Answer:
The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s
Explanation:
I'll assume that the correct question is
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?
mass of box = 51 kg
for the first 12 m, it is pulled with a constant force of 240 N
The acceleration of the box for this first 12 m will be
from F = ma
a = F/m
where F is the pulling force
m is the mass of the box
a is the acceleration of the box
a = 240/51 = 4.71 m/s^2
Since the body started from rest, the initial velocity u = 0
applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have
[tex]v^{2}= u^{2}+2as[/tex]
where v is the final velocity
u is the initial velocity which is zero
a is the acceleration of 4.71 m/s^2
s is the distance covered which is 12 m
substituting value, we have
[tex]v^{2}[/tex] = 0 + 2(4.71 x 12)
[tex]v^{2}[/tex] = 113.04
[tex]v = \sqrt{113.04}[/tex] = 10.63 m/s
For the final 10.5 m, coefficient of friction is 0.21
from f = μF
where f is the frictional force,
μ is the coefficient of friction = 0.21
and F is the pulling force of the box 240 N
f = 0.21 x 240 = 50.4 N
Net force on the box = 240 - 50.4 = 189.6 N
acceleration = F/m = 189.6/51 = 3.72 m/s^2
Applying newton's equation of motion
[tex]v^{2}= u^{2}+2as[/tex]
u is initial velocity, which in this case = 10.63 m/s
a = 3.72 m/s^2
s = 10.5 m
v = ?
substituting values, we have
[tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)
[tex]v^{2}[/tex] = 112.9 + 78.12
v = [tex]\sqrt{191.02}[/tex] = 13.82 m/s
A trash compactor can compress its contents to 0.350 times their original volume and 4 times denser than their original density. Neglecting the mass of air expelled, what factor is the old density of the rubbish
Answer:
2.8
Explanation:
Using p = m/v; (old density)
p' = m/v (new density)
=m/0.350 V
p'/p = (m/0.350V)/(m/v) = 1/0.350 = 2.86
Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?
Explanation:
It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.
We need to find the speed before and after the ball bounce.
Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,
[tex]\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s[/tex]
Let v is the final speed when it bounces to a height of 1.5 m. So,
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s[/tex]
So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.
Which is true about refraction from one material into a second material with a greater index of refraction when the incident angle is, say, 30º? At the interface, the ray bends toward the normal.
Answer:
Explanation:
Refraction is defined as the bending of light rays as an incident ray pass from one medium to another. If the incident ray is passing from the media with low refractive index to a greater refractive index, the refracted ray tends to bend away from the normal.
Refractive index is the ratio of the sin of angle of incidence to the sine of angle of refraction.
n = sin i/sin r
For us to have a greater index of refraction, the denominator must be lesser than the numerator. This means that the angle of refraction must be smaller and if the angle of refraction must get smaller, this means that the refracted ray must bend towards the normal
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that places them in series across normal household voltage. If each one is an incandescent bulb of fixed resistance, which statement about these bulbs is correct?
Answer:
When they are connected in series
The 50 W bulb glow more than the 100 W bulb
Explanation:
From the question we are told that
The power rating of the first bulb is [tex]P_1 = 100 \ W[/tex]
The power rating of the second bulb is [tex]P_2 = 50 \ W[/tex]
Generally the power rating of the first bulb is mathematically represented as
[tex]P_1 = V^2 R[/tex]
Where [tex]V[/tex] is the normal household voltage which is constant for both bulbs
So
[tex]R_1 = \frac{V^2}{P_1 }[/tex]
substituting values
[tex]R_1 = \frac{V^2}{100}[/tex]
Thus the resistance of the second bulb would be evaluated as
[tex]R_2 = \frac{V^2}{50}[/tex]
From the above calculation we see that
[tex]R_2 > R_1[/tex]
This power rating of the first bulb can also be represented mathematically as
[tex]P_ 1 = I^2_1 R_1[/tex]
This power rating of the first bulb can also be represented mathematically as
[tex]P_ 2 = I^2_2 R_2[/tex]
Now given that they are connected in series which implies that the same current flow through them so
[tex]I_1^2 = I_2^2[/tex]
This means that
[tex]P \ \alpha \ R[/tex]
So when they are connected in series
[tex]P_2 > P_1[/tex]
This means that the 50 W bulb glows more than the 100 \ W bulb
A parallel–plate capacitor is initially charged by connecting it to a battery. The battery is then disconnected. If the distance between the plates is increased, what happens to the charge on the capacitor and the voltage across it?
a. The charge remains fixed and the voltage decreases.
b. The charge decreases and the voltage remains fixed.
c. The charge remains fixed and the voltage increases.
d. The charge decreases and the voltage increases.
