[tex]\mu = 0.177[/tex]
Explanation:
Let's look at the forces on the two axes:
[tex]x:\:\:\:F - f_n = F - \mu N = 0\:\:\:\:\:\;(1)[/tex]
[tex]y:\:\:\:N - mg = 0\:\:\:\:\:\:\:\:\:(2)[/tex]
Substituting (2) into (1) and solving for [tex]\mu[/tex], we get
[tex]F = \mu mg[/tex]
[tex]\mu = \dfrac{F}{mg} = \dfrac{12.0\:\text{N}}{(6.9\:\text{kg})(9.8\:\text{m/s}^2)} = 0.177[/tex]
A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Answer:
13.3 pounds.
Explanation:
For a spring of constant K, with an attached object of mass M, the period can be written as:
T = 2*π*√(M/K)
Where π = 3.14
First, we know that the period is 4 seconds, then we have:
4s = (2*π)*√(M/K)
We know that if the mass is reduced by 10lb, the period becomes 2s.
Then the new mass of the object will be: (M - 10lb)
Then the period equation becomes:
2s = (2*π)*√((M-10lb)/K)
So we have two equations:
4s = (2*π)*√(M/K)
2s = (2*π)*√((M-10lb)/K)
We want to solve this for M.
First, we need to isolate K in one of the equations.
Let's isolate K in the first one:
4s = (2*π)*√(M/K)
(4s/2*π) = √(M/K)
(2s/π)^2 = M/K
K = M/(2s/π)^2 = M*(π/2s)^2
Now we can replace it in the other equation.
2s = (2*π)*√((M-10lb)/K)
First, let's simplify the equation:
2s/(2*π) = √((M-10lb)/K)
1s/π = √((M-10lb)/K)
(1s/π)^2 = ((M-10lb)/K
K*(1s/π)^2 = M - 10lb
Now we can use the equation: K = M*(π/2s)^2
then we get:
K*(1s/π)^2 = M - 10lb
(M*(π/2s)^2)*(1s/π)^2 = M - 10lb
M/4 = M - 10lb
10lb = M - M/4
10lb = (3/4)*M
10lb*(4/3) = M
13.3 lb = M
Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.
Answer:
thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.
A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency is 890 Hz, the wavelength is .10m, and the amplitude is 6.5 mm. The tension in the line, in SI units, is closest to
Answer:
T = 712.9 N
Explanation:
First, we will find the speed of the wave:
v = fλ
where,
v = speed of the wave = ?
f = frequency = 890 Hz
λ = wavelength = 0.1 m
Therefore,
v = (890 Hz)(0.1 m)
v = 89 m/s
Now, we will find the linear mass density of the wire:
[tex]\mu = \frac{m}{L}[/tex]
where,
μ = linear mass density of wie = ?
m = mass of wire = 90 g = 0.09 kg
L = length of wire = 1 m
Therefore,
[tex]\mu = \frac{0.09\ kg}{1\ m}[/tex]
μ = 0.09 kg/m
Now, the tension in wire (T) will be:
T = μv² = (0.09 kg/m)(89 m/s)²
T = 712.9 N
A 77 turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.18 T field, starting with the normal of the plane of the coil perpendicular to the field. Assume that the positive max emf is reached first.
a. What is the peak emf?
b. At what time is the peak emf first reached?
c. At what time is the emf first at its most negative?
d. What is the period of the AC voltage output?
Answer:
a) fem = 5.709 V, b) t = 0.196 s, c) t = 0.589 s, d) T = 0.785 s
Explanation:
This is an exercise in Faraday's law
fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]
fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]
The magnetic field and the area are constant
fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]
fem = - N B A (-sin θ) [tex]\frac{d \theta}{dt}[/tex]
fem = N B (π d² / 4) sin θ w
fem= [tex]\frac{\pi }{4}[/tex] N B d² w sin θ
with this expression we can correspond the questions
a) the peak of the electromotive force
this hen the sine of the angle is 1
sin θ = 1
fem = [tex]\frac{\pi }{4}[/tex] 77 1.18 0.10² 8.0
fem = 5.709 V
b) as the system has a constant angular velocity, we can use the angular kinematics relations
θ = w₀ t
t = θ/w₀
Recall that the angles are in radians, so the angle for the maximum of the sine is
θ= π/2
t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]
t = 0.196 s
c) for the electromotive force to be negative, the sine function of being
sin θ= -1
whereby
θ = 3π/ 2
t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]
t = 0.589 s
d) This electromotive force has values that change sinusoidally with an angular velocity of
w = 8 rad / s
angular velocity and period are related
w = 2π / T
T = 2π / w
T = 2π / 8
T = 0.785 s
A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?
