Answer:
There is no closed-loop path for the current to flow through the circuit.When the switch is closed,the light bulb operates since the current flows through the circuit.The bulb glows at its full brightness since its receives its full 120 volts and has the design current flow.
Explanation:
The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small compared to the distance from the slits to the screen, the distance between adjacent fringes near the center of the interference pattern _______.
Final answer:
The distance between adjacent fringes near the center of the interference pattern in a double-slit experiment decreases by the same factor when the distance between the slits is decreased by a factor of 2.
Explanation:
The distance between adjacent fringes near the center of the interference pattern in a double-slit experiment is determined by the wavelength of the light and the distance between the slits. When the distance between the slits is decreased by a factor of 2, the distance between adjacent fringes near the center of the interference pattern also decreases by the same factor. This is because the interference pattern is directly related to the slit separation.
Which is true about the potential energy stored in an inductor?
Answer:
It depends on the current
Explanation:
The formula for the energy stored in an inductor is W =1/2 *L*I2
We can state that an electric current flowing in inductor , there is energy stored in the magnetic field. let consider the pure inductor L, and the power which is supplied to initiate the current in inductor.
consider the equation P = i*v
write the equation in terms of inductor.
P = L*i di/dt
P*dt = L*i di
rewrite the Energy equation W = \int_{0}^{t}P*dt
W = \int_{0}^{t} L*i di
by solving we will get the W = 1/2 *L*I2
From this information and formula we can state that he energy stored in an inductor depends on the current.
Answer:
The complete option for the question
a.it depends on the self induced emf
b it depends on the inverse of the self induced emf
C.it depends on the current
d.it depends on the inverse of current
Answer is.it depends on the current
Explanation:
The energy stored in an inductor is W=1/2LI^2
From this equation it can be seen that energy in inductor L depends on current L
Four long, parallel conductors carry equal currents of I = 5.00 A. The direction of the current is into the page at points A and B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square with an edge length 0.200 m.
Answer:
Explanation:
The magnetic field due to a straight wire carrying current is given by
B = μo• I / 2πr
Where,
μo = permeability of free space
μo = 4π×10^-7 Tm/A
I is the current in wire
I = 5A
r Is the distance of point from the wire.
The distance between the parallel conductors and the point is r/2
R = 0.283/2 = 0.1415m
Check attachment on how I used Pythagoras theorem to find the diagonal of the square..
Hence, the magnetic field at point P is
B = μo•I / 2πR
B = 4π × 10^-7 × 5 / 2π × 0.1415
B = 7.07 × 10^-6 T.
What bodie is outside the solar system
Answer:
Oumuamua is outside the solar system which is in cigar shaped.
Hope it will help you :))
Every star, except the Sun, is outside the solar system.
An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator is holding a 5.1 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates
Answer: Tension = 53.6N
Explanation:
Given that
Height h = 1 m
Time t = 1.7 s.
Mass m = 5.1 kg
From the equation of the motion we can get the acceleration of the elevator:
h = X0+ V0t + at2/2;
Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus
a = 2h/t2 = 2 × 1/1.7^2
a = 0.69 m/s2.
Then we can find the tension in the cord by using the formula
T = mg + ma
= 5.1 (9.8 + 0.69)
= 5.1 × 10.5
= 53.6N
If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?
The mass that should be displaced 20.0 cm should be 20000 N/m.
Calculation of the mass:At the time When a spring should be stretched or compressed its length so it changed by an amount x from its equilibrium length due to which the restoring force is exerted.
here,
spring constant is [tex]k = 1.00 * 10^3 N/m[/tex]
mass is x = 20.0 cm
Now
As per Hooke's law
We know that
F = - kx
[tex]= - 1.00 *10 ^3 * 20.0[/tex]
F = 20000 N/m
Hence, The mass that should be displaced 20.0 cm should be 20000 N/m.
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To find the restoring force, use Hooke's Law: F = -kx. With a spring constant of 1.00 × 10³ N/m and a displacement of 20.0 cm (0.200 m), the restoring force is 200 N.
