Answer:
a) [tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex], b) [tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex], c) [tex]S_{gen} = 4.106\,\frac{J}{K}[/tex], d) Due to irreversibilities due to temperature differences.
Explanation:
a) The change in entropy of the hot reservoir is:
[tex]\Delta S_{in} = \frac{2400\,J}{860\,K}[/tex]
[tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex]
b) The change in entropy of the cold reservoir is:
[tex]\Delta S_{out} = \frac{2400\,J}{348\,K}[/tex]
[tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex]
c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:
[tex]\Delta S_{in} - \Delta S_{out} + S_{gen} = 0[/tex]
[tex]S_{gen} = \Delta S_{out} - \Delta S_{in}[/tex]
[tex]S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}[/tex]
[tex]S_{gen} = 4.106\,\frac{J}{K}[/tex]
d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.
Four long wires are each carrying 6.0 A. The wires are located
at the 4 corners of a square with side length 9.0 cm. All of
these wires are carrying current out of the page. The
magnetic field (in T) at one corner of the square is:
Answer:
[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]
Explanation:
To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:
[tex]B=\frac{\mu_oI}{2\pi r}[/tex]
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current = 6.0 A
r: distance to the wire in which magnetic field is measured
In this case, you have four wires at corners of a square of length 9.0cm = 0.09m
You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.
If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:
[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m
[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]
Then you have:
[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, how many more revolutions will it rotate through in the next 5.00 s?
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
E = -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E = -4556.18 N/m
The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.
What is electric field?The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.
Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.
The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)
The horizontal distance traveled by particle is
x = ucosθt
t = x/ucosθ
The force in an electric field is F = qE...................(1)
where, q is charge , E is the strength of electric field
From, newton 2nd law of motion, Force F = ma.................(2)
Equating both the equations, we get
ma = qE
a = qE/m..................(3)
The vertical distance, y =usinθt - 1/2at²
From equation 3, we have
y = usinθt - 1/2 (qE/m) t²
if y = 0, t = 2musinθ/(qE) = x / (ucosθ)
The electric field is represented as
Also, E = 2mu²×sinθ×cosθ/(xq)
Plug the values, we get
E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)
E = -4556.18 N/m
Thus, the electric field of the particle is -4556.18 N/m.
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5.
The solar system coalesced due to rotational forces and
gravity.
heat.
radioactivity.
solar wind.
Answer:
Gravity
Explanation:
The solar system is held together by rotational forces and gravity. This can be seen from billions of years ago when the solar system was a cloud of dust and gas. This cloud of dust and gas is known as the solar nebula. All of these dust and gas were brought together by the rotational movement as well as the action of gravity which brought all the particles together to form a larger one. This alone brought about the sun's formation in the center of the nebula as well the formation of other planetary bodies, etc.
Cheers.
a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released
Answer:
The final speed of the ball is 9 m/s.
Explanation:
We have,
A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :
[tex]v=u+at[/tex]
u = 0 (at rest), a = g
[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]
So, the final speed of the ball is 9 m/s.
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
a = 1.72 m/s²
When an electromagnetic wave falls on a white, perfectly reflectingsurface, it exerts a force F on that surface. If the surfaceis now painted a perfectly absorbing black, the force that the samewave would exert on the surface is:___________.
A) F
B) F/2
C) F/4
D) 2F
E) 4F
Answer:
B. F/2
Explanation:
The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave
i.e Prad = u
For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.
i.e Prad = 2u.
Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2
An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelength. Calculate the current and voltage output when the detector is used in the photo-conductive and photovoltaic modes respectively. The reverse saturation current (Is) is 9 nA.
