First: CT Scan
Second: Fluoroscopy
Explanation:
Both correct
A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.
Explanation:In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.
In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.
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horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, walk clockwise around the platform along its edge at the speed of "1.05" m/s with respect to the platform. Your 20.5 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 18.5 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 90.7 kg and radius 1.91 m. Calculate the total angular momentum of the system
Answer:
The total angular momentum is 292.59 kg.m/s
Explanation:
Given that :
Rotation of the horizontal circular platform [tex]\omega[/tex] = 0.919 rad/s
mass of the platform (m) = 90.7 kg
radius (R) = 1.91 m
mass of the poodle [tex]m_p[/tex] = 20.5 kg
Your mass [tex]m'[/tex] = 73.5 kg
speed v = 1.05 m/s with respect to the platform
[tex]V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}[/tex]
r = 0.955
Mass of the mutt [tex]m_m[/tex] = 18.5 kg
[tex]r' = \frac{3}{4} \ R[/tex]
Your angular momentum is calculated as:
Your angular velocity relative to the platform is [tex]\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s[/tex]
[tex]Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s[/tex]
[tex]I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2[/tex]
[tex]L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s[/tex]
For poodle :
[tex]Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s[/tex]
Actual [tex]\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s[/tex]
[tex]I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2[/tex]
[tex]L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s[/tex]
[tex]I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2[/tex]
[tex]L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s[/tex]
Disk [tex]I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2[/tex]
[tex]L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s[/tex]
Total angular momentum of system is:
L = [tex]L_D +L_Y+L_P+L_M[/tex]
= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s
= 292.59 kg.m/s
The drawing shows three identical rods (A, B, and C) moving in different planes. A constant magnetic field of magnitude 4.50 T is directed along the +y axis. The length of each rod is L = 1.3 m, and the speeds are the same, vA = vB = vC = 2.6 m/s. For each rod, find the magnitude of the motional emf, and indicate which end (1 or 2) of the rod is positive. rod A V ---Select--- End 1 is positive. End 2 is positive. No emf in rod. rod B V ---Select--- End 1 is positive. End 2 is positive. No emf in rod. rod C V ---Select--- End 1 is positive. End 2 is positive. No emf in rod.
Answer:
A)
The emf is zero because the velocity of the rod is parallel to the direction of the magnetic field, so the charges experience no force.
B)
The emf is vBL
= (2.6 m/s)(4.50 T)(1.3 m)
= 15.21 V.
The positive end is end 2.
C)
The emf is zero because the magnetic force on each charge is directed perpendicular to the length of the rod.
2. In a series circuit, all resistors have identical currents.
a) What is the relationship between the power and resistance of these resistors?
b) In a parallel circuit, all resistors have identical voltages. What is the relationship
between the power and resistance of these resistors?
Answer:
Explanation:
In series connection of resistors, same current flows in the circuit.
Power dissipated by a resistor is
P = i²R
And since same current flows in them, it implies that the power is directly proportional to Resistance, so the higher the resistance of the resistor the higher the power dissipated in the series connection. Also, the lower the resistance, the lower the power dissipated.
2. In parallel connection, same voltage is applied across the resistor.
So, power dissipated by each resistor is
P = V² / R
So, since the same voltage is applied across parallel connection, then, power dissipated in each resistor is inversely proportional to the resistance.,
So, the higher the resistance, the lower the power and the lower the resistance, the higher the power.
In what direction must a force be applied so that the forces on the 1 kg object are balanced
Answer:
towards the object
Explanation:
Final answer:
To balance the forces on the 1 kg object, a force should be applied in the downward direction.
Explanation:
The forces on the 1 kg object can be balanced by applying a force in the opposite direction to the net force acting on the object. In this case, the net force is the sum of the weight of the object and the tension in the string. Since the weight acts downward and the tension in the string acts upward, the force should be applied in the downward direction to balance the forces on the 1 kg object.