To answer this, we use Hooke's Law, which states:
F = -kx
Where:
F indicates the restoring force
k denotes the spring constant
x indicates the displacement
Given the spring constant k = 1.00 × 10³ N/m and displacement x = 20.0 cm (which we convert to meters, so x = 0.200 m), we substitute these values into the equation:
F = - (1.00 × 10³ N/m) × (0.200 m)
This simplifies to:
F = -200 N
The negative sign indicates that the force is directed opposite to the direction of displacement, which is typical for restoring forces. Therefore, the restoring force is 200 N.
how many days are in a year on mars
The layer of the sun that radiates most of the light that reaches earth is the
Answer:
photosphere
Explanation:
Photosphere, visible surface of the Sun, from which is emitted most of the Sun’s light that reaches Earth directly. Since the Sun is so far away, the edge of the photosphere appears sharp to the naked eye, but in reality the Sun has no surface, since it is too hot for matter to exist in anything but a plasma state—that is, as a gas composed of ionized atoms.
Source :
Photosphere astronomy
Written By:. Harold Zirin
Provide 2 examples in sport or rehab where instantaneous velocity is more important than average velocity and 2 examples where average velocity is more important. Provide rationale.
Answer:
In high jump as well as long jump, instateneous velocity is more important than average velocity.
In relay races and 400m races, average is more important than instateneous velocity.
Explanation:
Instantaneous velocity is the velocity of anything in motion at a specific point in time. This is determined quite similarly to average velocity, however, we look at the period of time so that it approaches zero. If there is a standard velocity over a period of time, its average and instantaneous velocities may be the same. Instantaneous velocity is calculated as the limit as t approaches zero of the change in d over the change in t.
The range or length of long jump depends on the instantenous velocity of the jump and the height of high jump depends on the instantenous velocity of the height.
A person with greater average velocity wins a race. The average velocity of anything or object is referred to as its total displacement divided by the total time taken. That is to say, it is the rate at which an object changes its position from one place to another. Average velocity is also a Vector quantity. Meters per second is the SI unit. Although, any distance unit per any time unit can be used when necessary, such as miles per hour (mph) or kilometer per hour (kmph)
Instantaneous velocity is more important in specific instances like a tennis stroke or a football collision, while average velocity is more important for calculating average rates of change in situations like road trips or rehab.
Explanation:Instantaneous velocity is more important than average velocity in situations where we need to know the velocity at a specific instant in time. Examples of this include a tennis player's stroke in which they aim to hit the ball on the sweet spot of the racket for maximum velocity and minimal vibration, and in a collision in football where a player with the same velocity but greater mass has a greater impact due to their greater momentum.
On the other hand, average velocity is more important in situations where we need to calculate the average rate of change of position over a given time interval. Examples include calculating the average velocity of a car during a road trip to determine the time taken to reach a destination, and in rehab where the average velocity of a patient's movement is measured to track progress over time.
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the intensity at the central maximum of a double-slit interference pattern is 4I_1. The intensity at the first minimum is zero. at what fraction of the distance from the central maximum to the first minimum is the intensity I_1? Thank you.
The fraction of the distance from the central maximum to the first minimum where the intensity is I1 is 1/2.
Explanation:The intensity at the central maximum of a double-slit interference pattern is 4I1 and the intensity at the first minimum is zero. To find the fraction of the distance from the central maximum to the first minimum, we can use the formula for intensity in double-slit interference:
I = 4I1 cos2(πd/λ)
At the first minimum, the intensity is zero. Therefore, we have:
0 = 4I1 cos2(πd/λ)
Simplifying this equation, we find that cos2(πd/λ) = 0.
If cos2(πd/λ) = 0, then cos(πd/λ) = 0.
And if cos(πd/λ) = 0, then πd/λ = π/2.
Solving for d/λ, we get d/λ = 1/2. Therefore, the fraction of the distance from the central maximum to the first minimum where the intensity is I1 is 1/2.
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Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small sphere, B with a net charge of -3.0 x 10^-6, when the distance between their centers is 10.0 meters k= 8.99 x10^9
Answer:
F = 5.33*10^-4N
Explanation:
to find the electrostatic force you use the Coulomb's law, given by the formula:
[tex]F=k\frac{q_Aq_B}{r^2}[/tex]
k: Coulomb's constant = 8.89*10^9 Nm^2/C^2
q_a: charge of A = 2.0*10^{-6}C
q_B: charge of B = -3.0*10^{-6}C
r: distance between the spheres = 10.0m
By replacing all these values you obtain:
[tex]F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N[/tex]
hence, the forcebetween the spheres is about 5.33*10^-4N
Points Possible: 1, Points Correct: 0 Next If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed to listening for it standing by the track? In other words, how much faster will the sound travel through the iron rails than through the air? Assume the air temperature is 0°C. Sound will reach you seconds sooner.
