.005 \times 10^{-3} / 0.2\\\\F= 5.025 \times 10^{-3} \;\rm N[/tex]
Thus, the required magnitude of the force exerted on the fly is of [tex]5.025 \times 10^{-3} \;\rm N[/tex].
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Compare the weight of a 60 kg person on the earth with the weight of the same person on
the moon. Then, describe a quick (but very costly) way for dieters
at NASA to lose weight.
Answer:
Explanation:
The formula for weight is
W = mg, where
W = the weight of the object or person
m = mass of the object or person
g = acceleration due to gravity
Now, we're given the mass of the person to be 60 kd, and thus, the weight of that person would be
W = 60 * 9.81
W = 588.6 N
On the surface of the moon, the weight of the person would be
W = 60 * 1.625
W = 97.5 N
Therefore, the weight of the person on both surfaces are 588.6 and 97.5 respectively
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be when it is very far from the Earth
Answer:
The value is [tex]v = 2.3359 *10^{4} \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 2.05 *10^{4} \ m/s[/tex]
Generally the total energy possessed by the space probe when on earth is mathematically represented as
[tex]T__{E}} = KE__{i}} + KE__{e}}[/tex]
Here [tex]KE_i[/tex] is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
[tex]KE_i = \frac{1}{2} * m * u^2[/tex]
=> [tex]KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2[/tex]
=> [tex]KE_i = 2.101 *10^{8} \ \ m \ \ J[/tex]
And [tex]KE_e[/tex] is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as
[tex]KE_e = \frac{1}{2} * m * v_e^2[/tex]
Here [tex]v_e[/tex] is the escape velocity from earth which has a value [tex]v_e = 11.2 *10^{3} \ m/s[/tex]
=> [tex]KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2[/tex]
=> [tex]KE_e = 6.272 *10^{7} \ \ m \ \ J[/tex]
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as
[tex]KE_p = \frac{1}{2} * m * v^2[/tex]
Generally from the law energy conservation we have that
[tex]T__{E}} = KE_p[/tex]
So
[tex]2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2[/tex]
=> [tex]5.4564 *10^{8} = v^2[/tex]
=> [tex]v = \sqrt{5.4564 *10^{8}}[/tex]
=> [tex]v = 2.3359 *10^{4} \ m/s[/tex]
a proton is in a box of width L. what must the width of the box be for the ground level energy to be 5.0 MeV, a typical value for the energy with which the particles in a nucleus are bound
Answer:
Width of a box, [tex]l=6.41\times 10^{-15}\ m[/tex]
Explanation:
The ground level energy of a proton in a box is, E = 5 MeV
[tex]E =5\times 10^6\ eV\\\\=5\times 10^6\times 1.6\times 10^{-19}\\\\=8\times 10^{-13}\ J[/tex]
Energy in a box is given by :
[tex]E=\dfrac{n^2h^2}{8ml^2}[/tex]
For ground state, n = 1
m is mass of proton
h is Planck's constant
l is width of the box
[tex]l^2=\dfrac{n^2h^2}{8mE}\\\\l^2=\dfrac{1^2\times (6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times 8\times 10^{-13}}\\\\l=\sqrt{\dfrac{1^{2}\times(6.63\times10^{-34})^{2}}{8\times1.67\times10^{-27}\times8\times10^{-13}}}\\\\l=6.41\times 10^{-15}\ m[/tex]
So, the width of the bx is [tex]6.41\times 10^{-15}\ m[/tex].
The width of the box, for the ground level energy with which the particles in a nucleus are bound, is 6.41×10⁻¹⁵ m.
What is the energy in a box?The energy in a box can be calculated with the following formula.
[tex]E=\dfrac{n^2h^2}{8mL^2}[/tex]
Here, (E) is the energy at the nth state, (n) n is the quantum number, (h) is plank's constant and (L) is the width of the box.
The proton is in a box of width L. The width of the box be for the ground level energy to be 5.0 MeV, a typical value for the energy with which the particles in a nucleus are bound.
The energy of the box is at ground level. Then the value of nth state will be 1. It is known that the value of plank's constant is 6.63×10⁻³⁴ m²kg/s.
Put this values in the above formula as,
[tex]8\times10^{-13}=\dfrac{(1)^2(6.63\times10^{-34})^2}{8(1.67\times10^{27})(l)^2}\\l=6.41\times10^{-15}\rm\; m[/tex]
Thus, the width of the box, for the ground level energy with which the particles in a nucleus are bound, is 6.41×10⁻¹⁵ m.
