Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C. Now a valve is opened, and half of the mass of the gas is allowed to escape. if the final pressure in the tank is 2.2 atm. The final temperature in the tank is: Hint: make sure you convert the units of temperature and pressure to the proper units
Answer:
Final Temperature = 71 °C
Explanation:
In this case, the ideal gas equation is written as;
PV = mRT
Where;
P is pressure
V is volume
m is mass
R is gas constant
T is temperature
We will take the volume to be constant.
So, in the initial state, we have;
P1•V = m1•R•T1 - - - eq(1)
In the final state, we have;
P2•V = m2•R•T2 - - - - eq(2)
Combining eq (1) and eq(2),we have;
P1•m2•R•T2 = P2•m1•R•T1
Dividing both sides by R gives;
P1•m2•T2 = P2•m1•T1
Making T2 the subject gives;
T2 = (P2•m1•T1)/(P1•m2)
Now, we are given;
m1 = 2kg
m2 = ½*2 = 1kg
P1 = 4 atm
P2 = 2.2 atm
T1 = 40°C = 273 + 40 K = 313K
Plugging in this values into the T2 equation, we have;
T2 = (2.2 × 2 × 313)/(4 × 1)
T2 = 344 K
Converting to °C, we have;
T2 = 344 - 273 = 71 °C
What caused the disappearance of land bridges?
A. Volcanic outgassing
B. Shrinking of the polar ice caps
C. Beginning of an ice age
D. A mass extinction
Answer: B
Explanation:
I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges
Answer:B :)
Explanation:
A block is supported on a compressed spring, which projects the block straight up in the air at velocity VVoj The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block?
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answer:
the correct answer is B
Explanation:
We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.
For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.
Let's write the kinematic equation for the two bodies
The block. At the highest point of the path
y = - ½ g t2
The ball, in its vertical movement
y = vo t - ½ g t2
therefore the correct answer is B
A ship can float on water as long as it weighs less than water.
O A. True
O B. False
Answer:
It's true
Explanation:
Because the ship is mafe up of aluminium, which is a light metal.
Answer:
False
Explanation:
Took The Quiz
In a Venn diagram, the separate circles contain characteristics unique to each compared and the intersection contains characteristics that are common to both items being compared. This Venn diagram compares the inner and outer planets. What belongs in the center section?
a. -Revolve around the Sun
-Rotate on an axis
-Generally have rings
b. -Revolve around the Sun
-Rotate on axis
-Generally have moons
c. -Rotate around the Sun
-Revolve on an axis
-Generally have moons
d. -Rotate around the Sun
-Revolve on an axis
-Generally have rings
Answer:
B. revolve around the sun
rotate on an axis
generally have moons
Explanation:
edge 2021
an object's resistance to any change in motion is the_________ of the object.
An object's resistance to any change in motion is the Inertia of the object.
A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 990 kg ? Neglect the buoyant force on the cargo volume itself. Assume gases are at 0∘C and 1 atm pressure (rhoair = 1.29 kg/m3, rhohelium = 0.179 kg/m3).
Answer:
The mass of the cargo is [tex]M = 188.43 \ kg[/tex]
Explanation:
From the question we are told that
The radius of the spherical balloon is [tex]r = 7.40 \ m[/tex]
The mass of the balloon is [tex]m = 990\ kg[/tex]
The volume of the spherical balloon is mathematically represented as
[tex]V = \frac{4}{3} * \pi r^3[/tex]
substituting values
[tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]
[tex]V = 1697.6 \ m^3[/tex]
The total mass the balloon can lift is mathematically represented as
[tex]m = V (\rho_h - \rho_a)[/tex]
where [tex]\rho_h[/tex] is the density of helium with a value of
[tex]\rho_h = 0.179 \ kg /m^3[/tex]
and [tex]\rho_a[/tex] is the density of air with a value of
[tex]\rho_ a = 1.29 \ kg / m^3[/tex]
substituting values
[tex]m = 1697.6 ( 1.29 - 0.179)[/tex]
[tex]m = 1886.0 \ kg[/tex]
Now the mass of the cargo is mathematically evaluated as
[tex]M = 1886.0 - 1697.6[/tex]
[tex]M = 188.43 \ kg[/tex]
An LC circuit has a 6.00 mH inductor. The current has its maximum value of 0.570 A at t =0s. A short time later the capacitor reaches its maximum potential difference of 66.0 V. What is the value of the capacitance?
