Final answer:
The volume of the gas at 408 K and 24 N/cm² will be 1.76 liters, calculated using the combined gas law that integrates Boyle's law and Charles's law for the given changes in pressure and temperature.
Explanation:
The volume of a gas is dependent on both its pressure and temperature, following Boyle's law and Charles's law, respectively. The relationship is such that volume varies inversely with pressure and directly with temperature (in kelvins) when the amount of gas is constant. To find the new volume when both the pressure and temperature change, we use the combined gas law, which integrates both Boyle's and Charles's laws:
V₁/T₁ * P₁ = V₂/T₂ * P₂
Given that at P₁ = 16 N/cm², T₁ = 340 K, and V₁= 2.2 liters, we can find V₂ when P₂ = 24 N/cm² and T₂ = 408 K using the formula:
V₂ = (V₁ * T₂/T₁) * (P₁/P₂)
Plugging in the values:
V₂ = (2.2 liters * 408 K / 340 K) * (16 N/cm² / 24 N/cm²)
V₂ = (2.2 liters * 1.2) * (2 / 3)
V₂ = 2.64 liters * (2 / 3)
V₂ = 1.76 liters
So, the volume of the gas when the temperature is 408 K and the pressure is 24 N/cm² would be 1.76 liters.
What causes differences in air pressure around the Earth?
Wind is driven due to the differences in air pressure around the Earth.
To find the answer, we need to know about the wind flow.
How does the wind flow?Due to the temperature difference at different places on the earth surface, we get difference in pressures at various regions.For keeping a constant pressure in all regions, the wind flows from high to low pressure region.Thus, we can conclude that wind is driven due to the differences in air pressure around the Earth.
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About how far does the S wave travel through Earth in 13 minutes?
2,000 km
4,000 km
6,000 km
8,000 km
An elastic conducting material is stretched into a circular loop of 14.8 cm radius. It is placed with its plane perpendicular to a uniform 0.814 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 88.4 cm/s. What emf is induced in volts in the loop at that instant
Answer: 0.666 V
Explanation:
Given
Radius of the loop, r = 14.8 cm = 0.148 m
Magnetic field present, B = 0.814 T
Rate of shrinking, dr/dt = 88.4 cm/s = 0.884 m/s
emf = dΦ/dt , where Φ = BA
emf = d(BA)/dt, where A = πr²
emf = d(Bπr²)/dt
if B is constant, then
emf = Bπ d(r²)/dt, on differentiating, we have,
emf = Bπ * 2r dr/dt
emf = 2πrB dr/dt, now if we substitute the values, we have
emf = 2 * 3.142 * 0.148 * 0.814 * 0.884
emf = 6.284 * 0.106
emf = 0.666 V
A block of iron quickly sinks in water, but ships constructed of iron float. A solid cube of iron 1.90 m on each side is made into sheets. From these sheets, to make a hollow cube that will not sink, what should the minimum length of the sides be? (density of iron is 7860 kg/m³)
Answer: 3.78 m
Explanation:
Volume = 1.9 * 1.9 * 1.9
Volume = 6.859 m³
Density of iron assumed to be 7870 kg/m³
Recall,
Density = mass / volume, so
Mass = density * volume
Mass = (7870 * 6.859)
Mass = 53980.33 kg
This mass we calculated, is the mass of water the new cube would have to displace, absolute minimum, so that it can float.
Now density of fresh water is assumed to be (1000 kg/m³).
Thus, the volume of water that will be displaced has to be
Volume = mass / density
Volume = (53980.33 / 1,000)
Volume = 53.98 m³.
Cube root of 53.88 = 3.78 m
Therefore, the minimum side length is 3.78 m
Answer:
3.781488032m or greater
Explanation:
General Rule of thumb is that any thing that is less dense than water will float in it.
So
density of water is 997Kg/m^3 and density of iron on the other hand is 7860kg/m^3.
so to make iron less dense it has to be in larger volume.
mass of the given cube is 53911.74Kg and we have contain it in volume such that density is less than water, mathematically this translates to .
[tex]\frac{53911.74Kg}{V}\leq \frac{997Kg}{m^3}[/tex] left side is density of iron and right side is density of water, solving v in this inequality gives us.
[tex]v \geq 54.0739m^3[/tex] which means each side is [tex]s \geq 3.7814880m[/tex]. so the minum is just greater than that.
Tia needs to produce a solenoid that has an inductance of 3.01 μ H 3.01 μH . She constructs the solenoid by uniformly winding 1.13 m 1.13 m of thin wire around a tube. How long, in centimeters, should the tube be?