Answer:
I = 4.189 mA V = 0.338 V
Explanation:
In order to do this, we need to apply the following expression:
I = Is[exp^(qV/kT) - 1] (1)
However, as the junction of the diode is illuminated, the above expression changes to:
I = Iopt + Is[exp^(qV/kT) - 1] (2)
Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:
I ≅ Iopt
Therefore:
I ≅ I₀Aλq / hc (3)
Where:
I₀A = Area of diode (radiation)
λ: wavelength
q: electron charge (1.6x10⁻¹⁹ C)
h: Planck constant (6.62x10⁻³⁴ m² kg/s)
c: speed of light (3x10⁸ m/s)
Replacing all these values, we can get the current:
I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)
I = 4.189x10⁻³ A or 4.189 mA
Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:
I = Is[exp^(qV/kT) - 1]
k: boltzman constant (1.38x10⁻²³ J/K)
4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]
4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]
465,444.44 + 1 = exp(38.65V)
ln(465,445.44) = 38.65V
13.0508 = 38.65V
V = 0.338 V
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) . Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Complete Question
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .
Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Part B
If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
Answer:
Part A
[tex]R = 1081 \ J[/tex]
Part B
[tex]h = 169.7 \ m[/tex]
Explanation:
From the question we are told that
The mass of the blade is [tex]m_b = 0.65 \ kg[/tex]
The length is [tex]l = 0.55 \ m[/tex]
The angular speed is [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]
Generally the moment of inertia of the of this mower is mathematically evaluated as
[tex]I = \frac{m_b * l^2 }{12}[/tex]
substituting values
[tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]
[tex]I = 0.016 \ kg m^2[/tex]
Generally the rotational kinetic energy of the bland is
[tex]R = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]
[tex]R = 1081 \ J[/tex]
At point where the gravitational potential energy is equal to the rotational kinetic energy we have that
[tex]P = R = m_b * h * g[/tex]
Where P is the gravitational potential energy
substituting values
[tex]1081 = 0.65 * 9.8 * h[/tex]
=> [tex]h = 169.7 \ m[/tex]
Explain why it is necessary to have a high voltage
Answer:
SO THAT
EACH APPLIANCE CAN GET SUFFICIENT POTENTIAL DIFF. TO RUNAstronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?
Answer:
a. about 20 years younger
b. Malcolm sits around for 49.94 years
c. 2.268x[tex]10^{17}[/tex] m
Explanation:
light travels 3x[tex]10^{8}[/tex] m in one seconds
in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m
for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex] = 3.78x[tex]10^{17}[/tex] m
Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex] = 2.4x[tex]10^{8}[/tex] m/s
time that will pass for Malcolm will be distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]
= 1575000000 s = 49.94 years
the relativistic time t' will be
t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]
t' = 49.94 x 0.6 = 29.96 years this is the time that has passed for Flo
this means that Flo will be about 20 years younger than Malcolm when she returns
relativistic distance is
d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x 0.6
d' = 2.268x[tex]10^{17}[/tex] m this is how far it is to Flo
A block of mass 15.0 kg slides down a ramp inclined at 28.0∘ above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 5.50 m along the ramp, find the work done on the block by friction.
Answer:
Work is done by friction = -165 J
Explanation:
Given:
Mass of block (m) = 15 kg
Ramp inclined = 28°
Friction force (f) = 30 N
Distance (d) = 5.5 m
Find:
Work is done by friction.
Computation:
Work is done by friction = -Fd
Work is done by friction = -(30)(5.5)
Work is done by friction = -165 J
A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.
Answer:
a) -5.26 m/s
b) 27.91 m
Explanation:
a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.
average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s
The change in velocity in the first half of the interval is ...
Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s
So, the initial velocity (at the beginning of the last 1.85 s interval) is ...
v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)
v1 = -5.259 m/s
__
b) The velocity when the object hits the ground is ...
v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s
This is related to the distance traveled by ...
v² = 2dg . . . . . where g is the acceleration and d is the distance traveled
d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters
The object travels a total distance of about 27.911 meters.
_____
The attached graph shows height vs. time.
Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..
Answer:
* air resistance.
*the direction of the rotation of the Earth
rotation of the thrown body
Explanation:
The projectile launch is described by the expressions
x-axis x = v₀ₓ t
y-axis y = [tex]v_{oy}[/tex] t - ½ gt²
When the things that affect this movement are analyzed, in order of importance we have:
* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.
* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.
* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption
* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
EXPLANATION ⛔
A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is
A. 0.4N
B. 0.6N
C. 6N
D. 60N
Answer:b
Explanation:I’m just trynna get more money dude
The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its
A.climate.
b.weather.
C. global warming
d. seasons
Answer:
a. global warming
Explanation:
that's the definitain of global warming
Answer:
A climate
Explanation:
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimetre high hydroelectric dam would be required to charge battery
Complete question is;
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes
Answer:
Amount of water required to charge the battery = 7.35 m³
Explanation:
The formula for Potential energy of the water at that height = mgh
Where;
m = mass of the water
g = acceleration due to gravity = 9.8 m/s²
H = height of water = 50 cm = 0.5 m
We know that in density, m = ρV
Where;
ρ = density of water = 1000 kg/m³
V = volume of water
So, potential energy is now given as;
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Now, formula for energy of the battery is given as;
E = qV
We are given;
q = 50 A.min = 50 × 60 = 3,000 C
V = 12 V
Thus;
qV = 3,000 × 12 = 36,000 J
E = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery.
Thus;
(4900V) = (36,000)
4900V = 36,000
V = 36,000/4900
V = 7.35 m³
A light wave will *Blank* if it enters a new medium perpendicular to the surface.
Answer:
A light wave will not stop if it enters a new medium perpendicular to the surface.
Explanation:
A light wave will not have any deviation if it enters a new medium perpendicular to the surface.
What is meant by refraction ?Refraction is defined as an optical phenomenon by which the direction of a light wave gets changed when it travels from one medium to another. This is because of the change in speed.
Here,
The light wave is entering a new medium such that it enters perpendicular to the surface. Angle of incidence is the angle between the incident ray and the line perpendicular to the surface at the point of incidence. Since, here the light ray is incident normal to the surface that means the angle of incidence is 0.
According to Snell's law,
sin i = μ sin r
where i is the angle of incidence, r is the angle of refraction and μ is the constant called refractive index.
As i = 0, sin i = 0
So, μ sin r = 0
Since μ is a constant, we can say that sin r = 0 or the angle of refraction,
r = 0
This means that there is no refraction and hence the light wave won't get deviated when it enters the medium normally.
Hence,
A light wave will not have any deviation if it enters a new medium perpendicular to the surface.
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A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
A. 26
B. 35
C. 42
D. 22
E. 48
Answer:
35Explanation:
According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.
sini/sinr = n
n is the constant = refractive index
Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng
nw is the refractive index of water and ng is that of glass
sini/sinr = nw/ng
given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass
On substitution;
sin 30/sinr = 1.33/1.5
1.5sin30 = 1.33sinr
sinr = 1.5sin30/1.33
sinr = 0.75/1.33
sinr = 0.5639
r = arcsin0.5639
r ≈35°
angle the light leave the glass is 35°
You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied
Answer:
18 N
Explanation:
Force can be found using the following formula.
f= m*a
where m is the mass and a is the acceleration.
We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.
m= 36 kg
a= 0.5 m/s^2
Substitute these values into the formula.
f= 36 kg * 0.5 m/s^2
Multiply 36 and 0.5
f=18 kg m/s^2
1 kg m/s^2 is equivalent to 1 Newton, or N.
f= 18 Newtons
The force being applied is 18 kg m/s^2, Newtons, or N
Question 9 of 10
2 Powie
You are riding a bicycle. You apply a forward force of 100 N, and you and the
bicycle have a combined mass of 80 kg. What is the acceleration of the
bicycle?
A. 125 m/s
B. 1.5 m/s2
c. 1.8 m/s?
D. 0.8 m/s
Answer:
1.25 m/s^2
Explanation:
F = m*a ...... force = mass * acceleration
force = 100 N, mass = 80 kg
100 = 80 * a
100/80 = a = 1.25 m/s^2
Answer:
The acceleration is 1.25m/s².