Answer:
[tex]t_{air} - t_{iron} = 282.17 sec[/tex]
Explanation:
The distance covered = 100 km = 100,000 m
The speed of sound in iron, [tex]v_{iron}[/tex] = 5130 m/s
The speed of sound in air at 0° C, [tex]v_{air}[/tex]= 331.5°C
Speed = distance/time
Time = Distance /speed
[tex]t_{iron} = \frac{d_{iron} }{v_{iron} } \\t_{iron} = 100000/5130\\t_{iron} = 19.49 sec[/tex]
[tex]t_{air} = \frac{d_{air} }{v_{air} } \\t_{air} = 100000/331.5\\t_{air} =301.66 sec[/tex]
[tex]t_{air} - t_{iron} = 301.66 - 19.49\\ t_{air} - t_{iron} = 282.17 sec[/tex]
A parallel-plate capacitor has plates of area 0.30 m2 and a separation of 2.10 cm. A battery charges the plates to a potential difference of 190 V and is then disconnected. A dielectric slab of thickness 4 mm and dielectric constant 4.8 is then placed symmetrically between the plates.
(a) What is the capacitance before the slab is inserted?
(b) What is the capacitance with the slab in place?
(c) What is the free charge q before and after the slab is inserted?
(d) What is the magnitude of the electric field in the space between the plates and dielectric and in the dielectric itself?
(f) With the slab in place, what is the potential difference across the plates?
(g) How much external work is involved in inserting the slab?
Answer:
a) 1.26e^-10F
b) 1.47e^-10F
c) 2.39e^-8C 2.89e^-8C
d) E=4500.94N/C
e) E'=5254.23N/C
f) 100.68V
g) 1.65e^-10J
Explanation:
To compute the capacitance we can use the formula:
[tex]C=\frac{k\epsilon_o A}{d}[/tex]
where k is the dielectric constant of the material between the plates. d is the distance between plates and A is the area.
(a) Before the material with dielectric constant is inserted we have that k(air)=1. Hence, we have:
[tex]k=1\\A=0.30m^2\\d=0.021m\\e_o=8.85*10^{-12}C^2/(Nm^2)\\\\C=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{0.021m}=1.26*10^{-10}F[/tex]
(b) With the slab we have that k=4.8 and the thickness is 4mm=4*10^{-3}m. In this case due to the thickness of the slab is not the same as d, we have to consider the equivalent capacitance of the series of capacitances:
[tex]C=(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_1})\\\\\\C_1=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{8.5*10^{-3}m}=3.1*10^{-10}F\\\\C_2=\frac{(4.8)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{4*10^{-3}m}=3.186*10^{-9}F\\\\C=1.47*10^{-10}F[/tex]
(c)
The charge between the plates for both cases, with the slab is given by:
Q : without the slab
Q': with the slab
[tex]Q=CV=(1.26*10^{-10}F)(190V)=2.39*10^{-8}C\\\\Q'=C'V=(1.47*10^{-10F})(190V)=2.79*10^{-8}C\\[/tex]
(d) The electric field between the plate is given by:
[tex]E=\frac{Q}{2\epsilon_o A}[/tex]
E: without the slab
E': with the slab
[tex]E=\frac{2.39*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=4500.94N/C\\\\E'=\frac{2.79*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=5254.23N/C\\[/tex]
(f) We can assume the system as composed by V=V1+V'+V1 as in (c). By using the equation V=Ed we obtain:
[tex]V=2(4500.94)(8.85*10^{-3}m)+(5254.23)(4*10^{-3}m)=100.68V[/tex]
(g) External work is the difference between the energies of the capacitor before and after the slab is placed between the parallels:
[tex]\Delta E=\frac{1}{2}[(1.26*10^{-10}F)(120V)-(1.47*10^{-10})(100.6V)]=1.65*10^{-10}J[/tex]
A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s and that the radius of the loop-the-loop is 3.033 km. What is the apparent weight that the pilot feels (i.e., the force with which the pilot presses against the seat) at the bottom of the loop-the-loop?