Answer: 13 cm
Explanation:
Given
Inductance of the solenoid, L = 3.01•10⁻⁶ H
Width of the wire, x = 1.13 m
Length of the tube, z = ?
Now, we know that
L = μ₀N²A/z, where
N = x/2πr, making r subject of formula,
r = x/2πN
Also,
A = πr², on substituting for A, we have
A = πx²/4π²N²
Now finally, we substitute in the initial equation and solve
L = μ₀N²A/z
L = μ₀N²πx²/4π²N²z
L = μ₀x²/4z, making z subject of formula, we have
z = μ₀x²/4L
z = 4π*10⁻⁷ * 1.13² /4 * 3.01*10⁻⁶
z = (4π*10⁻⁷ * 1.2769) / (4 * 3.01*10⁻⁶)
z = 1.6*10^-6 / 1.2*10^-5
z = 0.13 m
Therefore, the length of the tube should be 13 cm
The length of the tube should be 469 centimeters.
To find the required length of the tube for Tia's solenoid, we can use the formula for the inductance of a solenoid:
[tex]\[ L = \frac{{\mu_0 \cdot N^2 \cdot A}}{l} \][/tex]
Where:
L = inductance of the solenoid (3.01 μH)
[tex]\( \mu_0 \)[/tex] = permeability of free space (4π × 10^-7 T*m/A)
N = number of turns of wire
A = cross-sectional area of the solenoid
[tex]\( l \)[/tex] = length of the solenoid
We're given [tex]\( L \)[/tex] and [tex]\( N \cdot l \)[/tex] (the total length of wire wound around the tube). We need to find [tex]\( l \)[/tex], the length of the tube.
First, let's rearrange the formula to solve for [tex]\( l \)[/tex]:
[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot A}}{{L}} \][/tex]
We know the cross-sectional area ( A ) can be represented as [tex]\( A = \pi r^2 \)[/tex], where ( r ) is the radius of the tube. Since the wire is uniformly wound, we can calculate ( A ) using the length of the wire and the number of turns.
[tex]\[ A = \frac{{\text{{Length of wire}}}}{{\text{{Number of turns}}}} \][/tex]
Given that the wire length is 1.13 m and there are [tex]\( N \)[/tex] turns, [tex]\( A = \frac{{1.13}}{{N}} \)[/tex].
Substituting this into the equation for [tex]\( l \)[/tex], we get:
[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot \left(\frac{{1.13}}{{N}}\right)}}{{L}} \][/tex]
[tex]\[ l = \frac{{\mu_0 \cdot 1.13}}{{L}} \][/tex]
Now, plug in the known values:
[tex]\[ l = \frac{{4\pi \times 10^{-7} \cdot 1.13}}{{3.01 \times 10^{-6}}} \][/tex]
[tex]\[ l = \frac{{4\pi \times 1.13}}{{3.01}} \][/tex]
[tex]\[ l = \frac{{4 \times 3.1416 \times 1.13}}{{3.01}} \][/tex]
[tex]\[ l = \frac{{14.1376}}{{3.01}} \][/tex]
[tex]\[ l = 4.69 \, \text{m} \][/tex]
Finally, convert this length to centimeters:
[tex]\[ l = 469 \, \text{cm} \][/tex]
So, the length of the tube should be 469 centimeters.
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration the gravitational pull of the moon?
Answer:
The magnitude of the acceleration of earth due to the gravitational pull of earth is a = Gm/r^2
Where r = the center to center distance between the earth and the moon,
m = mass of the moon, and,
G is the gravity constant.
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
[tex]a_e = \frac{Gm}{r^2}[/tex]
Explanation:
We assume that:
M to represent the mass of the earth
m to equally represent the mass of the moon
r should be the distance between the center of the earth to the center of the moon.
Then;
the expression for the gravitational force can be written as:
[tex]F = \frac{GMm}{r^2}[/tex]
Where [tex]a_e[/tex] is the acceleration produced by the earth; then:
[tex]F =M *a_e[/tex]
Then:
[tex]M*a_e = \frac{GMm}{r^2}[/tex]
[tex]a_e = \frac{GMm}{Mr^2}[/tex]
[tex]a_e = \frac{Gm}{r^2}[/tex]
Therefore, the magnitude of the acceleration of the earth due to the gravitational pull of the moon [tex]a_e = \frac{Gm}{r^2}[/tex]
An electron, moving south, enters a uniform magnetic field. Because of this uniform magnetic field, the electron curves upward. We can conclude that the magnetic field must have a component A. upward; B. downward; C. toward the north; D. toward the east; E. toward the south; F. toward the west.