Answer:
t the battery of potential difference V be used to charge the capacitor of capacitance C.
∴ the charge stored in the capacitor q=CV
Now the battery is disconnected, so the the charge of the capacitor becomes constant
i.e q=constant OR CV=constant .............(1)
Capacitance of parallel plate capacitor C=
d
Aϵ
o
So if the distance between the plates is increased, then the capacitance will decrease which is compensated by the increase in voltage across the capacitor according to equation (1).
Also the energy stored in the capacitor E=
2C
q
2
⟹E∝
C
1
(∵q=constant)
Thus energy will increase due to the decrease in capacitance.
Explanation:
A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and the pump efficiency is 85%, the power required by the pump is most nearly:
Answer:
The power required by the pump is nearly 230.588 kW
Explanation:
Flow rate of the pump Q = 1 m^3/s
the head flow H = 20 m
specific weight of water γ = 9800 N/m^3
efficiency of the pump η = 85%
First note that specific gravity of water is the product of the density of water and acceleration due to gravity.
γ = ρg
where ρ is density. For water its value is 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
The power to lift this water at this rate will be gotten from the equation
P = ρgQH
but ρg = γ
therefore,
P = γQH
imputing values, we'll have
P = 9800 x 1 x 20 = 196000 W
But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.
we can say that
196000 W is 85% of the power of the pump power needed, therefore
196000 = 85% of [tex]P_{p}[/tex]
where [tex]P_{p}[/tex] is the power of the pump needed
85% = 0.85
196000 = 0.85[tex]P_{p}[/tex]
[tex]P_{p}[/tex] = 196000/0.85 = 230588.24 W
Pump power = 230.588 kW
Suppose a space vehicle with a rest mass of 150 000 kg travels past the International Space Station at a constant speed of 2.6 x 108 m/s with respect to the I.S.S. When an observer on the I.S.S. measures the moving vehicle, her measurement of the space vehicle length is 25.0 m. Determine the relativistic mass of the space vehicle. Determine the length of the space vehicle as measured by an astronaut on the space vehicle.
Answer:
m = 300668.9 kg
L₀ = 12.47 m
Explanation:
The relativistic mass of the space vehicle is given by the following formula:
[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]
where,
m = relativistic mass = ?
m₀ = rest mass = 150000 kg
v = relative speed = 2.6 x 10⁸ m/s
c = speed of light = 3 x 10⁸ m/s
Therefore
[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]
m = 300668.9 kg
Now, for rest length of vehicle:
L = L₀√(1 - v²/c²)
where,
L = Relative Length of Vehicle = 25 m
L₀ = Rest Length of Vehicle = ?
Therefore,
25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]
L₀ = (25 m)(0.499)
L₀ = 12.47 m
It takes 144 J of work to move 1.9 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates
Answer:
151.58 V
Explanation:
From the question,
The work done in a circuit in moving a charge is given as,
W = 1/2QV..................... Equation 1
Where W = Work done in moving the charge, Q = The magnitude of charge, V = potential difference between the plates.
make V the subject of the equation
V = 2W/Q.................. Equation 2
Given: W = 144 J. Q = 1.9 C
Substitute into equation 2
V = 2(144)/1.9
V = 151.58 V
A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.
Answer:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
1. weight
2. gravitational potential energy to kinetic energy.
Explanation:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
work done by toboggan = weight × distance
W = mg and the distance is down the icy slope
By using law of conservation of energy, energy can neither be created nor destroyed, but can be conserve from one form to another in a closed system.