Answer:
19m
Explanation:
we have proper length L = 100m
the speed of the train v = 0.95
the speed of light is given as = 3x10⁸
length of the tunnel is given as = 50 meters
we can solve for the lenght contraction as
LX√1-v²/c²
= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )
= 31.22 metres
the train would be well seen at
50 - 31.22
= 18.78
= this is approximately 19 metres
we conclude tht the trains ends clears the ends of the tunnel by 19 meters.
thank you!
The best and most common way to measure the intensity of a cardiovascular exercise is to determine
O The person's heart rate
O The fatigue level of the person
O Amount of perspiration the person produces
The person's breathing rate
Answer:
the person's heart rate
The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars? Show your work
Answer:
log T = 3.72 to 3.77
Explanation:
Temperature range is
T = 5200 to 5900
Take the log
So,
log T = log 5200 to log 5900
log T = 3.72 to 3.77
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
A magnetic force acting on an electric charge in a uniform magnetic field what happend
Answer:
hgff
Explanation:
Answer:
The charge moves to equilibrium.
E.e = B.e.V
E is electric field force.
e is the charge.
B is magnetic field force.
V is acceleration voltage.
Which formula below correctly states Coulomb's Law?
A. F = kqq/r2
B. F = kqq/r
C. F = qq/kr2
D. F = kr2/qq
[tex]A. \: \: \: \: F = kqq/r^ 2 \\B. \: \: \: \: F = kqq/r \\C. \: \: \: \: F = qq/kr^ 2 \\D. \: \: \: \: F = kr^2 /qq \\ [/tex]
ANSWER:-F is directly proportional to the product of the charges
[tex]F∝qq[/tex]
F is inversely proportional to the square of the distance between them
[tex]F∝ \frac{1}{ {r}^{2} } [/tex]
from above 2 equation:-
we get:-
[tex]F∝ \frac{qq}{ {r}^{2} } [/tex]
To remove proportionality sign we use constant for this case we r using constant k
[tex]F = \frac{Kqq}{ {r}^{2} } [/tex]
So your answer is :-
OPTION A.
[tex]F = \frac{Kqq}{ {r}^{2} } [/tex]
A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement, [tex]\theta=12\ rad[/tex]
Final angular speed, [tex]\omega_f=17\ rad/s[/tex]
The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]
Put all the values,
[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]
So, the initial angular speed of the CD is equal to 14.73 rad/s.
What is the maximum wavelength, in nm, of light that can eject an electron from a metal with Φ =4.50 x 10–19 J?
[tex]4.4×10^{-7}\:\text{m}[/tex]
Explanation:
The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:
[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]
or
[tex]\dfrac{hc}{\lambda} = \phi[/tex]
Solving for the wavelength [tex]\lambda[/tex],
[tex]\lambda = \dfrac{hc}{\phi}[/tex]
[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]
[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]
Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.
The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.
To find the wavelength, the given values are,
Energy (E) = 4.50×10¯¹⁹ J
What is wavelength?The distance between two consecutive crests and troughs is called the wavelength of a wave.
Here, for the wavelength,
Energy (E) = 4.50×10¯¹⁹ J
Planck's constant (h) = 6.626×10¯³⁴ Js
Speed of light (v) = 3×10⁸ m/s
The wavelength of the light can be obtained as illustrated below:
E = hv / λ
Cross multiply λ,
E × λ = hv
Divide both sides by E,
λ = hv / E
Substituting all the values,
λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹
λ = 0.000000441733 m
λ = 441.73nm
λ - The maximum wavelength of light.
Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm
Learn more about wavelength,
https://brainly.com/question/13047641
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A car of mass 2100 kg collides with a motorcycle of mass 290 kg. After the collision, the car and motorcycle stick and slide together. The car's velocity just before the collision was<30,-10>m/s , and that of the motorcycle was <10,10>m/s. Determine the velocity of the stuck-together car and motorcycle just after the collision.