Answer:
Towards the west
Explanation:
Magnetic force is the interaction between a moving charged particle and a magnetic field.
Magnetic force is given as
F = q (V × B)
Where F is the magnetic force
q is the charge
V is the velocity
B is the magnetic field
V×B means the cross product of the velocity and the magnetic field
NOTE:
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
So, if the electron is moving southward, then, it implies that the velocity of it motion is southward, so the electron is in the positive z-direction
Also, the electron is curved upward due to the magnetic field, this implies that the force field is directed up in the positive y direction.
Then,
V = V•k
F = F•j
Then, apply the theorem
F •j = q ( V•k × B•x)
Let x be the unknown
From vector k×i =j.
This shows that x = i
Then, the magnetic field point in the direction of positive x axis, which is towards the west
You can as well use the Fleming right hand rule
The thumb represent force
The index finger represent velocity
The middle finger represent field
If a current flowing through a lightbulb is 0.75 ampere and the voltage difference across the lightbulb is 120 volts, how much resistance does the light bulb have
Answer: [tex]160 \Omega[/tex]
Explanation:
According to Ohm's law:
[tex]V=R.I[/tex]
Where:
[tex]V=120 V[/tex] is the voltage difference across the light bulb
[tex]R[/tex] is the resistance of the light bulb (the value we want to find)
[tex]I=0.75 A[/tex] is the electric current
Isolating [tex]R[/tex]:
[tex]R=\frac{V}{I}[/tex]
[tex]R=\frac{120 V}{0.75 A}[/tex]
Finally:
[tex]R=160 \Omega[/tex] This is the resistance of the light bulb
Final answer:
To calculate the resistance of a lightbulb with a current of 0.75 amperes and a voltage difference of 120 volts, use Ohm's law. The formula is R = V / I, which yields a resistance of 160 ohms for the lightbulb.
Explanation:
Calculating the Resistance of a Lightbulb
The student has asked how to calculate the resistance of a lightbulb when a current of 0.75 amperes flows through it and the voltage difference across it is 120 volts. To find the resistance, we can use Ohm's law, which states that the resistance (R) of a circuit is equal to the voltage (V) across it divided by the current (I) flowing through it, which can be written as:
R = V / I
Plugging in the given values, we get:
R = 120 V / 0.75 A
R = 160 ohms
Therefore, the resistance of the lightbulb is 160 ohms.
A ball is thrown upward from a height of 256 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Find the attachments for complete solution
The function giving the height of the ball at time t is s(t) = 96t - 16t^2 + 256. The ball takes 6 seconds to reach the ground and achieves a maximum height of 400 feet.
Explanation:The physics problem you've posed can be approached by first understanding that the height of the ball (s(t)) at time t can be found by integrating the velocity function, due to v(t) being the derivative of s(t). So, integrating v(t)=96-32t with respect to t gives s(t) = 96t - 16t^2 + 256, where 256 is the constant of integration corresponding to the initial height from the ground.
To find out how long will the ball take to reach the ground, we solve the function s(t)=0, yielding t = 6 seconds as the answer, as that's when the ball hits the ground. Now, for the maximum height reached by the ball, it's where the ball has a velocity of zero before it begins its descent, i.e. v(t)=0. Solving this gives t=3 seconds, substituting this back into the s(t) equation gives the maximum reached height as 400 feet.
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A circular swimming pool has a diameter of 18 m. The circular side of the pool is 3 m high, and the depth of the water is 1.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m 3 kg/m^3.)How much work (in Joules) is required to: i. pump all of the water over the side? ii. pump all of the water out of an outlet 2m over the side?
To calculate the work required to pump water out of the swimming pool, physics principles such as the formula for work, volume of a cylinder, and weight of water are used to derive the necessary values for the complete calculation.
Explanation:The question requires the use of physics concepts to calculate work done in pumping water out of a swimming pool. Since we are dealing with the force of gravity, density of water, height to which water is raised, and volume of water, we must use the formula for work done (work = force x distance) alongside formulas for the volume of a cylinder (Volume = πr^2h) and the weight of the water (weight = mass x gravity).
To solve the problem, first calculate the volume of water in the pool using the volume formula for a cylinder, taking into account the water's depth. Then, calculate the mass by multiplying the volume by the density of water. The weight of the water is found by multiplying the mass by the acceleration due to gravity. Finally, calculate the work done by multiplying the weight of the water by the height to which it needs to be lifted, considering both scenarios: pumping over the side and pumping through an outlet 2 meters over the side.