In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)
Answer:
v(t) = 21.3t
v(t) = 5.3t
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
Explanation:
When no sliding friction and no air resistance occurs:
[tex]m\frac{dv}{dt} = mgsin \theta[/tex]
where;
[tex]\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}[/tex]
Taking m = 3 ; the differential equation is:
[tex]3 \frac{dv}{dt}= 128*\frac{1}{2}[/tex]
[tex]3 \frac{dv}{dt}= 64[/tex]
[tex]\frac{dv}{dt}= 21.3[/tex]
By Integration;
[tex]v(t) = 21.3 t + C[/tex]
since v(0) = 0 ; Then C = 0
v(t) = 21.3t
ii)
When there is sliding friction but no air resistance ;
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta[/tex]
Taking m =3 ; the differential equation is;
[tex]3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}[/tex]
[tex]\frac{dv}{dt}= 5.3[/tex]
By integration; we have ;
v(t) = 5.3t
iii)
To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv[/tex]
The differential equation is :
= [tex]3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v[/tex]
= [tex]3 \frac{dv}{dt}=16 -\frac{1}{3}v[/tex]
By integration
[tex]v(t) = 48 + Ce ^{\frac{t}{9}[/tex]
Since; V(0) = 0 ; Then C = -48
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
An engineer is designing a contact lens. The material has is an index of refraction of 1.55. In order to yield the prescribed focal length, the engineer specifies the following dimensions: inner radius of curvature = +2.42 cm outer radius of curvature = +1.98 cm where the inner radius of curvature describes the surface that touches the eye, and the outer radius of curvature describes the surface that first interacts with incoming light. What is the focal length of this contact lens (in cm)?
Answer:20 cm
Explanation:
Given
Refractive index of material [tex]n=1.55[/tex]
Outer radius [tex]R_1=1.98\ cm[/tex]
Inner radius [tex]R_2=2.42\ cm[/tex]
using lens maker formula
[tex]\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})[/tex]
[tex]\frac{1}{f}=(1.55-1)(\frac{1}{1.98}-\frac{1}{2.42})[/tex]
[tex]f=\frac{10.895}{0.55}[/tex]
[tex]f=19.81\approx 20\ cm[/tex]
A wheel rotating with a constant angular acceleration turns through 22 revolutions during a 5 s time interval. Its angular velocity at the end of this interval is 12 rad/s. What is the angular acceleration of the wheel? Note that the initial angular velocity is not zero. Answer in units of rad/s 2 .
Answer:
0.52rad/s^2
Explanation:
To find the angular acceleration you use the following formula:
[tex]\omega^2=\omega_o^2+2\alpha\theta[/tex] (1)
w: final angular velocity
wo: initial angular velocity
θ: revolutions
α: angular acceleration
you replace the values of the parameters in (1) and calculate α:
[tex]\alpha=\frac{\omega^2-\omega_o^2}{2\theta}[/tex]
you use that θ=22 rev = 22(2π) = 44π
[tex]\alpha=\frac{(12rad/s)^2-(0rad/s)^2}{2(44\pi)}=0.52\frac{rad}{s^2}[/tex]
hence, the angular acceñeration is 0.52rad/s^2
2. Categorize each statement as true or false. A cylindrical capacitor is essentially a parallel-plate capacitor rolled into a tube.The dielectric constant indicates the distance by which the two plates of a capacitor are separated.The charge on a capacitor increases quickly at first, then much more slowly as the capacitor charges.The voltage across a capacitor in an RC circuit increases linearly during charging.One of the principle purposes of a capacitor is to store electric potential energy.A capacitor charges rapidly when connected to an RC circuit with a battery. True False
The true statement is A. A cylindrical capacitor is a parallel-plate capacitor rolled into a tube, C. The charge on a capacitor increases quickly at first, then much more slowly as the capacitor charges E. One of the principal purposes of a capacitor is to store electric potential energy. F. A capacitor charges rapidly when connected to an RC circuit with a battery and the false statement are B. The dielectric constant indicates that the distance by which the two plates of a capacitor are separated and D. The voltage across a capacitor in an RC circuit increases linearly during charging.
