Answer:
Raw materials are most times gotten from the earth through various forms of extraction procedures.
A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.
B) Cat litter comprises of ceramic products which is made up of clay.
C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.
D)Lithium batteries are made up of elements in the earth such as lithium and carbon.
E)Aluminum beverage cans are made up of aluminum extracted from the ground.
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction
Answer:
the pressure after contraction is 2×10^5 Pa
the speed after contraction is 15m/s
Explanation:
We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,
For us to calculate pressure we need to calculate the area first as;
Let initial Area = A₁
And Final area A₂
We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means
A₂= A₁/3.................
V₁ is the speed
the pressure and speed of the water after the contraction can be calculated using equation of continuity below
A₂V₂ = A₁V₁
But
If we substitute given value in the expresion we have
V₂ = (3A *5)/A
V₂ = 15m/s
Therefore, the speed after contraction is 15m/s
Now we can calculate the pressure using
Bernoulli's equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
But we know that the pipe is horizontal, then "h" terms cancel out then
p₁ + ½ρv₁² = p₂ + ½ρv₂²
Making P₂ subject of formula we have
p₂ = 0.5ρ( V ₁² - v₂² ) + P₁
P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5
=2×10^5 Pa
Therefore, the pressure after contraction is 2×10^5 Pa
(a) the final speed of the water after contraction is 15 m/s.
(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.
The given parameters;
initial pressure, P₁ = 3.5 x 10⁵ Painitial speed, v₁ = 5 m/sdensity of water, ρ = 1000 kg/m³Let the initial area of the pipe = A₁
Apply the continuity equation to determine the final speed of the water after contraction as follows;
[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]
The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;
[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]
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If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b
Answer:
Will be equal to alpha x r; less than UsN
A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F⃗ to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=αt2+βt3, where α = 0.210 m/s2 and β = 2.04×10−2 m/s3 .
A. Calculate the velocity of the object at time t = 4.50 s .
B. Calculate the magnitude of F⃗ at time t = 4.50 s .
Express your answer to three significant figures.
C. Calculate the work done by the force F⃗ during the first time interval of 4.50 s of the motion.
Express your answer to three significant figures.
Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
A) V_air = 1.295 L
B) Volume is not reasonable
Explanation:
A) Let;
m be total mass of the man
m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung
m_3 be the mass above the water with the empty lung
m_5 be the mass above the water with full lung
F_b be the buoyant force due to the air in the lung
V_a be the volume of air inside man's lungs
w_p be the weight that the buoyant force opposes as a result of the air.
Now, we are given;
m = 78.5 kg
m_3 = 3.2% × 78.5 = 2.512 kg
m_5 = 4.85% × 78.5 = 3.80725 kg
Now, m_p = m_5 - m_3
m_p = 3.80725 - 2.512
m_p = 1.29525 kg
From archimedes principle, we have the formula for buoyant force as;
F_b = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
F_b = w_p = 1.29525 × 9.81
F_b = 12.7064 N
As earlier said,
F_b = (ρ_water × V_air × g)
Thus;
V_air = F_b/(ρ_water × × g)
V_air = 12.7064/(1000 × 9.81)
V_air = 1.295 × 10^(-3) m³
We want to convert to litres;
1 m³ = 1000 L
Thus;
V_air = 1.295 × 10^(-3) × 1000
V_air = 1.295 L
B) From research, the average lung capacity of an adult human being is 6 litres of air.
Thus, the calculated lung volume is not reasonable
An unpolarized beam of light with an intensity of 4000 W/m2 is incident on two ideal polarizing sheets. If the angle between the two polarizers is 0.429 rad, what is the emerging light intensity
Answer:
The intensity is [tex]I_2 = 1654 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 4000 \ W/m^2[/tex]
The angle between the ideal polarizing sheet is [tex]\theta = 0.429 \ rad = 0.429 * 57.296 = 24.58^o[/tex]
Generally the intensity of light emerging from the first polarizer is mathematically represented as
[tex]I_2 = \frac{I_o}{2}[/tex]
substituting values
[tex]I_1 = \frac{4000}{2}[/tex]
[tex]I_1 = 2000 \ W/m^2[/tex]
Then the intensity of incident light emerging from the second polarizer is mathematically represented by Malus law as
[tex]I_2 = I_1 cos^2 (\theta )[/tex]
substituting values
[tex]I_2 = 2000 * [cos (24.58)]^2[/tex]
[tex]I_2 = 1654 \ W/m^2[/tex]
A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.
Calculate di/dt
di/dt = _________.