Final answer:
Calculation of work required to pump water out of a circular swimming pool at different heights.
Explanation:
The work required to pump all the water over the side of the pool is:
Work = mgh = density x volume x gravity x height = 1000 kg/m³ x (pi x (9m)² x 1.5m) x 9.8 m/s² x 3m = 1222540 J
For pumping all water out of the outlet 2m over the side:
Work = mgh = density x volume x gravity x height = 1000 kg/m³ x (pi x (9m)²x 1.5m) x 9.8 m/s² x 5m = 2037567 J
PLEASE HELP! It’s urgent... and please show your work!!
1. Calculate the following, and express the answer in scientific notation with
the correct number of significant figures:
(0.82 +0.042)(4.4 x 10°)
a) 3.8 x 10
b) 3.78 x 10
c) 3.784 x 103
d) 3784
The correct result of the given operation expressed in scientific notation and with the appropriate number of significant figures should be 3.4 x 10^3. However, there are no matching options provided. A possible mistake might exist in the question.
Explanation:To solve the problem, you should first add the two numbers in the brackets together. This gives us 0.82 + 0.042 = 0.862. Next, we need to multiply this result by the number outside the brackets, which is 4.4 x 103. Thus, our equation becomes 0.862 x 4.4 x 103 = 3.3928 x 103.
The result needs to be expressed in scientific notation and with the correct number of significant figures. The least precise number in our calculation is 4.4 (which has two significant figures), so our final answer should also have two significant figures. Therefore, we round 3.3928 x 103 to 3.4 x 103.
However, none of the options provided matches this result. Please double-check the question as there might be a typo.
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First, add 0.82 + 0.042 = 0.862. Second, multiply the sum 0.862 x 4.4 = 3.7928. Finally, round the result to two significant figures in scientific notation giving the answer as 3.8 x 10 or just 3.8.
Explanation:To solve this problem, first, we need to add the numbers in the parentheses, then multiply the sum by 4.4 x 10⁰, which is just 4.4 since any number raised to the power of zero equals one.
Step 1: Add 0.82 + 0.042 = 0.862
Step 2: Multiply 0.862 x 4.4 = 3.7928
Lastly, we need to express the answer in scientific notation with the correct number of significant figures. As the question gives the numbers 0.82 and 0.042 with two and three significant figures respectively, we should express our final answer to two significant figures, as you always round to the fewest number of significant figures.
So, round 3.7928 to two significant figures, gives you 3.8 x 10⁰, which can alternatively be written as just 3.8, making the correct answer (a) 3.8 x 10.
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A toroidal inductor has a circular cross-section of radius a a . The toroid has N turns and radius R. The toroid is narrow ( a≪R ), so the magnetic field inside the toroid can be considered to be uniform in magnitude. What is the self-inductance L of the toroid?
Answer:
L=N/I*magnetic flux
=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2
=mu(0)N^2a^2/2R
Explanation:
The self-inductance L of the toroid is mu(0)N^2a^2/2R.
Calculation of the self-inductance L:Since
The toroidal inductor has a circular cross-section of radius a. The toroid has N turns and radius R.
The toroid is narrow ( a≪R )
So,
L=N/I*magnetic flux
=N/I*BA=N*mu(0)NIA/2pi*R=mu(0)N^2/2pi*R)*Pi*a^2
=mu(0)N^2a^2/2R
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Does the box contain positive charge, negative charge, or no charge? Does the box contain positive charge, negative charge, or no charge? Positive charge, since there is net electric flux passing outward through the surface of the closed box, it contains positive charge. Negative charge, since there is net electric flux passing outward through the surface of the closed box, it contains negative charge. Positive charge, since there is net electric flux passing intward through the surface of the closed box, it contains positive charge. Negative charge, since there is net electric flux passing inward through the surface of the closed box, it contains negative charge. No charge, because the flux into the bos is canceled by the flux out of it.
Due to the net electric flux that is moving outward through the closed box's surface, the box has a negative charge.
Given:-
The electric field is constant over all the faces.
From Gauss law:
[tex]\int \vec E \cdot d\vec A = \dfrac{Q}{e_o}[/tex]
[tex]\vec E =[/tex] An electric field throughout the cube.
[tex]d \vec A =[/tex] Elemental area of the cube surface.
[tex]Q =[/tex] Electric charge around the cube.