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In order for convection to transfer heat, particles need to
Select one:
a.
absorb solar and fossil energy
b.
circulate and move within a liquid or gas
c.
make contact with the heat source
d.
transmit electromagnetic waves
Answer:
the answer is b mark me as the brqinlist
A 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Answer:
4.5 × 10^6 Nm
Explanation:
Given that a 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Let's first calculate the area covered by the heel shoe by using the area of a circle.
Convert cm to m
0.6 / 100 = 0.006
A = πr^2
A = 22/7 × 0.006^2
A = 1.131 × 10^-4 m^2
The force exerted by the woman = mg
Force exerted = 52 × 9.8
Force exerted = 509.6 N
Pressure = force / Area
Pressure = 509.6 / 1.131 × 10^-4
Pressure = 4505853.3 N/m
Therefore, she exerts pressure of 4.5 × 10^6 Nm approximately on the floor.
the maximum normal force a pilot can withstand is about eight times his weight. What is the maximum radius of curvature that a jet planes pilot, pulling out of a vertical dive
Complete Question
the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?
Answer:
The value is [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]
Explanation:
From the question we are told that
The weight of the pilot is [tex]W = mg[/tex]
The maximum force a pilot can withstand is [tex]F_{max} = 7 W = 7 (mg)[/tex]
The speed is [tex]v = 250 \ m/s[/tex]
Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e
[tex]F_c = F_{max} - mg[/tex]
Here N is the normal force acting on the pilot
Now
[tex]F_c = \frac{m v^2 }{r}[/tex]
So
[tex]\frac{m v^2 }{r} = 7(mg) - mg[/tex]
=> [tex]r = \frac{v^2 }{6g}[/tex]
=> [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]
=> [tex]r = 1063 \ m[/tex]
come in my free candy van located at Skellerupvej 43, 8600 Silkeborg, Denmark
Answer:
Did you think you could pull that off-
:I
I live in Japan anyway so.. ウーフ-
HELP WILL GIVE BRAINLIEST OF CORRECT
Answer:
B)
Explanation:
The answer is B)- The object's speed must be changing.
So will I get the Brainliest...
The distance between the object and its image formed by a plane mirror appears to be 24 cm. What is the distance between the mirror and the object?
Answer:
in a plane mirror distance of object from mirror is equal to distance of image
so it will be 24 ÷ 2 = 12 cm
Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car.
a. Joel does 75% more work than Jerry.
b. Joel does 25% more work than Jerry.
c. Jerry does 50% more work than Joel.
d. Joel does 50% more work than Jerry
e. None of the above .
Answer:
the answer is B
Explanation:
The intensity of sound is measured on the decibel scale, dB. The equation dB=10 log I represents the decibel level, where I is the ratio of the sound to the human hearing threshold. A noise is 150,000 times greater than the human hearing threshold. Which shows a valid step in the process of finding the decibel level of the noise?
a. 150,000 = 10 log I
b. 15,000 = log I
c. dB = 10 log 150,000
d. 10dB = log 150,000
e. 10/dB= log 150,000
Answer:
The correct option is c
Explanation:
From the question we are told that
The ratio of the noise to human hearing threshold is [tex]I = 150 000[/tex]
Generally from the equation given we have that
dB = 10 log I
So
dB = 10 log 150000
The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct
Given the equation for calculating the intensity of sound which is measured in decibel expressed as:
dB=10 log I
where;
I is the ratio of the sound to the human hearing threshold
Given that noise is 150,000 times greater than the human hearing threshold.
Substitute I = 150,000 into the expression above;
dB = 10 log 150,000
Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000
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A 4 m^3 metal block is suspended using a cable in salt water with density 1030 kg/m^3. Tension on the cable is 42600N. What is the weight of the block. What is the density of the block?
Answer:
The density of the block is 1113.216 kilograms per cubic meter.
The weight of the block is 43669.237 newtons.
Explanation:
According to the Archimedes' Principle, the drag force experimented by the metal block is equal to the weight of the volume of water displace by the block. Besides, the block has a weight that cannot be neglected and experiments a tension from the cable. Given that the metal block is suspended, then we could consider that block is at rest.
From Newton's Laws of Motion we obtain the following equation of equilibrium:
[tex]\Sigma F = T-\rho_{m}\cdot V_{m}\cdot g + \rho_{w}\cdot V_{m}\cdot g = 0[/tex] (1)
Where:
[tex]T[/tex] - Tension on the cable, measured in newtons.