Answer:
V = 29.49 m/s
Explanation:
Given that,
The mass of a car,[tex]m_c=2100\ kg[/tex]
The mass of a motorcycle, [tex]m_m=290\ kg[/tex]
The initial velocity of the car,[tex]v_c=30i-10j[/tex]
[tex]|v_c|=\sqrt{30^2+(-10)^2} =31.62\ m/s[/tex]
The initial velocity of the motorcycle,[tex]v_m=10i+10j[/tex]
[tex]|v_m|=\sqrt{10^2+10^2} =14.14\ m/s[/tex]
As they stick together. Let V is the speed. So, using the conservation of momentum,
[tex]m_cv_c+m_mv_m=(m_c+m_m)V\\\\V=\dfrac{m_cv_c+m_mv_m}{(m_c+m_m)}\\\\V=\dfrac{2100\times 31.62+290\times 14.14}{(2100+290)}\\\\V=29.49\ m/s[/tex]
So, the velocity of the stuck together car and the motorcycle after the collision is 29.49 m/s.
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A
Complete Question
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2
Answer:
[tex]h=1614m[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]
Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]
Density [tex]\rho=1.20kg/m^2[/tex]
Generally the equation for Height climbed is mathematically given by
[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]
[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]
[tex]h=1614m[/tex]
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes
I guess that we want to find how much money you get each week.
We know that the job pays $8.60 per hour.
We know that you work 20 hours per week.
Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:
20*$8.60 = $172.
Now we know that your employer witholds:
10% + 7.65% + 5% = 22.65%
Then your employer withholds 22.65% of your gross pay.
if the 100% of your gross pay is $172
Then the 22.65% will be:
(22.65%/100%)*$172 = 0.2265*$172 = $38.96
This means that your employer withholds $38.96 of your weekly gross pay.
Then each week you get:
$172 - $38.96 = $133.04
If you want to learn more, you can read:
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Two carts are involved in an inelastic collision. Cart A with mass 0.900 kg hits cart B with mass 0.550 kg (initially at rest). The two carts stick together after the collision and continue to move along together. Cart A has an initial velocity of 0.29 m/s.
a. What is the final velocity of the two-cart system?
b. What is the initial kinetic energy of cart A?
c. What is the initial kinetic energy of cart B?
d. What is the final kinetic energy of the system?
e. Is kinetic energy conserved for inelastic collisions?
f. Is momentum conserved for inelastic collisions?
Answer:
a) v = 0.18 m / s, b) K₀ₐ = 0.0378 J, c) K_{ob}= 0, d) K = 0.02349 J,
Explanation:
a) For this exercise we must define a system formed by the two cars, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the hole
p₀ = ma v₀ₐ
final intnate. After the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ = (mₐ + m_b) v
v = [tex]\frac{m_a}{m_a+m_b} \ v_{oa}[/tex]
let's calculate
v = [tex]\frac{0.900}{0.900+0.550} \ 0.29[/tex]
v = 0.18 m / s
in the same direction of the movement of carriage A
b) the initial kinetic energy car A
K₀ₐ = ½ m v₀ₐ²
K₀ₐ = ½ 0.900 0.29²
K₀ₐ = 0.0378 J
c) kinetic energy of carriage B
k_{ob} = 0
because the car is stopped
d) the kinetic energy of the system
K = ½ (mₐ + m_b) v²
K = ½ (0.900 + 0.550) 0.18²
K = 0.02349 J
E) we see that part of the kinetic energy is lost, therefore the scientific reeling is not conserved in inelastic collisions
F) and momentum is conserved since it is equal to the variation of the moment and this is conserved in all collisions
5 How does air get polluted?
Answer:
Explanation:
- Pollution from cars
- Burning fossil fuels
3. Define 1 standard kilogram?
Answer:
standard kilogram is the SI unit of mass
Answer:
The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
A series LR circuit contains an emf source of having no internal resistance, a resistor, a inductor having no appreciable resistance, and a switch. If the emf across the inductor is of its maximum value after the switch is closed, what is the resistance of the resistor
Answer:
b. 1.9 Ω
Explanation:
Here is the complete question
A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?
Solution
The voltage across the inductor V is
[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] where V₀ = emf of source = 14 V, R = resistance, L = inductance = 34 H and t = time
Given that V = 80% of its maximum value after 4.0 s, this implies that V = 80 % of V₀ = 0.8V₀ and t = 4.0 s
Since [tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] and V = 0.8V₀.