[tex]e_o =[/tex] Electric constant.
[tex]\dfrac{Q}{e_o} = (-15-20-15+20+15+10)\\ \\ = -5A[/tex]
According to Gauss's law, the box contains a negative charge.
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The box's charge type (being positive, negative, or neutral) is determined by the net electric flux's direction. Outward flux signifies positive charge, whilst inward flux indicates a negative charge. Boxes with no net electric flux have no charge.
Explanation:The determination of whether a closed box contains a positive charge, negative charge, or no charge is dependent on the direction of the net electric flux. If there is net electric flux passing outward through the surface of the box, this indicates the box contains a positive charge. Conversely, if the net electric flux is passing inward through the surface, the box contains a negative charge. Lastly, if there is no net electric flux (flux in cancelled by flux out), it suggests there is no charge in the box.
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Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their mass ratios were 1:4 and their respective speeds as they approached the intersection were both 13 m/s, find the magnitude and direction of the final velocity of the wreck.
Answer with Explanation:
Let mass of one vehicle =m
Mass of other vehicle=m'
[tex]\frac{m}{m'}=\frac{1}{4}[/tex]
[tex]m'=4m[/tex]
Velocity of one vehicle=[tex]v=13 im/s[/tex]
Velocity of other vehicle=[tex]v'=13jm/s[/tex]
In x- direction
By law of conservation of momentum
[tex]mv+0=(m+m')V_x[/tex]
[tex]13m=(m+4m)V_x[/tex]
[tex]13m=5mV_x[/tex]
[tex]V_x=\frac{13}{5}[/tex]
In y- direction
By law of conservation of momentum
[tex]0+m'v'=(m+m')V_y[/tex]
[tex]4m(13)=5mV_y[/tex]
[tex]V_y=\frac{52m}{5m}=\frac{52}{5}[/tex]
Magnitude of velocity of the wreck,V=[tex]\sqrt{V^2_x+V^2_y}=\sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}=10.72 m/s[/tex]
Direction:[tex]\theta=tan^{-1}(\frac{V_y}{V_x})[/tex]
[tex]\theta=tan^{-1}(\frac{\frac{52}{5}}{\frac{13}{5}})=75.96^{\circ}[/tex]
A physical pendulum consists of a vertical board of of mass 6.12 kg, length 191 cm, and width 12 cm hanging from a horizontal, frictionless axle. A bullet of mass 143 g and a purely horizontal speed v impacts the pendulum at the bottom edge of the board. The board then makes a complete circle.
(a) If the bullet passed through the board, reducing its speed by a factor of 1/3, what is the minimum speed of the bullet?
(b) If the bullet embedded itself in the board, what is the minimum speed of the bullet?
Note (hint from instructor): It appears you may use Conservation of Energy to find angular speed. For potential energy: use center of mass. For kinetic energy, use impulse. Then, to find initial speed, use conservation of angular momentum.
This question involves applying the principles of Conservation of Energy and Angular Momentum to calculate the minimum speed of a bullet impacting a physical pendulum. The solution requires us to consider different scenarios: when the bullet passes through and when it embeds itself into the board.
Explanation:This problem involves the concepts of Conservation of Energy and Conservation of Angular Momentum.
(a) If the bullet passes through the board, we consider that the board moves by gaining kinetic energy (KE) from the bullet’s impact. The bullet's final speed is (1/3)*v, so the KE lost by the bullet will be (4/9)*mv². This energy is used for the board to perform a complete circle, which means it acquires a potential energy of (2*M*g*H), where H is the height when the board makes a complete circle (H=L/2). Equating these energies (4/9)*mv²-2*M*g*H = 0, we solve for v.
(b) If the bullet embeds itself, the system's final angular momentum equals its initial, which gives (1/2*M*L)v = (M+m)Lω. Then we equate the initial kinetic energy and potential energy during the complete circle, getting (1/2)(M+m)L²ω² = (M+m)gL. Solving for v will give us the minimum speed.
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the triceps muscle in the back of the upper arm extends the forearm. this muscle in a professional boxer exerts a force of 2.00×103 n with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 121 rad/s2. what is the moment of inertia of the boxer's forearm?