Let's look at each statement one by one to categorize them as true or false.
a. True. A cylindrical capacitor can be thought of as a parallel-plate capacitor with the plates rolled into cylindrical shapes.
b. False. The dielectric constant is a measure of a material's ability to increase the capacitance of a capacitor, not a measure of the distance between the plates.
c. True. The charge on a capacitor follows an exponential curve, increasing rapidly at first and then more slowly as it approaches its maximum charge.
d. False. The voltage increases exponentially, not linearly, when charging a capacitor in an RC circuit.
e. True. Capacitors store electric potential energy in the electric field between their plates.
f. True. Initially, the capacitor in an RC circuit charges quickly, but the rate of charging decreases over time as it gets closer to full charge.
A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules
Answer:2250J
Explanation:
mass(m)=20kg
velocity(v)=15m/s
Kinetic energy=(m x v^2)/2
Kinetic energy =(20 x 15^2)/2
Kinetic energy =(20x15x15)/2
Kinetic energy=4500/2
Kinetic energy=2250J
The kinetic energy of the ball will be 2250 joules.
We have a 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second.
We have to determine its kinetic energy is joules.
What is the formula to calculate the kinetic energy of the body of mass 'm' moving with velocity 'v' ?The kinetic energy of the body is as follows -
K.E. = [tex]\frac{1}{2} mv^{2}[/tex]
According to the question -
Mass of ball = 20 kg
Velocity of ball = 15 m/s
Substituting the values in the above formula, we get -
K.E. = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2} \times 20\times 15\times 15[/tex] = 225 x 10 = 2250 joules.
Hence, the kinetic energy of the ball will be 2250 joules.
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A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the point source the z axis. When a sheet of cardboard in which a square hole 0.020 m on a side has been cut is placed between the point source and the screen, 0.50 m from the point source with the hole centered on the z axis, a bright square shows up on the screen. If, instead, a second sheet of cardboard with a similar square hole is placed between the point source and screen, 0.25 m from the point source with the hole centered on the z axis, the bright square it casts on the screen is identical to the bright square from the first sheet. What is the length of the side of the hole in this sheet?
Answer:
The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]
Explanation:
From the question we are told that
The distance of the point source from the screen is [tex]d = 1.0 m[/tex]
The length of a side of the first square hole is [tex]L_1 = 0.020 \ m[/tex]
The distance of the cardboard from the point source is [tex]D_1 = 0.50\ m[/tex]
The distance of the second cardboard from the point source is [tex]D_2 = 0.25 \ m[/tex]
Let take the [tex]\alpha_{max }[/tex] as the angle at which the light is passing through the edges of the cardboards square hole
Since the bright square casted on the screen by both square holes on the individual cardboards are then it means that
[tex]\alpha_{max} __{1}} = \alpha_{max} __{2}}[/tex]
This implies that
[tex]tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})[/tex]
Looking at this from the SOHCAHTOA concept
[tex]tan (\alpha_{max} __{1}}) = \frac{opposite}{Adjacent}[/tex]
Here opposite is the length of the side of the first cardboard square hole
and
Adjacent is the distance of the from the first cardboard square hole to the point source
And for
[tex]tan (\alpha_{max} __{2}}) = \frac{opposite}{Adjacent}[/tex]
Here opposite is the length of the side of the second cardboards square hole (let denote it with [tex]L_2[/tex])
and
Adjacent is the distance of the from the second cardboards square hole to the point source
So
[tex]tan (\alpha_{max} __{1}}) = \frac{0.020}{0.50}[/tex]
And
[tex]tan (\alpha_{max} __{2}}) = \frac{L_2}{0.25}[/tex]
Substituting this into the above equation
[tex]\frac{0.020}{0.50} = \frac{L_2}{0.25}[/tex]
Making [tex]L_2[/tex] the subject
[tex]L_2 = \frac{0.25 *0.020}{0.50}[/tex]
[tex]L_2 = 0.01m[/tex]
Since it is a square hole the sides are the same hence
The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]
A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
before it hits the ground. We can ignore air resistance.
Answer: Vertical displacement = -27.6m
And takes 2.375 s
A pumpkin thrown at a horizontal speed of 4.0 m/s and travels 9.5 m horizontally before hitting the ground will take 2.375 seconds to hit the ground.
