Answer:
The voltage is equal to the batteries terminal voltage
Explanation:
Explanation:
If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Answer:
B. mineral oil – density of 0.910 g/mL.
Explanation:
Hello,
In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.
In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.
Best regards.
The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change
B. A chemical change
Explanation:
I'm guessing ?
If the ac peak voltage across a 100-ohm resistor is 120 V, then the average power dissipated by the resistor is ________
Answer:
The average power dissipated is 72 W.
Explanation:
Given;
peak voltage of the AC circuit, V₀ = 120 V
resistance of the resistor, R = 100 -ohm
The average power dissipated by the resistor is given by;
[tex]P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}[/tex]
where;
[tex]V_{rms}[/tex] is the root-mean-square-voltage
[tex]V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V[/tex]
The average power dissipated by the resistor is calculated as;
[tex]P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W[/tex]
Therefore, the average power dissipated is 72 W.
Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).
Image is missing, so i have attached it
Answer:
19.04 × 10⁻⁴ T in the +x direction
Explanation:
We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.
Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)
We are given;
I = 2.4 A
R = 5 cm = 0.05 m
μ₀ is a constant = 4π × 10⁻⁷ H/m
B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)
B = 9.6 × 10⁻⁴ T
To get the direction of the field from each wire, we will use Flemings right hand rule.
From the diagram attached:
We can say the field at P from the top wire will point up/right
Also, the field at P from the bottom wire will point down/right
Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.
If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.
Let the angle be θ
Thus;
sin(θ) = (1.3/2)/5
θ = sin⁻¹(0.13) = 7.47⁰
The x component of each field would be:
9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T
Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction
The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".
Magnetic fieldThe region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.
According to the question,
Current, I = 2.4 A
Radius, R = 5 cm or,
= 0.05 m
Constant, μ₀ = 4π × 10⁻⁷ H/m
We know the relation,
The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]
By substituting the values in the above relation, we get
= [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]
= 9.6 × 10⁻⁴ T
Thus the above answer is appropriate.
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The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at time t = -0.0s, with a velocity of 30m/s. The acceleration due to gravity at the surface of the moon is 1.62m/s2 the time when the rock is ascending at a height of 180m is closest to:______.
a. 8s .
b. 12s.
c. 17s.
d. 23s.
e. 30s
For the previous situation, the height of the rock when it is descending with a velocity of 20m/s is closest to:_____.
A. 115m.
B. 125m.
C. 135m.
D. 145m
E. 155m.
Explanation:
Given that,
Initial speed of the rock, u = 30 m/s
The acceleration due to gravity at the surface of the moon is 1.62 m/s².
We need to find the time when the rock is ascending at a height of 180 m.
The rock is projected from the surface of the moon. The equation of motion in this case is given by :
[tex]h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2[/tex]
It is a quadratic equation, after solving whose solution is given by:
t = 7.53 s
or
t = 8 seconds
(e)If it is decending, v = -20 m/s
Now t' is the time of descending. So,
[tex]v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s[/tex]
Let h' is the height of the rock at this time. So,
[tex]h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m[/tex]
or
h' = 155 m
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.
The researchers need to compare those who contracted the disease to those who did not.
A 0.50-T magnetic field is directed perpendicular to the plane of a circular loop of radius 0.25 m. What is the magnitude of the magnetic flux through the loop
Answer:
The magnitude of the magnetic flux through the loop is 0.0982 T.m²
Explanation:
Given;
magnitude of magnetic field, B = 0.5 T
radius of the loop, r = 0.25 m
Area of the loop is given by;
A = πr²
A = 3.142 x (0.25)²
A = 0.1964 m²
The magnitude of the magnetic flux through the loop is given by;
Ф = BA
Where;
B is the magnitude of the magnetic field
A is area of the field
Ф = 0.5 x 0.1964
Ф = 0.0982 T.m²
Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²
A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black c. White d. Green
The color that is reflected when a red card is illuminated by red light is white.
The color an object is perceived to have, depends on the frequency of light it reflects.
If white light incidents on a red filter, red is transmitted while blue and green are absorbed.
Consequently, when a red card is illuminated by red light, the red card will reflect back almost all the incident light on it, causing it to appear brighter which creates an illusion of white color to the eyes.
Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.
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A double-slit experiment is performed with light of wavelength 620 nm. The bright interference fringes are spaced 2.3 mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 360 nm?