Answer: 0.512 kgm²
Explanation:
Given
Force, F = 2*10^3 N
Angular acceleration, α = 121 rad/s²
Lever arm, r(⊥) = 3.1 cm = 3.1*10^-2 m
τ = r(⊥) * F
Also,
τ = Iα
Using the first equation, we have
τ = r(⊥) * F
τ = 0.031 * 2*10^3
τ = 62 Nm
Now we calculate for the inertia using the second equation
τ = Iα, making I subject of formula, we have
I = τ / α, on substituting, we have
I = 62 / 121
I = 0.512 kgm²
Thus, the moment of inertia of the boxers forearm is 0.512 kgm²
Suppose you have Avogadro's number of mini marshmallows and use them to cover the state of South Dakota which has a land area of 7.588 × 10 4 mi 2 . Each mini marshmallow has a diameter of 0.635 cm and a height of 2.54 cm. Assuming the marshmallows are packed together so there is no space between them, to what height above the surface, in kilometers, will the mini marshmallows extend?
Answer: 2.44*10^3 km
Explanation:
NOTE: We would be solving this question in "cm" instead of the usual "m"
1 mi² = 2.59 km²
2.59 km² = 2.59*10^10 cm²
Given, area of South Dakota
A = 7.588*10^4 mi²
A = 7.588*10^4 * 2.59*10^10
A = 1.97*10^15 cm² is the area of South Dakota
S = πd²/4
S = 3.142 * 0.635² / 4
S = 3.142 * 0.4/4
S = 3.142 * 0.1
S = 0.3142 cm² is the area of 1 marshmallow
1.97*10^15 / 0.3142 = 6.27*10^15, thus is the number of marshmallows in one single layer to cover the state area
6.02*10^23 / 6.27*10^15 = 9.6*10^7, this is the number of layers
9.6*10^7 * 2.54 = 2.44*10^8, this is the height in cm
Height in km is 2.44*10^3 km
a celestial body moving in an ellipical orbit around a star
Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.
A strip of copper 130 µm thick and 4.40 mm wide is placed in a uniform magnetic field of magnitude B = 0.79 T, that is perpendicular to the strip. A current i = 26 A is then sent through the strip such that a Hall potential difference V appears across the width. Calculate V. (The number of charge carriers per unit volume for copper is 8.47 × 1028 electrons/m3.)
Answer:
V = 1.1658 × [tex]10^{-5}[/tex] V
Explanation:
given data
strip of copper thick = 130 µm
strip of copper wide = 4.40 mm
uniform magnetic field of magnitude B = 0.79 T
current i = 26 A
number of charge carriers per unit volume = 8.47 × [tex]10^{28}[/tex] electrons/m³
solution
we know that number density is express as
n = \frac{Bi}{Vle} ...............1
B is uniform magnetic field and i is current and V is hall potential difference and l is thickness and e is electron charge 1.6 × [tex]10^{-19}[/tex] C
so V will be as
V = \frac{iB}{nle} .....................2
so put here value and we get V
V = [tex]\frac{26 \times 0.79}{8.47\times 10^{28}\times 130\times10^{-6}\times1.6 \times10^{-19}}[/tex]
V = 1.1658 × [tex]10^{-5}[/tex] V
A 325-turn circular-loop coil 9.4 cm in diameter is initially aligned so that its axis is parallel to Earth’s magnetic field. In 2.48 ms the coil is flipped so that its axis is perpendicular to Earth’s magnetic field. If an average voltage of 0.17 V is thereby induced in the coil, what is the value of Earth’s magnetic field at that location? Answer in units of µT.
Answer:
[tex]B = 187\ \mu T[/tex]
Explanation:
Given,
Number of turns, N = 325
Diameter of coil, d = 9.4 cm
time,t = 2.48 m s
average voltage,ε = 0.17 V
Earth magnetic field, B = ?
We know that
[tex]\epsilon = \dfrac{NBA}{t}[/tex]
[tex]0.17= \dfrac{325\times B \times \pi (0.047)^2}{2.48\times 10^{-3}}[/tex]
[tex]B = 187 \times 10^{-6}\ T[/tex]
[tex]B = 187\ \mu T[/tex]
Magnetic field of the earth is equal to [tex]B = 187\ \mu T[/tex]
A circular loop of wire with area A Bz of B By of B Determine the component Bz B . Determine the component By Bx of B? k . Determine the component Bx j , and k^ i , j^ A , i^ I , A U = - Mu - B is negative. The magnitude of the magnetic field is B0=15D/IA Determine the vector magnetic moment of the current loop. Express your answer in terms of the variables I D is a positive constant, and for this orientation of the loop the magnetic potential energy, where B vector is given by -z -direction toward the origin, a current I is circulating clockwise around the loop. The torque produced by an external magnetic field B? z -axis looking in the xy -plane. As viewed along the z A lies in the xy
Find the given attachment
A particle with a charge of 34.6 $\mu C$ moves with a speed of 68.4 m/s in the positive $x$ direction. The magnetic field in this region of space has a component of 0.466 T in the positive $y$ direction, and a component of 0.876 T in the positive $z$ direction. What is the magnitude of the magnetic force on the particle?