How to find time?Given the following information:
Horizontal speed of the pumpkin = 4.0 m/s
Horizontal distance traveled before hitting the ground = 9.5 m
Ignoring air resistance
Use the following formula to calculate the time it takes the pumpkin to hit the ground:
time = horizontal distance / horizontal speed
time = 9.5 m / 4.0 m/s = 2.375 seconds
Therefore, it takes the pumpkin 2.375 seconds to hit the ground.
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A solid, cylindrical wire conductor has radius R = 30 cm. The wire carries a current of 2.0 A which is uniformly distributed over the cross-section of the wire (current density is constant). What is the magnitude of the magnetic field due to the current in the wire at a radial distance of r = 200 cm from the center axis of the wire? HINT: Use Ampere’s law, noting that B is tangential.
Answer:
Explanation:
The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.
Applying Ampere's law :-
∫ B dl = μ₀ I ; I is current passing through ampere's loop
B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ T.
A 16ft seesaw is pivoted in the center. At what distance from the center would a 200lb person sit to balance a 120lb person on the opposite end?
Answer:
9.6 ft
Explanation:
Distance is inversely proportional to weight
distance = k / (weight), where
k is a constant
or you could say,
distance * weight = k
In this scenario,
120 * 16 = 200 * distance
On rearranging, making, distance the subject of formula, we have
Distance = 120 * 16 / 200
Distance = 1920 / 200
Distance = 9.6 ft
So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw
Answer:
4.8 ft
Explanation:
torque = wt × distance
t1 = 120lb x 8 ft =960
t2 = 200lb x X ft
set them equal to each other.
120(8) = 200x
960 = 200x
x = 4.8 ft
In your research lab, a very thin, flat piece of glass with refractive index 1.50 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength lambda 0 in vacuum at normal incidence onto the surface of the glass. When lambda 0 = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. Use these measurements to calculate the thickness of the glass. Express your answer with the appropriate units. What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?
Answer:
The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ = 3472.
Explanation:
Refractive index of Glass (given) = 1.5
For the case of a constructive interference,
2nt = (m + 1/2) λ
For case 1,
2nt = (m + 1/2) 496 nm
For case 2,
2nt = (m +1+ 1/2) 386 nm
2nt = (m+3/2) * 386 nm
(m + 1/2) 496 nm = (m+3/2) * 386 nm
m = 3
Inserting the value of m in 1.
2nt = (m + 1/2) 496 nm
2*1.5t = (3 + 1/2) * 496 nm
t = ((3 + 1/2) * 496 nm)/ 3
t = 578.6 nm
The thickness of the glass, t = 578.6 nm
b)
It is generally known that for constructive interference,
2nt = (m + 1/2) λ
λ = 2nt / ((m + 1/2))
For Longest Wavelength, m = 0
λ = 2*1.5*578.6/ (1/2)
λ = 3472 nm
A 15.0-kg object and a m^2 =10.0-kg object are joined by a cord that passes over a pulley with a radius of R =10.0 cm and a mass of M = 3.00 kg. The cord has a negligible mass and does not slip on the pulley. The pulley rotates on its axis without friction. The objects are released from rest when they are 3.00m apart and are free to fall. Ignore air resistance. Treat the pulley as a uniform disk, and determine the speeds of the two objects as they pass each other.
Final answer:
The speed of the 15.0-kg object as it passes the 10.0-kg object is 13.3 m/s, and the speed of the 10.0-kg object as it passes the 15.0-kg object is -13.3 m/s.
Explanation:
To determine the speeds of the two objects as they pass each other, we can use the principle of conservation of mechanical energy. When the objects are released from rest, the potential energy of the system is converted into kinetic energy as the objects fall. The sum of the kinetic energies of the two objects will be equal to the initial potential energy of the system.
Using the formula for potential energy (PE=mgh), we can calculate the initial potential energy of the system. The 15.0-kg object will fall a distance of 3.00m, so its potential energy is (15.0 kg)(9.8 m/s^2)(3.00 m) = 441 J. The 10.0-kg object will rise a distance of the same amount, so its potential energy is -441 J (we take the negative sign because the object is moving in the opposite direction).