Final answer:
The apparent weight that the pilot feels at the bottom of the loop-the-loop is calculated by summing the gravitational force and the centripetal force, resulting in an apparent weight of 4012.9 N.
Explanation:
The student's question is asking to calculate the apparent weight that a pilot feels at the bottom of a loop-the-loop. To find the apparent weight, we must calculate the normal force on the pilot, which is the combination of the gravitational force and the centripetal force required to keep the pilot in a circular path at the bottom of the loop.
Firstly, we calculate the gravitational force (weight) acting on the pilot: W = m × g, where m is the mass of the pilot, and g is the acceleration due to gravity. For the pilot with a mass of 84.0 kg, W = 84.0 kg × 9.8 m/s² = 823.2 N.
Next, we determine the centripetal force required for circular motion at the bottom of the loop: Fc = (mv²) / R, where m is the mass, v is the constant speed, and R is the radius of the loop. With v = 345 m/s and R = 3.033 km (or 3033 m), Fc = (84.0 kg × (345 m/s)²) / 3033 m = 3189.7 N.
The apparent weight of the pilot at the bottom of the loop is the sum of the gravitational force and the centripetal force: Apparent weight = W + Fc = 823.2 N + 3189.7 N = 4012.9 N.
Therefore, the pilot feels an apparent weight of 4012.9 N at the bottom of the loop-the-loop.
A worker pushes a wheelbarrow with a horizontal force of 50 N on level ground over a distance of 5.0 m. If a friction force of 43 N acts on the wheelbarrow in a direction opposite that of the worker, what work is done on the wheelbarrow by the worker?a) 250 J
b) 215 J
c) 35 J
d) 10 J
e) None of these answers is correct.
Answer:
Work done will be equal to 35 J
So option (c) will be correct answer
Explanation:
We have given force F = 50 N
Frictional force f = 43 N
So net force will be equal to [tex]F_{net}=50-43=7N[/tex]
Distance covered on the ground d = 5 m
We have to find the work done
Work done is equal to [tex]W=F_{net}\times d=7\times 5=35J[/tex]
So option (C) will be the correct answer.
The correct option is A 250 J.
Given, Horizontal force is 50 N.
Frictional force is 43 N.
The distance between worker and pushing of wheelbarrow is 5 m.
We know that, Work is a physics term describing the amount of energy transferred when it is moved over a distance by an external force.
So Work = Force . Displacement
work = F D cos [tex]\Theta[/tex]......(eq 1)
Here [tex]\Theta[/tex] is 0 .
So, [tex]work = 50 \times 5 \times 1[/tex] ( cos 0 = 1 )
Work = 250 J
Hence 250 J of work is done on the wheelbarrow by the worker.
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A(n)______is a device intended for the protection of personnel that functions to deenergize a circuit or portion thereof within an established period of time when a current to ground exceeds the values established for a Class A device.
Answer:
The answer is GFCI
(ground fault circuit interrupter)
Explanation:
A ground fault circuit interrupter (GFCI) can help prevent electrocution.
If a person's body starts to receive a shock, the GFCI senses this and cuts off the power before he/she can get injured. GFCIs are generally installed where electrical circuits may accidentally come into contact with water.
What is a circuit breaker?
It is an automatically operated electrical switch designed to protect an electrical circuit from damage caused by excess current from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.
. If a substance has an excess number of electrons on its surface, what type of charge does it have?
A positive charge.
A negative charge.
A neutral charge.
A nuclear charge.
Answer:
A negative charge
Explanation:
If there is an excess of electrons in a substance, this will have a negative net charge. This is because electrons have negative charge
q = -1.67*10^{-19}
each electron will contribute with this charge to the total negative net charge of the substance.
hence, the answer is "A negative charge"
A substance with an excess number of electrons on its surface has a negative charge due to the negatively charged nature of electrons.