Answer:
C = 44.75 x 10⁻⁸ F
Explanation:
Assuming no loss of energy between capacitor and inductor
energy in inductor initially = 1/2 Li₀² where L is inductance and i₀ is peak current .
= .5 x 6 x 10⁻³ x .57²
= .97 x 10⁻³ J .
This energy is transferred to capacitor .
energy of capacitor = 1/2 CV²
= .5 x C x 66²
= 2178 C
2178C = .97 x 10⁻³
C = 44.75 x 10⁻⁸ F .
The magnetic energy stored in the inductor is transformed into electrical energy stored in the capacitor. The value of capacitance for the given circuit is 44.75 x 10⁻⁸ F
Finding the capacitance:According to the law of conservation of energy, the magnetic energy stored in the inductor will be gradually lost and this energy will be stored in the capacitor as electrical energy.
Initially, the energy in the inductor is:
E = 1/2 Li₀²
where L is inductance
and i₀ is peak current.
E = 0.5 × 6 × 10⁻³ × (0.57)²
E = 0.97 × 10⁻³J
This energy is transformed into electrical energy stored in the capacitor.
So the capacitor energy is:
E = 1/2 CV²
where C is the capacitance
E = 0.5 × C × 66²
E = 2178 C
0.97 x 10⁻³ = 2178 C
C = 44.75 x 10⁻⁸ F
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the heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water
(a) per unit mass
(b) per unit volume
Answer:
A. 4148 J/K/Kg
B. 4148 J/K/L
Explanation:
A. Heat capacity per unit mass is known as the specific heat capacity, c.
C = Heat capacity/mass(kg)
C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg
B. Volume of water = mass/density
Density of water = 1 Kg/L
Volume of water = 0.125 Kg/ 1Kg/L
Volume of water = 0.125 L
Heat capacity per unit volume = (523 J/K) / 0.125 L
Heat capacity per unit volume = 4148 J/K/L
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section), and the power fit of your Trendline equation,calculate the drag coefficient. Solve for it first (see video) and then plug in the values.
Answer:
The drag coefficient is [tex]D_z = 1.30512[/tex]
Explanation:
From the question we are told that
The density of air is [tex]\rho_a = 1.21 \ kg/m^3[/tex]
The diameter of bottom part is [tex]d = 0.15 \ m[/tex]
The power trend-line equation is mathematically represented as
[tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]
let assume that the velocity is 20 m/s
Then
[tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]
[tex]F_{\alpha } = 5.1453 \ N[/tex]
The drag coefficient is mathematically represented as
[tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]
Where
[tex]F_{\alpha }[/tex] is the drag force
[tex]\rho[/tex] is the density of the fluid
[tex]v[/tex] is the flow velocity
A is the area which mathematically evaluated as
[tex]A = \pi r^2 = \pi \frac{d^2}{4}[/tex]
substituting values
[tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]
[tex]A = 0.0176 \ m^2[/tex]
Then
[tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]
[tex]D_z = 1.30512[/tex]
Can someone help me with this question
Answer:
hypothesis , hope it helps
Explanation:
Answer:
Inference
Explanation:
Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.
Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?
Answer:
The velocity is [tex]v_2= 0.45 \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed of the hot water is [tex]v_1 = 0.85 \ m/s[/tex]
The pressure from the heater [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]
The height of the hot water before flowing is [tex]h_1 = 0 \ m[/tex]
The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]
The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Apply Bernoulli equation
[tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]
Substituting values
[tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]
=> [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]
=> [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]
=> [tex]v_2= 0.45 \ m/s[/tex]
1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts
experience on the Moon or on the space station? Explain
Answer:
by using it's buoyant or floating effect by Archimedes.
the buoyant force act on the astronauts body and make he/ she feels like in low gravity.
the buoyant force equation is
F = Density of liquid x earth gravitational field x volume of astronauts body and suit.
the Weight of astronauts in the pools will be less than in the land or air.
Weight in water = weight in air/land - buoyant force
so the astronauts will feel like in the outer space with low gravity.
An accident in a laboratory results in a room being contaminated by a radioisotope with a half life of 4.5 hours. If the radiation is measured to be 64 times the maximum permissible level, how much time must elapse before the room is safe to enter? The mass of Helium atom is 4.002602 u (where u = 1.66 x 10-27 kg) but the mass of 1 proton is 1.00730 u and 1 neutron is 1.00869 u. Calculate the binding energy per nucleon in MeV.