Toboggan converts gravitational potential energy (mgh) to kinetic energy(¹/₂mv²)
A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 2.60 cm thick flat piece of crown glass and back to air again. The beam strikes the glass at a 28.0° incident angle.
A) At what angles do the two colors emerge?
B) By what distance are the red and blue separated when they emerge?
Answer:
A: 28°
B. 1x10^-3M
Explanation:
See attached file
Suppose I am viewing light through a camera lens (i.e. a circular aperture). If I want a wider field of view I should _____ the diameter of the lens.
Answer:
Increase
Explanation:
Because For a given focal length, a lens with a larger front element will generally be faster. That is, it'll have a larger maximum aperture, allowing a shorter exposure time, But a larger aperture requires larger elements to maintain the same angle of view
The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.
Answer:
27°
Explanation:
The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)
So theta = arcsin(0.45)
=27°
The angle between the wire and the magnetic field is 27°.
Calculation of the angle:Since The magnetic force per meter on a wire is measured to be only 45 %
So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field
Therefore,
theta = arcsin(0.45)
=27°
Hence, The angle between the wire and the magnetic field is 27°.
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A solenoid of 200 turns carrying a current of 2 A has a length of 25 cm. What is the magnitude of the magnetic field at the center of the solenoid?
Answer:
Explanation:
For magnetic field in a solenoid , the formula is
B = μ₀ n I
Where n is number of turns per unit length and i is current
Putting the values
B = 4π x 10⁻⁷ x (200 / .25) x 2
= 2.00 x 10⁻³ T
wrench is to Hammer as
Answer:
Pencil is to pen
Step by step explanation:
They are similar items, as they are both tools, but are different as to how they function.
A block with a mass of 0.28 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.0 N on the block. When the block is released, it oscillates with a frequency of 1.2 Hz. How far was the block pulled back before being released?
Answer:
Explanation:
For spring
[tex]n=\sqrt{\frac{k}{m} }[/tex]
where n is frequency of oscillation and k is force constant and m is mass
Putting the values
[tex]1.2=\sqrt{\frac{k}{.28} }[/tex]
k = .4032 N/m
F= k x
where F is force , k is force constant and x is extension
Putting the given values
1 = .4032 x
x = 2.48 m
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and the axis of the second is rotated θ = 65° from the vertical.
Required:
a. What is the intensity of the light after it passes through the first polarizer in W/m2?
b. What is the intensity of the light after it passes through the second polarizer in W/m2?
Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
A Buchner funnel uses _______ when separating a(n) _______ by filtration.
Explanation:
A Buchner funnel uses perforatet glass plate when separating a(n) solide from liquid by filtration.
[tex]hope \: this \: helps[/tex]
A charged particle moves into a region of uniform magnetic field B (pointing out of the page), goes through half a circle, and exits the region. The particle is either a proton or an electron. It spends 130 ns in the region. (a) What is the magnitude of B
The figure is missing, so i have attached it
Answer:
Magnitude of B = 0.252 T
Explanation:
From the image, considering the point at which it enters the field-filled region, the velocity vector is pointing downwards. The field points out of the page so that; (v→) × (B→) points leftward, points leftward which indeed seems to be the direction it is pushed. Therefore q > 0 and thus it's a proton.
The equation for the period since it goes through half circle is;
T = 2t = 2πm/(e|B|)
Where;
m is mass of proton = 1.67 × 10^(-27) kg
e is electron charge = 1.60 x 10^(-19) Coulombs.
|B| is magnitude of magnetic field
t = 130 ns = 130 × 10^(-9) s
Making |B| the subject, we have;
|B| = πm/et
Thus, plugging in all relevant values, we have;
|B| = π(1.67 × 10^(-27))/(1.60 x 10^(-19) × 130 × 10^(-9)) = 0.252 T
The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This electron beam impinges on the inside of the picture tube screen.
How many electrons strike the screen each second?