[tex]\rho_{w}[/tex], [tex]\rho_{m}[/tex] - Densities of salt water and the metal block, measured in kilograms per cubic meter.
[tex]V_{m}[/tex] - Volume of the metal block, measured in cubic meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]T = 42600\,N[/tex], [tex]\rho_{w} = 1030\,\frac{kg}{m^{3}}[/tex], [tex]V_{m} = 4\,m^{3}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the density of the block is:
[tex]T+\rho_{w}\cdot V_{m}\cdot g = \rho_{m}\cdot V_{m}\cdot g[/tex]
[tex]\rho_{m} = \frac{T+\rho_{w}\cdot V_{m}\cdot g}{V_{m}\cdot g}[/tex]
[tex]\rho_{m} = \frac{42600\,N+\left(1030\,\frac{kg}{m^{3}} \right)\cdot (4\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(4\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\rho_{m} = 1113.216\,\frac{kg}{m^{3}}[/tex]
The density of the block is 1113.216 kilograms per cubic meter.
Lastly, the weight of the block ([tex]W[/tex]), measured in newtons:
[tex]W = \rho_{m}\cdot V_{m}\cdot g[/tex] (2)
[tex]W = \left(1113.216\,\frac{kg}{m^{3}}\right)\cdot (4\,m^{3}) \cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 43669.237\,N[/tex]
The weight of the block is 43669.237 newtons.
(c) The plates are moved farther apart with each plate maintaining the same net charge. In a coherent paragraph-length response, apply concepts of work and energy to explain how the electric potential difference between the plates changes, if at all, when the plates are moved farther apart.
Answer:
Energy is applied on the charge to do work.
Explanation:
Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..
Assuming perfect optics, if the smallest feature you can resolve when observing around 660 nm is of angular size 0.04 arcsec, what would the size (in arcsec) of the finest feature you can resolve if you make your observation around wavelength 1320 nm, assuming everything else is the same
Answer:
The value is [tex]\theta _2 = 0.08 \ arcsec[/tex]
Explanation:
From the question we are told that
The first wavelength is [tex]\lambda_1 = 660 \ nm = 660 *10^{-9 }[/tex]
The first angular size is [tex]\theta_1 = 0.04 \ arcsec[/tex]
The second wavelength is [tex]\lambda _2 = 1320 \ nm = 1320 *10^{-9 } \ m[/tex]
Generally according to Rayleigh Criterion we have that
[tex]\theta= 1.22 * \frac{\lambda }{D}[/tex]
given every other thing remains constant we have that
[tex]\theta = k * \lambda[/tex]
Here k represented constant so
[tex]k = \frac{\theta }{\lambda}[/tex]
=> [tex]\frac{\theta_1}{ \lambda_1} = \frac{\theta_2}{ \lambda_2}[/tex]
=> [tex]\frac{\theta_1}{ \theta_2} = \frac{\lambda_1}{ \lambda_2}[/tex]
So
[tex]\frac{ 0.04}{ \theta_2} =0.5[/tex]
=> [tex]\theta _2 = 0.08 \ arcsec[/tex]
A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.
Answer:
The value is [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
Explanation:
From the question we are told that
The force is [tex]F = 22.7 \ N[/tex]
The value of room temperature is [tex]T = 298 \ K[/tex]
Generally the rate at which its entropy changes as it stretches is mathematically represented as
[tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]
=> [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]
=> [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
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questions
Cause and Effect Graphic Organizers
Describe how T-charts and Star Charts can be used to organize information into cause and effect relationships.
I
Answer:
A T-chart is used comparing two sides of a topic for example pros and cons or cause and effect. A star diagram is used for organizing the characteristics of a single topic, a central space is used for displaying the topic, with each point showing or listing an attribute about the topic.
Which has more mass Helium or mass
given that there is no second choice i have to go with helium. if you get the second choice i will revise my answer
Thermal expansion of a gas causes a change in the ____ of the gas.
A. Radioactiveness
B. Density
C. Mass
D. Altitude
Answer:
B. Density
Explanation:
Thermal expansion of a gas will cause a definite change in the density of the gas.
It is product of an increase in the average kinetic energy of the system in which molecules are found.
Gaseous thermal expansion will increase the volume in which gases occupy. Mass of the gas will still remain the same. Since density is the mass per unit volume, as the gases expand, their density reduces more and more.Therefore, thermal expansion of gas will cause a change in the density of the gas.
If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?
Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm[/tex]
Magnification,
[tex]m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25[/tex]
Hence, the magnification of the image is 0.25.