Answer:
a) t = 27.00 h
b) B = 6.84 MeV/nucleon
Explanation:
a) The time can be calculated using the following equation:
[tex] R = R_{0}e^{-\lambda*t} [/tex]
Where:
R: is the radiation measured at time t
R₀: is the initial radiation
λ: is the decay constant
t: is the time
The decay constant can be calculated as follows:
[tex] t_{1/2} = \frac{ln(2)}{\lambda} [/tex]
Where:
t(1/2): is the half life = 4.5 h
[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 h} = 0.154 h^{-1} [/tex]
We have that the radiation measured is 64 times the maximum permissible level, thus R₀ = 64R:
[tex] \frac{R}{64R} = e^{-\lambda*t} [/tex]
[tex] t = -\frac{ln(1/64)}{\lambda} = -\frac{ln(1/64)}{0.154 h^{-1}} = 27.00 h [/tex]
b) The binding energy (B) can be calculated using the following equation:
[tex]B = \frac{(Z*m_{p} + N*m_{n} - M_{A})}{A}*931.49 MeV/u[/tex]
Where:
Z: is the number of protons = 2 (for [tex]^{4}_{2}He[/tex])
[tex]m_{p}[/tex]: is the proton mass = 1.00730 u
N: is the number of neutrons = 2 (for [tex]^{4}_{2}He[/tex])
[tex]m_{n}[/tex]: is the neutron mass = 1.00869 u
[tex]M_{A}[/tex]: is the mass of the He atom = 4.002602 u
A = N + Z = 2 + 2 = 4
The binding energy of [tex]^{4}_{2}He[/tex] is:
[tex]B = \frac{(2*1.00730 + 2*1.00869 - 4.002602)}{4}*931.49 MeV/u = 7.35\cdot 10^{-3} u*931.49 MeV/u = 6.84 MeV/nucleon[/tex]
Hence, the binding energy per nucleon is 6.84 MeV.
I hope it helps you!
A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio
Answer:[tex]8.062\ m/s[/tex]
Explanation:
Given
masss of football player [tex]M=110\ kg[/tex]
Velocity of football player [tex]u_1=8\ m/s[/tex]
mass of football [tex]m=0.41\ kg[/tex]
velocity of football [tex]u_2=25\ m/s[/tex]
Final velocity will be given by applying conservation of linear momentum
After catching the ball Player and ball moves with same velocity
[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]
[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]
[tex]\Rightarrow 880+10.25=110.41\times v[/tex]
[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]
So, final velocity will be [tex]8.062\ m/s[/tex]
A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.30 rev/s in 3.05 s . Part A What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s
Answer:
[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
Explanation:
Angular acceleration
[tex]\begin{aligned}
\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\
\omega_{i} &=0 \\
\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\
&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\
&=27.02 \mathrm{rad} / \mathrm{s} \\
\alpha &=\frac{(27.02-0)}{3.15} \\
&=8.57 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
a)Tangential acceleration
[tex]\begin{aligned}
a &=r \alpha \\
&=\frac{12}{2} \times 10^{-2} \times 8.57 \\
a &=0.51 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
This question involves the concepts of the equations of motion for angular motion.
The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".
First, we will use the first equation of motion for the angular motion to find out the angular acceleration:
[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
where,
[tex]\alpha[/tex] = angular acceleration = ?
[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s
[tex]\omega_i[/tex] = initial angular speed = 0 rad/s
t = time taken = 3.05 s
Therefore,
[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]
Now, the tangential acceleration can be given as follows:
[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]
a = 0.532 m/s²
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The attached picture shows the angular equations of motion.
If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.
- 2x
- 1/4
- 1/2
- 4x
Answer:
The correct option is;
- 4x
Explanation:
From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source
The inverse square law can be presented as follows;
[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]
As the distance, r, increases, the surface it covers also increases by the power of 2
Therefore, where the distance increases from r to 2·r, we have;
When, I, remain constant
[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]
The surface increases to 4·S by the inverse square law
Therefore, the correct option is 4 × x.
On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?
2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Answer:
Explanation:
Volume of asteroid = 4/3 x π x 160³
= 17.15 x 10⁶
mass = volume x density
= 17.15 x 10⁶ x 2600
= 445.9 x 10⁸ kg
kinetic energy
= 1/2 x 445.9 x 10⁸ x( 12.6 )² x 10⁶
= 35.4 x 10¹⁷ J .
2 )
energy of 15 megaton
= 4.184 x 10¹⁵ x 15 J
= 62.76 x 10¹⁵ J
No of bombs required
= 35.4 x 10¹⁷ / 62.76 x 10¹⁵
= 56.4 Bombs .