The electrons move with a velocity of 4.0\times10^7\;{\rm m/s}. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 {\rm mm}?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)
Answer:
A.3.13x10^14 electrons
B.330A/m²
C.9.11x10^5N/C
D. 0.23W
.pls see attached file for explanations
Three point charges (some positive and some negative) are fixed to the corners of the same square in various ways, as the drawings show. Each charge, no matter what its algebraic sign, has the same magnitude. In which arrangement (if any) does the net electric field at the center of the square have the greatest magnitude?
Answer:
The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.
A negatively charged object is located in a region of space where the electric field is uniform and points due north. The object may move a set distance d to the north, east, or south. Write the three possible movements by the change in electric potential energy (Ue) of the object.
Answer:
the three possible movements by the change in electric potential energy (Ue) of the object are NORTH EAST SOUTH
Explanation:
This is because When the object moves south, the force is in the direction of the displacement, and positive work is done with decreasing electric potential energy.
The opposite is true if the particle moves north—that is, negative work is done with increasing electric potential energy.
No work is done and the electric potential energy is constant if the motion is perpendicular to the electric field.
Suppose a 225 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m. How high can it coast up the hill, if you neglect friction in m?
a) m = 180 kg
b) v = 29 m/s
c) h = 32 m
Answer:
It can coast uphill 6.2m
Explanation:
See attached file pls
15pts! brainliest to who answers!If Kyla picks up a grocery bag, using 10 N of force to lift it 1.5 m off the floor, how much work did Kyla do on the bag?
Explanation:
work = force x distance
w = 10 x 1.5 = 15Nm
The work done on the bag is the product of the force applied on it and the displacement of the bag. The work done to lift the bag up to 1.5 m by applying a force of 10 N is 15 J.
What is work done ?When a force applied to an object make a displacement of the body or stopes its motion, the force is said to be work done on the object. Thus, work done can be taken as the product of force and displacement.
Work done like force is a vector quantity thus characterized with magnitude and direction. Work done is equivalent to the energy required to make the object displaced.
Given the force = 10 N
displacement = 1.5 m
work done = force × displacement
w = 10 N × 1.5 m = 15 J.
Therefore, the work done on the car is 15 J.
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Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y2 (measured in coulombs per square meter). Find the total charge on the disk.
Answer:
the total charge on the disk 256pi Coulombs
Explanation:
Pls see attached file
A small solid conductor with radius a is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius b. The inner andouter conductors carry equal currents i in oppositedirections.
Required:
a. Use Ampere's Law to find the magnetic field at any pointin the volume between the conductors.
b. Write the expression for the flux dΦB through anarrow strip of length l parallel to the axis , of width dr, at a distancer from the axis of the cableand lying in a plane containing the axis.
c. Integrate your expression from part B over the volumebetween the two conductors to find the total flux produced by acurrent i in the central conductor.
d. Use equation U=(1/2)LI2 to calculate the energy stored in the magnetic field for alength l of the cable.
Answer:
Pls see attached file
Explanation:
Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. For stars orbiting at distances of about 1014 m from the object, the orbital velocities are about 106 m/s. Assume the orbits are circular, and estimate the mass of the object, in units of the mass of the sun (MSun = 2x1030 kg). If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be a supermassive black hole.
Answer:
The mass of the object is 745000 units of the sun
Explanation:
We know that the centripetal force with which the stars orbit the object is represented as
[tex]F_{c}[/tex] = [tex]\frac{mv^{2} }{r}[/tex]
and this centripetal force is also proportional to
[tex]F_{c}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]
where
m is the mass of the stars
M is the mass of the object
v is the velocity of the stars = 10^6 m/s
r is the distance between the stars and the object = 10^14 m
k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
We can equate the two centripetal force equations to give
[tex]\frac{mv^{2} }{r}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]
which reduces to
[tex]v^{2}[/tex] = [tex]\frac{kM}{r}[/tex]
and then finally
M = [tex]\frac{rv^{2} }{k}[/tex]
substituting values, we have
M = [tex]\frac{10^{14}*(10^{6})^{2} }{6.67*10^{-11} }[/tex] = 1.49 x 10^36 kg
If the mass of the sun is 2 x 10^30 kg
then, the mass of the the object in units of the mass of the sun is
==> (1.49 x 10^36)/(2 x 10^30) = 745000 units of sun
When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery
Answer:
The internal resistance is [tex]r = 0.5 \ \Omega[/tex]
Explanation:
From the question we are told that the resistance of
The resistance of the resistor is [tex]R = 20.0\ \Omega[/tex]
The voltage is [tex]V = 12.0 \ V[/tex]
The magnitude of the voltage fall is [tex]e = 0.300\ V[/tex]
Generally the current flowing through the terminal due to the voltage of the battery is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
substituting values
[tex]I = \frac{12.0 }{20 }[/tex]
[tex]I = 0.6 \ A[/tex]
The internal resistance of the battery is mathematically represented as
[tex]r = \frac{e}{I}[/tex]
substituting values
[tex]r = \frac{0.300}{ 0.6 }[/tex]
[tex]r = 0.5 \ \Omega[/tex]
The internal resistance of the battery is 0.5 ohms.