Explanation:
You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :
[tex]f = m \times a[/tex]
Let F = 100,
Let m = 80,
[tex]100 = 80 \times a[/tex]
[tex]100 = 80a[/tex]
[tex]a = 100 \div 80[/tex]
[tex]a = 10 \div 8[/tex]
[tex]a = 1.25[/tex]
Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?
Answer:
a. 63.89° in the north-southward manner
b. 2.2 sec
Explanation:
The goose is flying at 100 km/h
Air from east to west is 44 km/h
angle relative to the north-south direction for the bird to travel south will be
cos∅ = 44/100 = 0.44
∅ = [tex]cos^{-1}[/tex]0.44 = 63.89° in the north-southward manner
Speed south relative to the ground will be v
Tan 63.89 = v/100
2.04 = v/100
v = 2.04 x 100 = 204 km/hr
to cover a distance of 450 m from north to south at this speed time will be
t = d/v = 450/204 = 2.2 sec
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 3.5 mV/m. At what rate is the magnetic field changing?
Answer
The rate at which the magnetic field is changing is [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]
Explanation
From the question we are told that
The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]
The radius is [tex]r = 1.5 \ m[/tex]
The rate of change of the magnetic field is mathematically represented as
[tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]
Where [tex]dl[/tex] is change of a unit length
[tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]
Where A is the area which is mathematically represented as
[tex]A = \pi r^2[/tex]
So
[tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]
[tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]
where L is the circumference of the circle which is mathematically represented as
[tex]L = 2 \pi r[/tex]
So
[tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]
[tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]
[tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]
substituting values
[tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]
[tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]
Carbon is added to iron to make steel. Steel is stronger than either carbon or iron by itself.
What does this example show?
Answer:
This example shows that alloys are stronger than either of it's parent materials by themselves.
Explanation:
Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.
This is because an alloy is a mixture of two or more elements, where at least one element is a metal.
Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.
Archimedes and Heron are playing on a seesaw. Archimedes weighs 75 kg and Heron weighs 150 kg. If Heron is sitting 2 meters from the fulcrum, how many meters does Archimedes need to sit from the fulcrum?
Answer:
4metresExplanation:
Using the principle of moment to solve the problem. The principle states that the sum of clockwise moments is equal to the sum of anticlockwise moment.
Moment = force *perpendicular distance
Moment of Archimedes about the fulcrum = 75 * x ... 1
x is the distance of Archimedes from the fulcrum
Moment of Heron about the fulcrum = 150 * 2 = 300kgm... 2
Equation 1 and 2 according to principle of moment to get x we have;
75x = 300
x = 300/75
x = 4metres
Archimedes need to sit 4m from the fulcrum
A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.
Answer:
C) less than 129 lb.
Explanation:
Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards with magnitude a .
If R be the reaction force
For the elevator is going downwards with acceleration a
mg - R = ma
R = mg - ma
R measures its apparent weight . Spring scale will measure his apparent weight.
So its apparent weight is less than 129 lb .
Nuclear fusion in our Sun happens when
- hydrogen atoms combine to make helium atoms and release energy
- uranium atoms break apart and release energy
- hydrogen atoms are burned and release energy
- helium atoms break apart and release energy
Answer:
A
Explanation:
Fussion occurs when elements of lower atomic mass combines to form that of a larger atomic mass, releasing energy in the process .
Hydrogen has a lower atomic mass than Helium.
The “turning effect of a force” (T = F * r) is:
(a) determined as the product of force and the moment of inertia.
(b) generated by concentric forces.
(c) equivalent to the angular momentum.
(d) determined as a product of torque and moment arm.
(e) called “moment” or “torque”.
Answer:
b and e
Explanation:
r x F is the formula for torque.
The "turning effect" or torque happens when concentric forces rotate an object along said center.
a) False because T = Fr = Ia (a = angular acceleration)
b) True
c) False. L = Iw (w = angular velocity), which does not equal Ia
d) False. It is torque, not the product of torque and something else
e) True.
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s ²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)