If a substance has an excess number of electrons on its surface, it will have a negative charge. This is because electrons have a negative charge, and having more electrons than protons results in a net negative charge for the substance. By contrast, if a substance had more protons than electrons, it would have a positive charge. Substances are typically electrically neutral when the number of electrons equals the number of protons, and this balance means there is no net electric charge.
An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Conversely, when an atom loses one or more electrons, it becomes a positively charged atom, also known as a cation. For instance, when a neutral sodium atom loses one electron, it becomes a sodium cation with a 1+ charge. Conversely, an oxygen atom that gains two electrons becomes an anion with a 2- charge.
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 126-mile trip in a typical midsize car produces about 2.99 x 109 J of energy. How fast would a 45.8-kg flywheel with a radius of 0.512 m have to rotate to store this much energy? Give your answer in rev/min.
ω = ?
mass = 45.8kg
r = 0.512m
E = 2.99*10⁹J
Kinetic Energy of rotation = I * ω²
K.E = I * ω²
I = ½ m*r²
I = ½ * 45.8 * (0.512)²
I = 6.0kgm²
K.E = ½ * I * ω²
ω = √(2K.E / I )
ω = √[( 2* 2.99*10⁹) / 6]
ω = 3.157*10⁴ rad/s
What are some of the phenotypic plant traits that scientists are investigating? Why do you think these traits were chosen?
The term phenotype in a more specific context to describe the collective expression of the genotype in conjunction with the environment on a plant's observable characteristics.
Explanation:
Phenotype, any of the measurable features of an organism arising from its genotype association (total genetic inheritance) with the environment. Behaviour, biochemical properties, colour, form, and scale are examples of measurable characteristics.An allele is another type of one gene. A phenotype triggered by the expression of a gene or group of genes is the external manifestation of a human trait. A trait is an attribute unique to an organism. The genotype's external expression is called the phénotype.Some of the phenotypic traits includes Height, color, shape.
What is a phenotype?Phenotype is a physical properties or expression of a gene in an individual.
Phenotype are features that is observable or can be seen. This could include the physical appearance where we have height, color, shape.
Phenotypic traits are the observable features such as plant height and eye color, hair color.
The genetic make up of an individual and environment determines the phenotype.Scientific investigation is based on the interest and expected output or result. For a trait or many traits to be investigated it must be related to the research to get the expected outcome.
Therefore, some of the phenotypic traits that scientist can investigate are height, shape and color.
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Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction n2. If total internal reflection occurs at the boundary that separates medium 1 and medium 2, then we know which of the following? a) n1= n2 b) n1< n2 c) n1> n2 d) n1 ≥ n2
Answer:
option C
Explanation:
Given,
Refractive index of medium 1 = n₁
Refractive index of medium 2 = n₂
For total internal reflection to take place light should move from denser medium to the rarer medium.
Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2
n₁ > n₂
The correct answer is option C
A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform in all directions, how much acoustic power is emitted by the whistle? The threshold of human hearing is 1.0 × 10-12 W/m2.
Answer:
Emitted power will be equal to [tex]7.85\times 10^{-5}watt[/tex]
Explanation:
It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m
Threshold of human hearing [tex]I=10^{-12}W/m^2[/tex]
We have to find the emitted power
Emitted power is equal to [tex]P=I\times A[/tex]
[tex]P=I\times 4\pi R^2[/tex]
[tex]P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt[/tex]
So emitted power will be equal to [tex]7.85\times 10^{-5}watt[/tex]
Tall towers support power lines h = 59 m above the ground and 19 m apart that run from a hydroelectric plant to a large city, carrying 60 Hz alternating current with amplitude 4 104 A. That is, the current in both of the power lines is 1 = (4 × 104 A) sin(2n-60 Hz . t). out in Base
(a) calculate the amplitude largest magnitude and direction of the magnetic field produce by the power lies at ne base of the tower, when a current of x 104 A in the left power line is headed out of the page, and a current of 4 x 104 A in the right power line is headed into the page. magnitude direction up
(b) This magnetic field is not large compared to the Earth's magnetic field, but it varies in time and so might have different biological effects than the Earth's steady field. For a person lying on the ground at the base of the tower, approximately what is the maximum emf produced around the perimeter of the body which is about 2 m long by half a meter wide? mw
Answer:
a) 4.25 x 10∧-5 T
b) 16.03 mv
Explanation:
the solution is shown in the pictures attached
The maximum magnetic field that could be induced at the base of the power line towers would be around 1.67 µT, directed upwards. For a person lying down near the base of these power lines, the approximate maximum emf produced around their perimeter would be about 5.6 mV. The magnetic field strength is quite weak and its likelihood to cause biological impact is low.