Answer:
[tex]\frac{di}{dt} = 7.31 \ A/s[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 450 \ turns[/tex]
The radius is [tex]r = 1.17 \ cm = 0.0117 \ m[/tex]
The position from the center consider is x = 3.45 cm = 0.0345 m
The induced emf is [tex]e = 8.20 *10^{-6} \ V/m[/tex]
Generally according to Gauss law
[tex]\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A[/tex]
=> [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A[/tex]
Where A is the cross-sectional area of the solenoid which is mathematically represented as
[tex]A = \pi r ^2[/tex]
=> [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2[/tex]
=> [tex]\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}[/tex]ggl;
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]
=> [tex]\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}[/tex]
=> [tex]\frac{di}{dt} = 7.31 \ A/s[/tex]
The value of di/dt from the given values of the solenoid electric field is;
di/dt = 7.415 A/s
We are given;
Number of turns; N = 450 per m
Radius; r = 1.17 cm = 0.0117 m
Electric Field; E = 8.2 × 10⁻⁶ V/m
Position of electric field; r' = 3.45 cm = 0.0345 m
According to Gauss's law of electric field;
∫| E*dl | = |-d∅/dt |
Now, ∅ = BA = μ₀niA
where;
n is number of turns
i is current
A is Area
μ₀ = 4π × 10⁻⁷ H/m
Thus;
E(2πr') = (d/dt)(μ₀niA) (negative sign is gone from the right hand side because we are dealing with magnitude)
Since we are looking for di/dt, then we have;
E(2πr') = (di/dt)(μ₀nA)
Making di/dt the subject of the formula gives;
di/dt = E(2πr')/(μ₀nA)
Plugging in the relevant values gives us;
di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)
di/dt = 7.415 A/s
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A 2100 kg truck traveling north at 38 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck's kinetic energy
Answer:
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
Explanation:
Given:
Mass of truck(m) = 2,100 kg
Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s
Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s
Find:
Change in kinetic energy (ΔKE)
Computation:
Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]
Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]
Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]
Change in kinetic energy (ΔKE) = 1,050[121.9648]
Change in kinetic energy (ΔKE) = 128063.04
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
A car is going 8 meters per second on an access road into a highway
and then accelerates at 1.8 meters per second squared for 7.2
seconds. How fast is it then going?
Answer:
20.96 m/s^2 (or 21)
Explanation:
Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.
At first, we know a car is going 8 m/s, that is its initial velocity.
Then, we know the acceleration, which is 1.8 m/s/s
We also know the time, 7.2 second.
Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.
(final velocity - initial velocity) = time * acceleration
final velocity = time*acceleration + initial velocity
After plugging the found values in, we get 20.96 m/s/s, or 21 m/s
A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference, this resistor will dissipate:
Answer:
0.056 WExplanation:
[tex]Power = IV[/tex]
From ohms law we know that
[tex]V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}[/tex]
Given data
P1 = 0.5 Watt
P2 = ?
V1= 3 Volts
V2= 1 Volt
Thus we can solve for the power dissipated as follows
[tex]P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}[/tex]
[tex]\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W[/tex]
The resistor will dissipate 0.056 Watt
(a) If electrons were used (electron microscope), what minimum kinetic energy would be required for the electrons
Answer:
K = 1.6 10⁻¹⁵ J
Explanation:
In an electron microscope, electrons are used to form images, these electrons are accelerated in electric fields so that they have a kinetic energy that allows obtaining a good amplification with the microscope.
electrical potential energy is converted to kinetic energy
U = K
e V = ½ m v²
v = √2eV /m
the wavelength of these electrons we obtain from the de Broglie equation
λ = h / p
p = mv
λ = h / mv
λ = h / mra 2eV / m
λ = h / ra 2eVm
where we can see that as the potential energy increases, it electrifies the shorter the wavelength of the electrons and consequently the greater the magnification of the microscope
in general these microscopes use from 10000X onwards therefore for this saponification
K = e V
K = 1.6 10⁻¹⁹ 10000
K = 1.6 10⁻¹⁵ J
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have
Answer:
The planet that is farther from the star must have a time period greater.
Explanation:
We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex] (1)
Where:
r: is the radius of the circular orbit of the planet and the star
T: is the period
G: is the gravitational constant
M: is the mass of the planet
From equation (1) we have:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex] (2)
Where k is a constant
From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.
I hope it helps you!
Equal currents of magnitude I travel into the page in wire M and out of the page in wire N. The direction of the magnetic field at point P which is at the same distance from both wires is
Answer:
The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction
Answer:
Explanation:
Using the Continuity equation
v X A = v' xA'
so if A is 1/2of A' then A velocity must be 2 times the A'
after-contraction v = 2 x 5.0m/s = 10m/s
Using the Bernoulli equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
, the "h" terms cancel
3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²
p₂ = 342500pa
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .
Answer:
[tex]\theta=10.20^{\circ}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Explanation:
First of all, we analyze the system of blocks before starting to move.
[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]
[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]
[tex]tan(\theta)=0.18[/tex]
[tex]\theta=arctan(0.18)[/tex]
[tex]\theta=10.20^{\circ}[/tex]
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.
[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]
Where:
[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]
[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]
[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
(a) The inclined angle for which both blocks begin to slide is 10.3⁰.
(b) The compression of the spring is 0.22 ft.
The given parameters;
mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ftThe normal force on block A and B:
[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]
The frictional force on block A and B:
[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]
The net force on the blocks when they starts sliding;
[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]
The change in the energy of the blocks is the work done in compressing the spring;
[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]
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Polarized light passes through a polarizer. If the electric vector of the polarized light is horizontal what, in terms of the initial intensity I0, is the intensity of the light that passes through a polarizer if the polarizer is tilted 22.5° from the horizontal?
Answer: I0*0.853
Explanation:
Ok, the Malus's law says that:
If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.
In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°
Then:
I(22.5°) = I0*cos^2(22.5°) = I0*0.853
A ball travels with velocity given by [21] [ 2 1 ], with wind blowing in the direction given by [3−4] [ 3 −4 ] with respect to some co-ordinate axes. What is the size of the velocity of the ball in the direction of the wind?
Answer:
2/5 m/s
Explanation:
There are two vectors v and w . Let θ be angle b/w the two vector.
[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]
velocity of the ball in direction of the the wind
[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]
The size of the velocity of the ball in the direction of the wind is 2/5 ms.
Calculation of the size of velocity:Since there are two vectors v and w
Also, here we assume θ be angle b/w the two vector.
So
Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)
= 2/5√5
Now the velocity of the ball should be
= √(2^2 + 1^2) 2 ÷ 5√(5)
= 2 /5
hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.
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Two 75 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb
Answer:
300 W
Explanation:
power of each bulb P = 75 W
voltage in the circuit = 120 V
we know that electrical power P = IV ....1
and V = IR
we can also say that I = V/R
substituting for I in equation 1, we have
P = [tex]V^{2}/R[/tex] ....2
The total total power in the circuit = 75 x 2 = 150 W
from equation 2, we have
150 = [tex]120^{2} /R[/tex]
R = [tex]120^{2}/150[/tex] = 96 Ω this is the resistance of the whole circuit.
This resistance is due to the two light bulbs, for each light bulb since they are arranged in series
R = 96/2 = 48 Ω
From P = [tex]V^{2}/R[/tex]
for each light bulb, power is
P = [tex]120^{2} /48[/tex] = 300 W
When an ideal gas undergoes a slow isothermal expansion, A : the work done by the environment is the same as the energy absorbed as heat. B : the increase in internal energy is the same as the work done by the environment. C : the work done by the gas is the same as the energy absorbed as heat. D : the increase in internal energy is the same as the heat absorbed. E : the increase in internal energy is the same as the work done by the gas.
Explanation:
When an ideal gas undergoes a slow isothermal expansion, following phenomenon occur
1. Work done bu the gas = Energy absorbed as heat.
2. Work done by environment = Energy absorbed as heat.
3. Increase in internal energy= Heat absorbed= work done by gas = work done by environment.
Hence all option are correct.
Increase in internal energy is equal to the heat absorbed or work done by gas or environment. All the statements are correct.
If an ideal gas undergoes a slow isothermal expansion,
Work done by the gas is directly proportional energy absorbed as heat.
Work done by environment is directly proportional energy absorbed as heat.
Increase in internal energy is equal to the heat absorbed or work done by gas or environment.
To know more about the Ideal gas,
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Based on the passage, why is it important that different ethnic groups worked together on the strike? The groups needed to avoid speaking to one another because they wouldn’t understand. The different ethnic groups believed in being separate. The groups needed to trick the owners. They needed to be able to unite even though they spoke different languages.
Answer:D
Explanation:I got it right
Answer:
They needed to be able to unite even though they spoke different languages.
Explanation:
1. What was the Michelson-Morley experiment designed to do?2. When was the Michelson-Morley experiment done?3. What was the ether?4. What does the speed of a wave depend on?5. How many light beams are used in Michelson’s interferometer?6. What sort of problems did Michelson have with his first interferometer?7. How many times more sensitive was Michelson’s second interferometer?8. What did the new interferometer float on?9. What was the surprising outcome of the Michelson-Morley experiment?10. What were the implications of the experiment?11. What is the principle behind relativity?12. Who became the first American to win the Nobel Prize?13. Did Einstein base his Theory of Relativity on the Michelson-Morley experiment?