Now, we can equate the sum of the kinetic energies of the two objects to the initial potential energy of the system. Let v1 be the speed of the 15.0-kg object and v2 be the speed of the 10.0-kg object. The kinetic energy of the 15.0-kg object is (1/2)(15.0 kg)(v1^2) and the kinetic energy of the 10.0-kg object is (1/2)(10.0 kg)(v2^2). Setting the sum of these two kinetic energies equal to 441 J, we can solve for v1 and v2.
441 J = (1/2)(15.0 kg)(v1^2) + (1/2)(10.0 kg)(v2^2)
Simplifying the equation, we have 441 J = (7.5 kg)(v1^2) + (5.0 kg)(v2^2). Since the objects are joined by a cord and the pulley does not slip, the speeds of the two objects will be equal in magnitude but opposite in direction. So we can write v2 = -v1 and substitute into the equation. We can then solve for v1:
441 J = (7.5 kg)(v1^2) + (5.0 kg)(-v1^2)
Simplifying further, 441 J = (2.5 kg)(v1^2)
Solving for v1,
v1^2 = 176.4 m^2/s^2
v1 = 13.3 m/s
Therefore, the speed of the 15.0-kg object as it passes the 10.0-kg object is 13.3 m/s, and the speed of the 10.0-kg object as it passes the 15.0-kg object is -13.3 m/s.
The specific heat of a liquid x is 2.09 cal/g°c. A sample amount of grams of this liquid at 101 k is heated to 225 k. the liquid absorbs 5.23 kcals. what is the sample of liquid in grams? (round off decimal in the answer to nearest tenths)
Answer: 20 grams
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by liquid = 5.23 kcal = 5230 cal (1kcal=1000cal)
C = heat capacity of liquid = [tex]2.09cal/g^0C[/tex]
Initial temperature of the liquid = [tex]T_i[/tex] = 101 K
Final temperature of the liquid = [tex]T_f[/tex] = 225 K
Change in temperature ,[tex]\Delta T=T_f-T_i=(225-101)K=124K[/tex]
Putting in the values, we get:
[tex]5230=m\times 2.09cal/g^0C\times 124K[/tex]
[tex]m=20g[/tex]
Thus the sample of liquid in grams is 20
X rays of wavelength 0.00758 nm are directed in the positive direction of an x axis onto a target containing loosely bound electrons. For Compton scattering from one of those electrons, at an angle of 145°, what are (a) the Compton shift, (b) the corresponding change in photon energy, (c) the kinetic energy of the recoiling electron, and (d) the angle between the positive direction of the x axis and the electron's direction of motion? The electron Compton wavelength is 2.43 × 10-12 m.
A rectangular coil of wire with a dimension of 4 cm x 5 cm and 10 turns is located between the poles of a large magnet that produces a uniform magnetic field of 0.75 T. The surface of the coil which is originally parallel to the field is rotated in 0.10 s, so that its surface is perpendicular to the field. Calculate the average induced emf across the ends of coil as the coil rotates.
Answer:0.15 V
Explanation:
Given
Dimension of coil [tex]4cm\times 5cm[/tex]
Area of coil [tex]A=4\times 5=20\ cm^2[/tex]
Magnetic field [tex]B=0.75\ T[/tex]
Time of rotation [tex]t=0.1\ s[/tex]
No of turns [tex]N=10[/tex]
Initial flux associated with the coil
[tex]\phi_i=N(B\cdot A)[/tex]
[tex]\phi_i=N(BA\cos \theta )[/tex]
where [tex]\theta [/tex]=angle between magnetic field and area vector of coil
[tex]\phi_i=N(BA\cos 90 )[/tex]
Finally when coil is perpendicular to the field
[tex]\phi_f=N(B\cdot A)[/tex]
[tex]\phi_i=N(BA\cos 0 )[/tex]
and induced emf is given by
[tex]e=-\frac{d\phi }{dt}[/tex]
[tex]e=-\frac{\phi_1-\phi_2}{t-0}[/tex]
[tex]e=-\frac{(0-10\times 0.75\times 20\times 10^{-4})}{0.1}[/tex]
[tex]e=0.15\ V[/tex]
the gravitational pull of the moon is much less than the gravitational pull of earth, which two statements are true for an object with a mass of 20 kilograms that weighs 44 pounds on earth
Answer:
b
Explanation:
the earths mass is more than the moon .