A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 m to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?
Answer:
a. 42N
b. 11.8m/s
c. 1.69s
d. 160N
Explanation:
a) The tension of the rope is 130.66 N.
b) The speed of the bucket while strike the water = 4.64 m/s.
c) The time of fall is = 4.303 second.
d) While the bucket is falling, what is the force exerted on the cylinder by the axle is 130.66 N.
Mass of the water bucket; M = 15.0 kg
Mass of the cylinder; m = 12.0 kg
Height of the bucket; h = 10.0 m.
They are connected by a rope and a pivots.
So, acceleration of them is same and let it be a.
So equation of motion of both of them be:
Mg - T = Ma
and, T - mg = ma
Hence, a = g(M-m)/(M+m)
= 9.8(15-12)/(15+12)
= 1.08 m/s²
And, T = m(g+a)
= 12.0(9.8+1.08)
= 130.66 N.
a) so tension of the rope is 130.66 N.
b) speed of the bucket while strike the water = √2ah =√(2×1.08×10.0) m/s = 4.64 m/s.
c) The time of fall is = √2h/a = √(2×10/1.08) second = 4.303 second.
d) While the bucket is falling, what is the force exerted on the cylinder by the axle is tension of the rope, that is, 130.66 N.
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Help with this answer please
Answer:
Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL
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In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.
Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]
Answer:
Explanation:
The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]
What is equivalent resistance?The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.
What is equivalent resistance in series?Resistors are in series whenever the current flows through the resistors sequentially. It is given by
[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]
What is equivalent resistance in parallel?Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.
The equivalent resistance is the total resistance measured in a parallel. It is given by
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]
Given:
Resistor, [tex]R_{1} = 100 ohm[/tex]
Resistor, [tex]R_{2} = 200 ohm[/tex]
Voltage, [tex]V = 10 Volt[/tex]
Since, resistors are connected in parallel, the equivalent resistor is given by,
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]
[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]
[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]
Hence, the equivalent resistor is [tex]66.66 ohm[/tex].
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A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins
Answer:
Cube temperature = 526.83 K
Explanation:
Volume of the cube and sphere will be the same.
Now, volume of cube = a³
And ,volume of sphere = (4/3)πr³
Thus;
a³ = (4/3)πr³
a³ = 4.1187r³
Taking cube root of both sides gives;
a = 1.6119r
Formula for surface area of sphere is;
As = 4πr²
Also,formula for surface area of cube is; Ac = 6a²
Thus, since a = 1.6119r,
Then, Ac = 6(1.6119r)²
Ac = 15.5893r²
The formula for radiant power is;
Q' = eσT⁴A
Where;
e is emissivity
σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k
T is temperate in kelvin
A is Area
So, for the cube;
(Qc)' = eσ(Tc)⁴(Ac)
For the sphere;
(Qs)' = eσ(Ts)⁴(As)
We are told (Qc)' = (Qs)'
Thus;
eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)
eσ will cancel out to give;
(Tc)⁴(Ac) = (Ts)⁴(As)
Since we want to find the cube's temperature Tc,
(Tc)⁴ = [(Ts)⁴(As)]/Ac
Plugging in relevant figures, we have;
(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²
r² will cancel out to give;
(Tc)⁴ = [556⁴ × 4π]/15.5893
Tc = ∜([556⁴ × 4π]/15.5893)
Tc = 526.83 K
In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 35.6° and then moves down the incline with constant speed when the angle is reduced to 30.8°. From these data, determine the coefficients of static and kinetic friction for this experiment.
Answer:
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
Explanation:
The Free Body Diagram associated with the experiment is presented as attachment included below.
Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.
Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:
[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]
Where:
[tex]W[/tex] - Weight of the eraser, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]\theta[/tex] - Angle of the incline, measured in degrees.
The maximum allowable static friction force is:
[tex]f = \mu_{s} \cdot N[/tex]
Where:
[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
Likewise, the kinetic friction force is described by the following model:
[tex]f = \mu_{k} \cdot N[/tex]
Where:
[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:
[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]
[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]
But [tex]N = m\cdot g \cos \theta[/tex], so that:
[tex]\mu = \tan \theta[/tex]
Now, coefficients of static and kinetic friction are, respectively:
[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]
[tex]\mu_{s} \approx 0.716[/tex]
[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]
[tex]\mu_{k} \approx 0.596[/tex]
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
The self-referencing effect refers to ________.
Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new pipe which has a diameter half that of the first section of pipe. Determine the speed and pressure of the water in the new, reduced in size pipe.
Answer:
The velocity is [tex]v_2 = 6.8 \ m/s[/tex]
The pressure is [tex]P_2 = 204978 Pa[/tex]
Explanation:
From the question we are told that
The speed at which water is travelling through is [tex]v = 1.7 \ m/s[/tex]
The pressure is [tex]P_1 = 205 k Pa = 205 *10^{3} \ Pa[/tex]
The diameter of the new pipe is [tex]d = \frac{D}{2}[/tex]
Where D is the diameter of first pipe
According to the principal of continuity we have that
[tex]A_1 v_1 = A_2 v_2[/tex]
Now [tex]A_1[/tex] is the area of the first pipe which is mathematically represented as
[tex]A_1 = \pi \frac{D^2}{4}[/tex]
and [tex]A_2[/tex] is the area of the second pipe which is mathematically represented as
[tex]A_2 = \pi \frac{d^2}{4}[/tex]
Recall [tex]d = \frac{D}{2}[/tex]
[tex]A_2 = \pi \frac{[ D^2]}{4 *4}[/tex]
[tex]A_2 = \frac{A_1}{4}[/tex]
So [tex]A_1 v_1 = \frac{A_1}{4} v_2[/tex]
substituting value
[tex]1.7 = \frac{1}{4} * v_2[/tex]
[tex]v_2 = 4 * 1.7[/tex]
[tex]v_2 = 6.8 \ m/s[/tex]
According to Bernoulli's equation we have that
[tex]P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}[/tex]
substituting values
[tex]205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}[/tex]
[tex]P_2 = 204978 Pa[/tex]
One car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. During their periods of travel, the cars definitely had the same
Answer:
They had the same speed.
Explanation:
It won't be velocity, because velocity is a vector quantity. Speed is scalar.
Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.
What is Velocity?Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.
Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,
Distance = 40 meters
Time = 5 seconds
Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec
Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,
Distance = 64 meters
Time = 8 seconds
Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec
Hence, During their periods of travel, the cars definitely had the same velocity.
Learn more about Velocity here:
https://brainly.com/question/18084516
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Which of the following statements is true of a gas?
It has a fixed volume, but not a fixed shape
It has closely packed molecules
It can change into a liquid by adding heat
It takes the shape and size of a container
Answer:
it takes the shape and size of the container that it is in
Explanation:
Answer:
it takes the shape and size of a container
A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
[tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]
- We will plug in the given parameters in the equation of motion given above:
[tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2
uring a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? 26.7 N 16.7 N 13.3 N 107 N 40.0 N
Answer:
107 N, option d
Explanation:
Given that
mass of the ball, m = 0.2 kg
initial velocity of the ball, u = 20 m/s
final velocity of the ball, v = -12 m/s
time taken, Δt = 60 ms
Solving this question makes us remember "Impulse Theorem"
It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"
Mathematically, it is represented as
FΔt = m(v - u), where
F = the average force
Δt = time taken
m = mass of the ball
v = final velocity of the ball
u = initial velocity of the ball
From the question we were given, if we substitute the values in it, we have
F = ?
Δt = 60 ms = 0.06s
m = 0.2 kg
v = -12 m/s
u = 20 m/s
F = 0.2(-12 - 20) / 0.06
F = (0.2 * -32) / 0.06
F = -6.4 / 0.06
F = -106.7 N
Thus, the magnitude is 107 N
A mass of 5.00 kg pulls down vertically on a string that is wound around a rod of radius 0.100 m and negligible moment of inertia. The rod is fixed in the center of a disk. The disk has mass 125 kg and radius 0.2 m. They turn freely about a fixed axis through the center. What is the angular acceleration of the rod, in radians/s 2
Answer:
0.981 rad/sec^2
Explanation:
mass that pulls on string = 5 kg
weight due to mass = 5 x 9.81 = 49.05 N
radius of rod = 0.1 m
torque produced by this force on the rod = force x radius
torque = 49.05 x 0.1 = 4.905 N-m
mass of disk = 125 kg
radius of disk = 0.2 m
moment of inertia of the disk I = m[tex]r^{2}[/tex]
I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2
from the equation, T = Iα
where T is torque
I is moment of inertia
α is angular acceleration
imputing values,
4.905 = 5α
α = 4.905/5 = 0.981 rad/sec^2