Answer:
1.34 mm
Explanation:
A double slit experiment is conducted with a light which has a wavelength of 620 nm
The fringes are separated 2.3 mm apart
The light is changed to a wavelength length of 360 nm
Let x represent the fringe spacing as a result of the change in wavelength
Therefore,the fringe spacing can be calculated as follows
2.3mm/x= 620nm/360nm
Multiply both sides
x × 620= 2.3×360
620x= 828
x= 828/620
x= 1.34 mm
Suppose a 58-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.10 T? magnitude V direction ---Select--- †\
Answer:
95.7v
Explanation
Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop
So it is given as
E= Ndစ/dt
E= N BA-0/ deta t
Given that
N = 58turns
B = 1.10T
A = 0.150m^²
Deta t= 0.1s
now we have
E = 58(1.10x0.150)/0.1
= 95.7v
Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE
Explanation:
Light of wavelength 520 nm is incident a on a diffraction grating with a slit spacing of 2.20 μm , what is the angle from the axis for the third order maximum?
Answer:
θ = 45.15°
Explanation:
We need to use the grating equation in this question. The grating equation is given as follows:
mλ = d Sin θ
where,
m = order number = 3
λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m
d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m
θ = angle from the axis = ?
Therefore,
(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ
Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)
Sin θ = 0.709
θ = Sin⁻¹(0.709)
θ = 45.15°
A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
Improved balance is a primary benefit of regular cardiovascular exercise .
Answer:
Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.
Explanation:
Cardiovascular exercise is activity increase heart rate and raises oxygen large muscle group of the body.Cardiovascular exercise is that contain cardio improve to the health mental health, heart health.Cardiovascular exercise such as walking, swimming, running is that exercise is benefit to the health.Cardiovascular exercise to the internal body organs that the healthy heart for the function and performance of the heart.Cardiovascular exercise that having involve feet of the ground this type of activity is called high impact of cardio.Cardio is a good and maintaining exercise for the lungs and heart or healthy bones.Cardio exercise is performed that to a water in reduce to the gravity of that pull on the body weight.Cardiovascular daily to build the stronger muscle and that control the blood pressure.A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11
Explanation:
Change in Velocity = final velocity - initial velocity
Change in velocity = 30km/h - (- 45km/h )
= 75 km/h due west
A bucket filled with water has a mass of 23 Kg and is attached to a rope, which in turn is wound around a 0.050 m radius cylinder at the top of a well. What torque does the weight of water and bucket produce on the cylinder if the cylinder is ont permitted to rotate? (g= 9.8 m/s2)
Answer:
The torque is 11.27 N m
Explanation:
Recall that torque is the vector product of the force times the distance to the pivoting point. So in our case, the distance to the pivoting point is the radius of the cylinder (0.05 m), and the force is given by the weight of the bucket full of water (W = 9.8 * 23 N = 225.4 N)
Then the torque is: 0.05 * 225.4 N m = 11.27 N m
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 1.5. Light incident parallel to the central axis is focused at a point _____________
Answer:
36cm from the surfaceExplanation:
Equation of refraction of a lens is expression according to the formula given below;
[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]
R is the radius of curvature of the convex refracting surface = 12cm
v is the image distance from the refracting surface
u is the object distance from the refracting surface
n₁ and n₂ are the refractive indices of air and the medium respectively
Given parameters
R = 12 cm
u = [tex]\infty[/tex] (since light incident is parallel to the axis)
n₁ = 1
n₂ = 1.5
Required
focus point of the light that is incident and parallel to the central axis (v)
Substituting this values into the given formula we will have;
[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]
Cross multiply
[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]
Hence Light incident parallel to the central axis is focused at a point 36cm from the surface
Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The resultant vector + is given by
Answer:
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Explanation:
First, each vector is determined in terms of absolute coordinates:
6-meter vector with direction: 30º north of east.
[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]
[tex]\vec A = 5.196\,i + 3\,j[/tex]
4-meter vector with direction: 30º east of north.
[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]
[tex]\vec B = 2\,i + 3.464\,j[/tex]
The resultant vector is obtaining by sum of components:
[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Explain why the two plates of a capacitor are charged to the same magnitude when a battery is connected to the capacitor.
Answer:
This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.
Explanation:
In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are taken to the negatively charged plate.
A neutron star has a mass of between 1.4-2.8 solar masses compressed to the size of:
A. Earth
B. The state of Oregon
C. North America
D. An average city
The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
The bar magnet is pushed toward the center of a wire loop. Looking down from the top view (would appear the magnet is coming up toward the observer); Which is true? A. There is no induced current in the loop B. There is a counterclockwise induced current in the loop C. There is not enough information to correctly answer the question D. There is a clockwisee induced current in the loop
Answer:
Explanation:
B. There is a counterclockwise induced current in the loop
Explanation:
This in line with the right hand grip rule,
The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.
A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the screen. In terms of f, what is the smallest value d can have for an image to be in focus on the screen?
Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
Near Point(cm) Far Point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]
Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.
1. Berenice
2. Avishka
3. Francesca
4. Edouard
Answer:
1. Berenice = 0.67 D
2. Avishka = 1.50 D
3. Francesca = 2.00 D
4. Edouard = 2.75 D
Explanation:
The farsighted people are those with near point greater than 25 cm.
They include Avishka, Berenice, Edouard and Francisca.
A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative
From the lens formula, 1/f = 1/v 1/u; but v is negative
Therefore, 1/f = 1/u - 1/v
But, power of a lens = 1/f in meters.
Therefore, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
P = 1/ 0.25 - 1/0.4 = 1.5 D
For Berenice, u = 0.25 m, v = 0.3 m
P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v = 0.80 m
P = 1/0.25 - 1/0.80 =2.75 D
For Francesca, u = 0.25 m, v = 0.50 m
P = 1/0.25 - 1/0.5 = 2.0 D
When The farsighted people are those with a near point greater than 25cm.
Berenice is = 0.67 D
Avishka is = 1.50 D
Francesca is = 2.00 D
Edouard is = 2.75 D
What is Hyperopic Vision?
When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.
Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.
Also, From the lens formula, 1/f = 1/v 1/u; but v is negative
Thus, 1/f = 1/u - 1/v
But, when the power of a lens = 1/f in meters.
Thus, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
After that, P = 1/ 0.25 - 1/0.4 = 1.5 D
Then, For Berenice, u = 0.25 m, v = 0.3 m
Now, P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v is = 0.80 m
Then, P = 1/0.25 - 1/0.80 is =2.75 D
Now, For Francesca, u = 0.25 m, v is = 0.50 m
Therefore, P = 1/0.25 - 1/0.5 = 2.0 D
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A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light
Answer:
b
Explanation:
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.
Answer:
ΔT = 1ºC , 2ºCand 3ºC
Explanation:
In this exercise they indicate the specific heat of lithium
let's calculate the temperature increase as a function of the heat introduced
Q = m [tex]c_{e}[/tex] ΔT
ΔT = Q / m c_{e}
calculate
for Q = 3.5 J
ΔT = 3.5 / (1 3.5)
ΔT = 1ºC
For Q = 7.0 J
ΔT = 7 / (1 3.5)
ΔT = 2ºC
for Q = 10.5 J
ΔD = 10.5 / (1 3.5)
ΔT = 3ºC
we see that this is a straight line, see attached
A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
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Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.
The correct answer is C. The inner planets are also called terrestrial planets.
Explanation:
Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".
Answer:
The answer is c.) The inner planets are also called terrestrial planets.
Explanation:
A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement
Answer:
displacements are 0.776m, 0.114m
Explanation:
We were given mass of 26-g rifle bullet , then we can convert to Kg since
Momentum is conserved here.
The initial momentum before impact = (Mi * Vi)
Where Mi= initial given mass
Vi=initial velocity given
= 0.026 * 220 = 5.72 kgm/s
The final momentum after impact is (Mf * Vf )
Mf= final mass
5.72=( 3.82* Vf )
= 5.72/ 3.82
= 1.497 m/s
the speed of the pendulum bob with bullet afterwards= 1.497 m/s
the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.
M = 3.82 kg the mass of the bob containing the bullet
E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J
When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before
E(total) = Mgh + 0 = Mgh = 4.280J
solving for h and substituting,
h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m
Since the height is found,we the angle of the pendulum at the top of the swing can also be determined
A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees
Since A is known, the displacement along the horizontal axis can be calculated as
x = 2.7* sin(A) = 0.776m
therefore, displacement is 0.776m, 0.114m
the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m
If 1.7 kg of 238Pu is required to power the spacecraft's data transmitter, for how long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
The question is incomplete. Here is the complete question.
The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.
(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.
(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) 91 years.
Explanation:
(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:
Q(t) = [tex]0.5Q_{0}[/tex]
[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]
[tex]0.5=e^{-87.7k}[/tex]
[tex]ln(0.5)=ln(e^{-87.7k})[/tex]
[tex]ln(0.5)=-87.7k[/tex]
[tex]k = \frac{ln(0.5)}{-87.7}[/tex]
k = 0.0079
Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) Using function:
[tex]Q(t)=3.5e^{-0.0079t}[/tex]
[tex]1.7=3.5e^{-0.0079t}[/tex]
[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]
[tex]e^{-0.0079t}=0.4857[/tex]
[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]
[tex]-0.0079t=-0.7221[/tex]
[tex]t = \frac{-0.7221}{-0.0079}[/tex]
t = 91.41
t ≈ 91 years
Scientists will be able to receive data for approximately 91 years.