Answer:
The object's weight would be less on the moon.
The Object's mass would be the same on the moon
Explanation:
A fellow astronaut passes by you in a spacecraft traveling at a high speed. The astronaut tells you that his craft is 21.1 m long and that the identical craft you are sitting in is 17.3 m long. (a) According to your observation, how long is your craft? m (b) According to your observation, how long is the astronaut's craft? m (c) According to your observation, what is the speed of the astronaut's craft relative to your craft?
Answer:
A) 21.1 m
B) 17.3 m
C) 3.267x10^7 m/s
Explanation:
This is a case of special relativity.
Let the relative speed of astronauts ship to my ship be v.
According to my observation,
My craft is 21.1 m long, according to my observation, astronauts craft is 17.3 m long.
If we fix the reference frame as my ship, then the rest lenght of our identical crafts is 21.1 m and the relativistic lenght is 17.3 m
l' = 21.1 m
l = 17.3 m
From l = l'(1 - p^2)^0.5
Where p is c/v, and c is the speed of light
17.3 = 21.1 x (1 - p^2)^0.5
0.82 = (1 - p^2)^0.5
Square both sides
0.67 = 1 - p^2
P^2 = 0.33
P = 0.1089
Revall p = v/c
v/c = 0.1089
But c = speed of light = 3x10^8 m/s
Therefore,
v = 3x10^8 x 0.1089 = 3.267x10^7 m/s
Following are the response to the given points:
a) Its ship travels to the distance of [tex]21.1\ m[/tex]
b) The astronaut's craft would be at a range of [tex]17.3\ m[/tex]
c) Relativity's use of length contraction:
[tex]\to L=L_0(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to \frac{L}{L_0}=(\sqrt{1-\frac{v^2}{c^2}})[/tex]
Here,
[tex]\to \frac{L}{L_0}=\frac{17.3}{21.1}=0.81[/tex]
Hence
[tex]\to 0.81=(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to 0.6561=1-\frac{v^2}{c^2}\\\\\to \frac{v^2}{c^2} =1- 0.6561\\\\\to \frac{v^2}{c^2} =0.3439\\\\\to \frac{v}{c} =0.58\\\\\to v= 0.58 c[/tex]
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On earth, what is a child’s mass if the force of gravity on the child’s body is 100 N
Answer: 10.2 kg if g = 9.8, 10 if g = 10.
Explanation:
Weight or the "force of gravity" on a person is simply defined by the equation: F = ma. In this case, the acceleration is g, which is 9.8 but can be rounded up to 10. Based on this, we have:
F = mg
100 = m*9.8
m = 10.2(or 10 if we set g to 10).
The child's mass on the earth if the force of gravity on the child’s body is 100 N will be equal to 10.2 kg.
What is gravity?The fundamental force of attraction operating on all matter is recognized as gravity, also spelled gravity, in mechanics.
It has no impact on identifying the interior properties of common matter because it is the weakest force known to exist in nature.
The formation and growth of planets, galaxies, and the universe are all under the influence of this long-range, cosmic force, which further determines the trajectories of objects throughout the universe and the entire universe.
As per the given information in the question,
Weight, w = 100 N
Use the formula,
W = m × g
100 N = m × 9.8
m = 100/9.8
m = 10.2 kg
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In your own words, describe how dog breeds today came from wolves. In other words, describe selective breeding.
Pleaseeeeeeeeeeeeeeeeeeeeee HELPPPPPPPP!!!FASTTT!!!
Answer:
Dogs were probably domesticated by accident, when wolves began trailing ancient hunter-gatherers to snack on their garbage. Docile wolves may have been slipped extra food scraps, the theory goes, so they survived better, and passed on their genes. Eventually, these friendly wolves evolved into dogs
Selective breeding, also known as artificial selection, is a process used by humans to develop new organisms with desirable characteristics. Breeders select two parents that have beneficial phenotypic traits to reproduce, yielding offspring with those desired traits.