Answer:
1) designed to measure the difference in speed of light in different directions , 1887
Explanation:
1) This experiment was designed to measure the difference in speed of light in different directions and therefore find the speed of the ether.
2) was made in 1887
3) At that time it was assumed that it was the medium in which light traveled and it is everywhere
4) the speed of the wave depends on the characteristics of the medium where it travels,
for the one in a string depends on the tension and density
for an electromagnetic wave of the permittivity and permeability of the vacuum
5) In this type of interferometer the beam is divided into two rays
6) In his interrupter, he had to accurately measure the displacement of the fringes in a telescope, for which he had to minimize vibrations, he had problems in the movement of one of the arms, changes in temperature
7) In Michelsom's second experiment, the apparatus could measure 0.01 fringes by increasing the length of the arms by 11 m
8) The new interferometer floated on a bed of mercury
9) Couldn't measure any difference in speed of light in different directions
10) Physics was forced to eliminate the concept of ETHER
11) One of the principles of relativities that the speed of light is constant in all inertial efficiency systems
12) Michelson in 1907
13) It seems that Einstein did not know the results of this experiment
Two ice skaters push off against one another starting from a stationary position. The 45.0-kg skater acquires a speed of 0.375 m/s. What speed does the 60.0-kg skater acquire in m/s
Answer:
0.2812
Explanation:
Given that
mass of skater 1, m1 = 45 kg
mass of skater 2, m2 = 60 kg
speed of skater 1, v1 = 0.375 m/s
To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.
Thus, momentum p = mv
Law of conservation of momentum infers that,
m1v1 = m2v2
Now we proceed to substitute our values into the formula.
45 * 0.375 = 60 * v2
v2 = 16.875 / 60
v2 = 0.2812 m/s
Therefore the speed of the second skater has to be 0.2812 m/s
A string of holiday lights has 15 bulbs with equal resistances. If one of the bulbs
is removed, the other bulbs still glow. But when the entire string of bulbs is
connected to a 120-V outlet, the current through the bulbs is 5.0 A. What is the
resistance of each bulb?
Answer:
Resistance of each bulb = 360 ohms
Explanation:
Let each bulb have a resistance r .
Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.
[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]
[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]
R(equivalent) = r/15
Now, As per Ohms Law :
V = I * R(equivalent)
120 V = 5 A * r/15
r = 360 ohms
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 6.76 s. How much time does the driver of the car measure for his trip between the poles
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?
Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]
So, the force has a magnitude of 4212 N
A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the platform is m = 137 kg
The radius is r = 1.53 m
The mass of the person is [tex]m_p = 68.7 \ kg[/tex]
The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]
The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]
The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]
Generally the moment of inertia of the system is mathematically represented as
[tex]I = I_1 + I_2 + I_3[/tex]
Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as
[tex]I_1 = \frac{m * r^2}{2}[/tex]
substituting values
[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]
[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]
Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as
[tex]I_2 = m_p * d_c^2[/tex]
substituting values
[tex]I_2 = 68.7 * 1.19^2[/tex]
[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]
Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as
[tex]I_3 = m_d * d_d^2[/tex]
substituting values
[tex]I_3 = 25.9 * 1.45^2[/tex]
[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]
Thus
[tex]I= 160.35 + 97.29 + 54.45[/tex]
[tex]I= 312.09 \ kg \cdot m^2[/tex]
The Milky Way has a diameter (proper length) of about 1.2×105 light-years. According to an astronaut, how many years would it take to cross the Milky Way if the speed of the spacecraft is 0.890 c?
Answer:
t = 134834.31 years
Explanation:
First we find the speed of the ship:
v = 0.890 c
where,
v = speed of the ship = ?
c = speed of light = 3 x 10⁸ m/s
Therefore, using the values, we get:
v = (0.89)(3 x 10⁸ m/s)
v = 2.67 x 10⁸ m/s
Now, we find the distance in meters:
Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)
s = 11.35 x 10²⁰ m
Now, for the time we use the following equation:
s = vt
t = s/v
t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)
t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)
t = 134834.31 years
A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA
An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody emitters. For each star she makes a plot of the radiation intensity per unit wavelength as a function of wavelength. She notices that the curve for star A has a maximum that occurs at a shorter wavelength than does the curve for star B. What can she conclude about the surface temperatures of the two stars
Answer:
Star A has a higher surface temperature than star B.
Explanation:
The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:
[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)
Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.
A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.
Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.