To calculate the internal resistance of the battery, we use the formula below
Formula:
(V/R)r = V'............. Equation 1Where:
V = Voltage across the terminal of the batteryR = Resistance connected across the batteryr = internal resistance of the batteryV' = voltage drop of the battery.Make r the subject of the equation
r = V'R/V............ Equation 2From the question,
Given:
V = 12 VR = 20 ohmsV' = 0.3 VSubstitute these values into equation 2
r = (0.3×20)/12r = 6/12r = 0.5 ohms.Hence, The internal resistance of the battery is 0.5 ohms.
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What do Equations 1 and 2 predict will happen to the single-slit diffraction pattern (intensity, fringe width, and fringe spacing) as the slit width is increased.
Equation 1:
Sinθ = mλ/ω
Equaiton 2:
I= Io [Sinθ (πωλ/πωλ/Rλ)
Answer:
the firtz agrees with the expression for the shape of the curve of diracion of a slit
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
where a is the width of the slit, t is the angle from the center of the slit, l is the wavelength and m is an integer that corresponds to the maximum diffraction.
the previous equation qualitatively describes the curve of the diffraction phenomenon the equation takes the form
I = I₀ [(sin ππ a y / R λ) / π a y / Rλ]²
I = I₀ ’[sin π a y /Rλ]²
I₀ ’= I₀ / (π a y /Rλ)²
By reviewing the two expressions given
equation 1
w sin θ = m λ
where w =a w is the slit width
we see that the firtz agrees with the expression for the shape of the curve of diracion of a slit
Equation 2
the squares are missing
A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?
Answer:
31
Explanation:
An aluminum cup of mass 150 g contains 800 g of water in thermal equilibrium at 80.0°C. The combination of cup and water is cooled uniformly so that the temperature decreases by 1.50°C per minute. At what rate is energy being removed by heat? Express your answer in watts.
Answer:
Heat Flow Rate : ( About ) 87 W
Explanation:
The heat flowing out of the system each minute, will be represented by the following equation,
Q( cup ) + Q( water ) = m( cup ) [tex]*[/tex] c( al ) [tex]*[/tex] ΔT + m( w ) [tex]*[/tex] c( w ) [tex]*[/tex] ΔT
So as you can see, the mass of the aluminum cup is 150 grams. For convenience, let us convert that into kilograms,
150 grams = .15 kilograms - respectively let us convert the mass of water to kilograms,
800 grams = .8 kilograms
Now remember that the specific heat of aluminum is 900 J / kg [tex]*[/tex] K, and the specific heat of water = 4186 J / kg [tex]*[/tex] K. Therefore let us solve for " the heat flowing out of the system per minute, "
Q( cup ) + Q( water ) = .15 [tex]*[/tex] ( 900 J / kg [tex]*[/tex] K ) [tex]*[/tex] 1.5 + .8 [tex]*[/tex] ( 4186 J / kg [tex]*[/tex] K ) [tex]*[/tex] 1.5,
Q( cup ) + Q( water ) = 5225.7 Joules
And the heat flow rate should be Joules per minute,
5225.7 Joules / 60 seconds = ( About ) 87 W