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The key discovery about Cepheid variable stars that led in the 1920s to the resolution of the question of whether spiral "nebulae" were separate and distant galaxies or part of the Milky Way Galaxy was the direct relationship between the pulsation period and the absolute brightness or luminosity of the Cepheid variables. A measurement of ____ brightness of a variable star could then be used to determine the distance to the "nebula" containing it.
Answer:
Apparent
The key discovery about Cepheid variable stars that led in the 1920s to the resolution of the question of whether spiral "nebulae" were separate and distant galaxies or part of the Milky Way Galaxy was the direct relationship between the pulsation period and the absolute brightness or luminosity of the Cepheid variables. A measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.
Explanation:
A variable star is a star with changing apparent brightness. The changes can occur over years or in a fraction of seconds. For example the sun whose energy output varies by approximately 0.1 percent of its magnitude, over an 11-year solar cycle. This variable(apparent brightness) can be used to determine how far a variable star is (distance). Therefore, a measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.
Calcula el peso aparente de una bola de aluminio de 50 cm3, cuando se encuentra totalmente sumergida en alcohol. Datos: la densidad del aluminio es 2,7 g / cm3 y la densidad del alcohol es 0,8 g / cm3
Answer:
W_apparent = 93.1 kg
Explanation:
The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.
W_apparent = W - B
The push is given by the expression of Archimeas
B = ρ_fluide g V
ρ_al = m / V
m = ρ_al V
we substitute
W_apparent = ρ_al V g - ρ_fluide g V
W_apparent = g V (ρ_al - ρ_fluide)
we calculate
W_apparent = 980 50 (2.7 - 0.8)
W_apparent = 93100 g
W_apparent = 93.1 kg
Allie is flying from New York to London. Her plane will most likely fly in the:
stratosphere
thermosphere
troposphere
mesosphere
Answer:
troposphere
Explanation:
because troposphere is the layer of atmosphere closest to the earth . air is thicker at lower altitudes requiring more energy to push themselves to the sky . however , the air is thinner caused flight more fuel efficient .
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire fully lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. The magnetic force on the wire is *
Answer:
The force is "19 µN".
Explanation:
The lane's j-component is meaningless, as the current is flowing in the -j line.
Therefore the power is now in the direction of + z (out of the page if x and y are in the page plane) and has the magnitude.
[tex]\ formula: \\\\\ Forec (F) = mA \\\\ \ F \ = 2.0mA \times \int {0.3} \ y \ dy \rightarrow \ from\ 0 \ to \ 0.25 \\\\\ F \ = \ 2.0mA \times 0.15 * 0.25^{2} m\cdot T \\\\ F = 19 \µN[/tex]
A particle of positive charge ???? is assumed to have a fixed position at P. A second particle of mass m and negative charge −q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent in the second particle in order to increase the radius of the circle of motion, centered at P, to r2.
Answer:
[tex]W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]
Explanation:
To find the work W to put the negative charge in the new orbit you can use the following formula:
[tex]W=\Delta K\\\\K=\frac{1}{2}mv^2[/tex]
That is, the total work is equal to the change in the kinetic energy of the negative charge. Then you calculate the speed of the electron, by using the second Newton Law and the expression for the electrostatic energy:
[tex]F=ma_c\\\\-k\frac{(q_1)(q_2)}{r_1^2}=m\frac{v^2}{r_1}\\\\v^2=k\frac{q_1q_2}{mr_1}[/tex]
r1: radius of the first orbit
m: mass of the negative charge
v: velocity of the charge
k: Coulomb's constant
q1: charge of the fixed particle at point P
q2: charge of the negative charge
Hence, the velocity of the charge in a new orbit with radius r2 is:
[tex]v'^2=k\frac{q_1q_2}{mr_2}[/tex]
Finally the work required to put the charge in the new orbit is:
[tex]W=\Delta K =\frac{1}{2}m[v'^2-v^2]\\\\W=\frac{1}{2}m[k\frac{q_1q_2}{mr_2}-k\frac{q_1q_2}{mr_1}]\\\\W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]
Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water.