Answer:
The magnitude of the magnetic force on the particle is 67.86 N.
Explanation:
Given that,
Charge, [tex]q=34.6\ \mu C=34.6\times 10^{-6}\ C[/tex]
Speed of particle, v = 68.4 m/s in +x direction
Magnetic field, [tex]B=(0.466 j+0.876 k)\ T[/tex]
We need to find the magnitude of the magnetic force on the particle. The magnetic force is given by :
[tex]F=q(v\times B)\\\\F=34.6\times 10^{-6}(68.4i+0+0\times (0+0.466j+0.876 k))\\\\F=\begin{pmatrix}0&-59.9184&31.8744\end{pmatrix}\\\\F=(-59.9184j+31.8744k)\ N[/tex]
The magnitude of magnetic force on the particle is :
[tex]|F|=\sqrt{(-59.9184)^2+(31.8744)^2} \\\\|F|=67.86\ N[/tex]
So, the magnitude of the magnetic force on the particle is 67.86 N.
The left end of a long glass rod 7.00 cmcm in diameter has a convex hemispherical surface 3.50 cmcm in radius. The refractive index of the glass is 1.60. The glass rod is immersed in oil (nn = 1.45). An object placed to the left of the rod on the rod's axis is to be imaged 1.15 mm inside the rod.
How far from the left end of the rod must the object be located to form the image?
Answer:
n1/p + n2/q = (n2-n1)/R
1.45/p + 1.6/1.15 = (1.6 - 1.45)/0.035 (m)
1.45/p + 1.39 = 4.28
p = 0.50 m (50 cm)
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is connected to a battery producing a current of 4.0 A in the wire. (a) What is the magnitude of the magnetic dipole mo- ment of this device? (b) At what axial distance d will the mag- netic field have the magnitude 5.0 mT (approximately one-tenth that of Earth’s magnetic field)?
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
A cylinder with a piston contains 0.300 molmol of oxygen at 2.50×105 PaPa and 350 KK . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.
Part A
Find the work done by the gas during the initial expansion.
--
Part B
Find the heat added to the gas during the initial expansion.
--
Part C
Find internal-energy change of the gas during the initial expansion.
--
Part D
Find the work done during the final cooling;
--
Part E
Find the heat added during the final cooling;
--
Part F
Find the internal-energy change during the final cooling;
--
Part G
Find the internal-energy change during the isothermal compression.
Answer:
Explanation:
find the solution below
The work, heat, and internal energy changes are calculated for different stages of the gas expansion and cooling processes in a cylinder with a piston containing oxygen as an ideal gas.
Explanation:Part A: The work done by the gas during the initial expansion can be calculated using the formula:
Work = Pressure * Change in Volume
Since the expansion is isobaric (constant pressure) and the volume doubles, the change in volume is 2 times the initial volume. Therefore, the work done is 2 times the initial pressure times the initial volume.
Part B: The heat added to the gas during the initial expansion can be calculated using the formula:
Heat = Pressure * Change in Volume, since the expansion is isobaric.
Part C: The internal energy change of the gas during the initial expansion is equal to the heat added minus the work done.
Part D: The work done during the final cooling is zero because the process is isochoric (constant volume) and no work is done.
Part E: The heat added during the final cooling can be calculated using the equation:
Heat = Change in Internal Energy, since no work is done.
Part F: The internal energy change during the final cooling is equal to the negative of the heat added.
Part G: The internal energy change during the isothermal compression is also equal to the negative of the heat added, as no work is done during an isothermal process.
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he electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are R1 and R2 in parallel and in series - and you can't separate these two resistor combinations. You separately connect each resistor network to a battery with emf 57.0 V and negligible internal resistance and measure the power P supplied by the battery in both cases. For the series combination, P = 48.0 W; for the parallel combination, P = 256 W. You are told that R1>R2. Calculate R1.
Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
R₂ = 16.92 Ω
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
R₁ = 50.77 Ω
Final answer:
To calculate R1, first, find the total resistance for series (R1 + R2) and parallel (1/R1 + 1/R2) connections using the power formula P = V^2 / R, with provided power values for each case. Then solve the simultaneous equations.
Explanation:
The problem can be solved by using the formulas for resistances in series and parallel, as well as the formula for the power supplied by the battery.