Since we need to find R, we make R subject of the formula, we have
[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex]
[tex]V/V_{0}= e^{-\frac{Rt}{L} }[/tex]
taking natural logarithm of both sides, we have
㏑(V/V₀) = -Rt/L
R = -L㏑(V/V₀)/t
Substituting the values of the variables into the equation, we have
R = -34㏑(0.8V₀/V₀)/4.0 s
R = -34㏑(0.8)/4.0 s
R = -34 × -0.2231/4.0 s
R = 7.587/4
R = 1.896 Ω
R ≅ 1.9 Ω
So, B is the answer
Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?
a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features
Answer:
c). Easier setup
Explanation:
As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.
A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is not negligible.
Identify the forces on the falcon.
a. Kinetic friction
b. Weight w
c. Static friction
d. Drag D
e. Normal force n
f. Thrust
g. Tension T
Answer:
Explanation:
When a falcon is hovering, the force of up thrust is balanced by the weight.
When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.
The displacement x of a particle varies with time t as x = 4t 2 -15t + 25. Find the position,
velocity and acceleration of the particle at t = 0. When will the velocity of the particle becomes
zero? Can we call the motion of the particle as one with uniform acceleration?
Answer:
x = 4 t^2 - 15 t + 25 displacement of particle
dx / dt = 8 t - 15 velocity of particle
d^2x / dt^2 = 8 acceleration of particle
If 8 t -15 = o then t = 8 / 15
Since acceleration is a constant 8 then motion has uniform acceleratkon
what is the gravitational potential in a field produced by an object of mass 2000 kg at a distance of 10 km
Answer:
196 megajoules
Explanation:
Since you are talking about the gravitational potential I am assuming 10km is the height of the object in free fall.
PEg = mgh 2000kg×9.8m/s²×10000m = 196 megajoules
any four difference between velocity and acceleration
Answer:
https://physicsabout.com/acceleration-and-velcoity/
A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.
Answer:
[tex]T=3*10^-3 N/m[/tex]
Explanation:
From the question we are told that:
Radius :
[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]
Work [tex]W=1.5 * 10^{-4}[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=TdA[/tex]
Where
[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]
[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]
[tex]dA=0.050m^2[/tex]
Therefore
[tex]W=TdA[/tex]
[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]
[tex]T=3*10^-3 N/m[/tex]
A frictionless spring with a 9-kg mass can be held stretched 1.8 meters beyond its natural length by a force of 80 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after tt seconds. meters
Answer:
the required solution is; x(t) = 0.675sin( 2.222t )
Explanation:
Given the data in the question;
Using both Newton's and Hooke's law;
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5
given that mass m = 9 kg
[tex]x[/tex] = 1.8 m
k is F / x
hence
k = F / x
given that, F = 80 N
we substitute
k = 80 / 1.8
k = 44.44
so
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,
we input
9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,
[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0
so auxiliary equation is,
r² + 4.9377 = 0
r² = -4.9377
r = √-4.9377
r = ±2.222i
hence, the solution will be;
x(t) = A×cos( 2.222t ) + B×sin( 2.222t )
⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )
using initial conditions
x(0) = 0
⇒ 0 = A
[tex]x^t[/tex](t) = 1.5
1.5 = 2.222B
so
B = 1.5 / 2.222 = 0.675
Hence, the required solution is; x(t) = 0.675sin( 2.222t )
In what kind of reaction is water (H20) broken down into hydrogen gas (H2) and oxygen gas (O2)?
A. Combination
B. Decomposition
C. Displacement
D. Combustion
Answer:
Answer is B (Decomposition)
Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(
(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 J of useful work while metabolizing 500 kcal of food energy? % (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%? kcal
Answer:
a) The energy efficiency of the out-of-condition professor is 9.082 %.
b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
Explanation:
a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:
[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)
Where:
[tex]W[/tex] - Useful work, in joules.
[tex]E[/tex] - Food energy, in joules.
If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:
[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]
[tex]\eta = 9.082\,\%[/tex]
The energy efficiency of the out-of-condition professor is 9.082 %.
b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:
[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]
[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]
[tex]E = 7.60\times 10^{5}\,J[/tex]
[tex]E = 181.644\,kcal[/tex]
The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