If the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)
Answer:
0.05m
Explanation:
Density of water = ρ(w) = 1000 kg/ m³ ;
Density of Mercury = ρ(m) = 13628.95 kg/ m³
Total pressure at bottom of cylinder=1.1atm
Therefore, pressure due to water and mercury =1.1-1 =atm
0.1atm=10130pa
The pressure at the bottom is given by,
ρ(w) x g[0.4 - d] + ρ(m) x g x d = 10130
1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130
3924 - 9810d + 133416d= 10130
123606d= 6206
d= 6202/123606
d= 0.05m
Depth of mercury alone = d = 0.05m
We have that for the Question " determine the depth of the mercury."
It can be said that
The depth of the mercury = [tex]4.333*10^{-2}[/tex]
From the question we are told
the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)
Therefore,
Absolute pressure at the bottom of the container =
[tex]P = 1.1 atm = 1.1 * (1.013*105) Pa\\\\= 1.1143 * 10^5 Pa[/tex]
Where,
Height of the cylinder = H = [tex]0.4 m[/tex]
Height of the water in the cylinder = [tex]H_1[/tex]
Height of the mercury in the cylinder = [tex]H_2[/tex]
Therefore,[tex]H = H_1 + H_2\\\\H_1 = H - H_2\\\\P = P_{atm} + \rho_1gH_1 + \rho_2gH_2\\\\P = P_{atm} + \rho_1g(H - H_2) + \rho_2gH_2\\\\1.1143*10^5 = 1.013*10^5 + (1000)(9.81)(0.4 - H_2) + (1.36*10^4)(9.81)H_2\\\\1.013*10^4 = 3924 - 9810H_2 + 133416H_2\\\\143226H_2 = 6206\\\\H_2 = 4.333*10^{-2} m[/tex]
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Peggy is an astronaut and volunteers for the first manned mission to Alpha Centauri, the nearest star system to the Solar System. Her spacecraft will travel at 80%80% of the speed of light, and the trip there and back will take over 1010 years. Her twin sister Patty is an astronomer and will remain on Earth, studying Alpha Centauri using telescopes. When Peggy returns from her trip, how will their ages compare?
Answer:
If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time-dilation.
Explanation:
Peggy and Patty are sisters. Peggy is an astronaut and Patty is an astronomer.
Peggy goes for mission to Alpha Centauri, the nearest star system to the Solar System at 80% of the speed of light, and will come back after 1010 years.
If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time dilation.
This is due to the fact that time moves slower in Alpha Centauri because of its massive gravitational force which bends space time. Moreover, It is known that Peggy's spacecraft moves at 80% of the speed of light, it will result in velocity time dilation since time moves slow if you travel at a speed near to the speed of light.
Final answer:
Peggy, the astronaut twin traveling at 80% of the speed of light, will experience less time due to time dilation, and upon her return will be younger than her Earth-bound twin sister, Patty.
Explanation:
The question is about the relativistic effects that occur when one twin travels at significant fraction of the speed of light while the other remains on Earth.
According to the theory of relativity, time dilation will cause the traveling twin, Peggy, to age more slowly compared to her twin sister, Patty, who remains on Earth.
If Peggy travels to Alpha Centauri, which is 4.3 light years away, at 80% of the speed of light, and assuming the round trip takes 10 years for the Earth-bound twin, we can calculate that the moving twin will experience less than 10 years of elapsed time due to the effects of time dilation.
This happens because the faster Peggy travels, the more pronounced the effect of time dilation will be. This is a well-known result predicted by Einstein's special theory of relativity and has been confirmed through experiments involving high-speed particles and precise clocks.
Thus, when Peggy returns, she will be younger than her twin sister Patty, who has experienced the full 10 years on Earth.
A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 250 to 350 mT. (a) Determine the emf induced in the loop (in V). (Enter the magnitude.) V (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
Answer:
Magnitude of induced emf is 0.00635 V
Explanation:
Radius of circular loop r = 45 mm = 0.045 m
Area of circular loop [tex]A=\pi r^2[/tex]
[tex]A=3.14\times 0.045^2=0.00635m^2[/tex]
Magnetic field is increases from 250 mT to 350 mT
Therefore change in magnetic field [tex]dB=250-350=100mT[/tex]
Emf induced is given by
[tex]e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}[/tex]
[tex]e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V[/tex]
Magnitude of induced emf is equal to 0.00635 V