In series, the resistances add up: Rtotal = R1 + R2. The power provided by the battery to the series circuit is given by P = V2 / Rtotal, where P is 48.0 W and V is 57.0 V. Rearranging to solve for Rtotal, we get Rtotal = V2 / P.
For parallel resistances, we have 1/Rtotal = 1/R1 + 1/R2. The power in the parallel case is P = V2 / Rtotal. With V and P given as 57.0 V and 256 W, respectively, we can solve for Rtotal.
With both Rtotal values calculated, we can set up two equations with two unknowns (R1 and R2), knowing that R1 > R2. Solving these simultaneous equations gives us the value of R1.
Suppose you are in a spaceship traveling at 99% of the speed of light past a long, narrow space station. Your direction of travel is parallel to the length of the station. If you measure lengths of objects on the station and also how time is passing on the station, what results will you get?
A) Lengths will appear shorter and time will appear to pass slower.
B) Lengths will appear shorter and time will appear to pass faster.
C) Lengths will appear shorter and time will appear to pass faster.
Answer:
B) Lengths will appear shorter and time will appear to pass faster.
Explanation:
This is in line with the laws of relativity.
Answer:
A. Lengths will appear shorter and time will appear to pass slower.
Explanation:
From theory of relativity, we know that:
L₀ = length of object, measured in stationary frame of reference
L = length of object measured from a frame moving with respect to object, called ‘relativistic length’
v = relativistic speed between observer and the object
c = speed of light
then,
L = L₀ √(1-v²/c² )
Hence, the length of the object decreases with the increase in its relativistic speed "v"
t₀ = time measured by clock at rest with respect to event.
t = time measured by clock in motion relative to the event
v = relativistic speed between observer and the object
c = speed of light
then,
t = t₀/√(1-v²/c² )
Hence, the time increases with the increase in in relativistic speed and as a result it appears to pass slower.
Hence, the correct option is:
A. Lengths will appear shorter and time will appear to pass slower.
3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed and acceleration of her swing. She has a device that measures the linear velocity of the bat during the swing. Her bat is 0.85 m long. On her first swing, the linear velocity of the bat just prior to the swing is 0 m/s, and 0.15 s later, just prior to ball contact is 16 m/s. What is the average angular acceleration of the end of her bat during the swing?
Answer:
The average angular acceleration is [tex]\alpha =125.487 rad /s^2[/tex]
Explanation:
From the question we are told that
From the question we are told that
The length of the bat is [tex]l = 0.85m[/tex] \
The initial linear velocity is [tex]u = 0 m/s[/tex]
The time is [tex]t = 0.15s[/tex]
The velocity at t is [tex]v = 16 m/s[/tex]
Generally average angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_o}{t}[/tex]
Where [tex]w_f[/tex] is the finial angular velocity which is mathematically evaluated as
[tex]w_f = \frac{v}{l}[/tex]
[tex]w_f = \frac{16}{0.85}[/tex]
[tex]= 18.823 rad/s[/tex]
and [tex]w_o[/tex] is the initial angular velocity which is zero since initial linear velocity is zero
So
[tex]\alpha = \frac{18.823 - 0}{0.15}[/tex]
[tex]\alpha =125.487 rad /s^2[/tex]
g If the primary coil of wire on a transformer is kept the same and the number of turns of wire on the secondary is increased, how will this affect the voltage observed at the secondary? There is no voltage measured on the secondary coil since it is only connected to a resistor and not connected to a battery. The voltage will stay the same. The voltage will increase. The voltage will decrease.
Answer:
The voltage will increase.
Explanation:
Increason the number of coil in the secondary will increase the voltage in a transformer.
A convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 2. Light incident parallel to the central axis is focused at a point: Group of answer choices 3 cm from the surface 6 cm from the surface 24 cm from the surface 12 cm from the surface 18 cm from the surface
Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 2. Light incident parallel to the central axis is focused at a point 24cm from the surface
What is refraction?When a light ray passes through from one medium to another medium ,due to the difference in medium index, the light deflect from its original path .
This deflection from its path is called refraction and The phenomenon of refraction can be seen in sound and water waves.
Cornea reflect the light before light reaches to the retina. Change in the speed of light when it travels from one medium to other medium then due to this there is change in the path of light, this phenomenon is known as refraction of light.
Equation for refraction
n2/v- n1/u = n2-n1/R
= u = infinity ( parallel)
2/v= (2-1)/R
V = 2R
V= 24cm
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