Explanation:First, for (a), the field is the result of both sources, with one of the currents in opposite direction. So, we can just take the double of it, giving us that the magnitude of magnetic field is B = (2μ₀I)/(2πr). Plugging the known values in, we get B = 2 x (4π x 10⁻⁷ Tm/A) x (4x10⁴A) / (2πx19m) approximately equal to 1.67 µT directed upward.
For (b), we will consider the person as a rectangular coil. In this case, we can use Faraday's Law of Electromagnetic Induction which states that emf = -N ∆Φ/∆t, where ∆Φ is the change in magnetic flux and N is the amount of loops or turns. Here, N=1 as the person is considered as a single loop.
The flux changes from B*A to -B*A, so ∆Φ = 2BA. Resulting, emf = (2BA Δt) with Δt = 1/120 second (as it's 60 Hz, so frequency = 2 per every 1/60 seconds). Plugging all the known values, we get emf approximately equal to 0.0056 V or 5.6 mV. Even though it changes over time, this magnetic field strength is still very weak and thus, is less likely to cause biological effects.
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In my solar system, we have a planet that is the innermost to our star that is exactly like the innermost planet in your solar system. The planet has a very large impact crater with a basin that covers a large region of the planet's surface, but few smaller craters have formed on top of it. What can one conclude from this?
Answer:
1. The planet doesn't have a thick enough atmosphere.
2. There have been multiple impacts on the planet.
Explanation:
As the planet is very close to the star, there is high possibility that it will not have an atmosphere. Just like Mercury doesn't have one. Presence of a very large crater with basin indicates that in the past a huge body had hit the planet and thus creating the crater with basin. Also, it must be very old.
Second observation that is given is the presence of smaller craters in the basin. This indicates impact craters created by smaller objects. If the planet had an atmosphere, these smaller objects would not be able to penetrate and reach the surface. Thus presence of these smaller crater indicate towards the planet not having any atmosphere.
Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0 kg cart repels one way with a velocity of 0.12 m/s. The 2.0 kg cart goes the other way. At what velocity does it travel?
Answer:
V = -0.3 m/sec.
Explanation:
5.0 x 0.12 + 2.0 x v = 0. Which means that V = -0.3 m/sec.
The -ve sign shows it moves in the opposite direction.
Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Explanation:
Let us assume that the mass of a pitched ball is 0.145 kg.
Initial velocity of the pitched ball, u = 47.5 m/s
Final speed of the ball, v = -51.5 m/s (in opposite direction)
We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :
[tex]\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s[/tex]
So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.
Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :
[tex]J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s[/tex]
Hence, this is the required solution.
The magnitude of the change in momentum of the ball is 99 m/s. The magnitude of the impulse applied to the ball by the bat is also 99 m/s.
Explanation:To find the magnitude of the change in momentum of the ball, we can use the equation:
Magnitude of change in momentum = magnitude of final momentum - magnitude of initial momentum
Given that the initial velocity of the ball is 47.5 m/s and the final velocity of the batted ball is 51.5 m/s in the opposite direction, the magnitude of the change in momentum is:
Magnitude of change in momentum = 51.5 m/s - (-47.5 m/s) = 99 m/s
The magnitude of the impulse applied to the ball by the bat is equal to the magnitude of the change in momentum. Therefore, the magnitude of the impulse applied to the ball is 99 m/s.
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The idea that little children are continually coming up with ideas and testing them is called the ______ theory.
atomic
theory
moral foundations
general
Answer:
The idea that little children are continually coming up with ideas and testing them is called the general theory.
Which situations contain unbalanced forces? Check all that apply.
Answer:
Explanation:
Without seeing the options i can tell you this:
A balanced force is equal on both sides
The friction would equal the force
The amount on the left (friction side) would equal the amount on right (force side)
Or the net force would equal zero
You should post the answer options for more help
Answer: A book falls to the floor.
A car skids to a stop.
A foam ball launches
Explanation:
You drop an irregular piece of metal into a container partially filled with water and measure that the water level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?
Answer:
Density = 1464.8kg/m3
the density of the metal is 1464.8kg/m^3
Explanation:
Given;
Mass m = 450g
Density = Mass/Volume = m/V
Volume V = change in height × base area = ∆h × A
∆h = 4.8cm
A = 8×8 = 64cm^2
V = 4.8×64 = 307.2cm^3
Density = 450g/307.2cm^3
Density = 1.4648g/cm^3
Density = 1.4648 × 1000kg/m^3
Density = 1464.8kg/m3
the density of the metal is 1464.8kg/m^3
Answer:
1464.84 kg/m³
Explanation:
Density = mass/volume.
D = m/v................. Equation 1
But from Archimedes principle,
every object immersed in water will displaced an amount of water equal to its own volume
Therefore,
v = v'................... Equation 2
Where v = volume of the irregular object, v' = volume of water displaced.
Since the base of the container is a square,
Then,
v' = L²(d)...................... Equation 3
Where L = length of the square base of the container, d = rise in water level.
Substitute equation 3 into equation 1
D = m/L²d......................... Equation 4
Given: m = 450 g = 0.45 kg, L = 8 cm = 0.08 m, d = 4.8 cm = 0.048 m
Substitute into equation 4
D = 0.45/(0.08²×0.048)
D = 0.45/0.0003072
D = 1464.84 kg/m³
In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is
Fext=dp/dt=m dv/dt+v dm/dt
Suppose the mass of the raindrop depends on the distance x that it has fallen. Then m = kx, where k is a constant, and dm/dt=kv
dm/dt=kv This gives, since Fext=mg
Fext=mg,
mg=m dv/dt+v(kv)
Or, dividing by k,
xg=x dv/dt+v2
This is a differential equation that has a solution of the form v = at, where a is the acceleration and is constant. Take the initial velocity of the raindrop to be zero.
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
The acceleration of the raindrop is equal to the acceleration due to gravity. The distance the raindrop has fallen in 3.00 seconds is 44.1 meters. The mass of the raindrop at 3.00 seconds is 88.2 grams.
Explanation:To find the acceleration, we can use the given proposed solution for velocity, v = at. Taking the derivative of this equation with respect to time gives us dv/dt = a. Substituting this into the differential equation, we have xg = x(a) + v^2. Since the initial velocity is zero, v = 0 and the equation simplifies to xg = xa. Dividing by x, we get g = a. Therefore, the acceleration is equal to the acceleration due to gravity, g.
To find the distance the raindrop has fallen in t = 3.00 s, we can use the equation x = (1/2)at^2. Since we know the acceleration is g, we plug in the values into the equation: x = (1/2)(g)(3.00 s)^2 = (1/2)(9.8 m/s^2)(9.00 s^2) = 44.1 meters.
To find the mass of the raindrop at t = 3.00 s, we can use the given equation m = kx. Plugging in the values, m = (2.00 g/m)(44.1 m) = 88.2 grams.
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A 2.05 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m. The spring has force constant 830 N/m. The coefficient of kinetic friction between the floor and the block is 0.380. The block and spring are released from rest and the block slides along the floor.What is the speed of the block when it has moved a distance of 0.0200 m from its initial position?
Answer:
0.552 m/s
Explanation:
Given,
Mass of the block = 2.05 Kg
initial compression, x₁ = 0.03690 m
Spring constant, k = 830 N/m
coefficient of friction of the block, μ = 0.380
distance moved by the block,x₂ = 0.20 m
Speed, v = ?
Using conservation of energy
Initial spring energy + Work done by friction = Final spring energy + kinetic energy
[tex]\dfrac{1}{2}kx_1^2 - \mu mg x_2 = \dfrac{1}{2}kx_2^2 + \dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}\times 830\times 0.039^2 - 0.38\times 2.05\times 9.81\times 0.02 = \dfrac{1}{2}\times 830 \times 0.02^2 + \dfrac{1}{2}\times 2.05\times v^2[/tex]
v² = 0.3047
v = 0.552 m/s
Hence, the speed of the block is equal to 0.552 m/s