The question is incomplete. The complete question is :
A long, straight wire lies along the z-axis and carries a 3.90-A current in the + z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mm segment of the wire centered at the origin.
A) x=2.00m,y=0, z=0
Bx,By,Bz = ? T
Enter your answers numerically separated by commas.
B) x=0, y=2.00m, z=0
C) x=2.00m, y=2.00m, z=0
D) x=0, y=0, z=2.00m
Solution :
The expression of the magnetic field using the Biot Savart's law is given by :
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
a). The position vector is on the positive x direction.
[tex]$\vec r = (2 \ m) \ \hat i$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat i }{4 \pi \times (2)^3}$[/tex]
[tex]$d \vec B=(5.85 \times 10^{-11} \ T)\hat j$[/tex]
The magnetic field is [tex]$(0, \ 5.85 \times 10^{-11} \ T, \ 0).$[/tex]
b). The position vector is in the positive y-direction.
[tex]$\vec r = (2 \ m) \ \hat j$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat j }{4 \pi \times (2)^3}$[/tex]
[tex]$d \vec B=(5.85 \times 10^{-11} \ T)(-\hat{i})$[/tex]
The magnetic field is [tex]$(- 5.85 \times 10^{-11} \ T, \ 0, \ 0).$[/tex]
c). The position vector is :
[tex]$\vec r = (2)\hat i + (2)\hat j$[/tex]
[tex]$|\vec r| = \sqrt{(2)^2+(2)^2}$[/tex]
[tex]$=2.828 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times ((2)\hat i + (2) \hat j) }{4 \pi \times (2.828)^3}$[/tex]
[tex]$=(4.13\times 10^{-11})\hat j+(4.13\times 10^{-11})(-\hat i)$[/tex]
The magnitude of the magnetic field is :
[tex]$|d\vec B|=\sqrt{(4.13\times 10^{-11})^2+(4.13\times 10^{-11})^2}$[/tex]
[tex]$=5.84 \times 10^{-11} \ T$[/tex]
Therefore, the magnetic field is [tex]$(-4.13 \times 10^{-11}\ T, \ 4.13 \times 10^{-11}\ T, \ 0 )$[/tex]
d). The position vector is in the positive y-direction.
[tex]$\vec r = (2 \ m) \ \hat k$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat k }{4 \pi \times (2)^3}$[/tex]
= 0 T
The magnetic field is (0, 0, 0)
Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has tripled
Answer:
Explanation:
It is a case of adiabatic expansion .
[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]
[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]
1.552 = 292 / T₂
T₂ = 188 K .
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon? A. Emission spectra of atoms B. Blackbody radiation C. Interference patterns of electrons D. Photoelectric effect
Answer:
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon is:
C. Interference patterns of electrons.
Answer:
C. Interference patterns of electrons
which statement regarding the idealized model of motion called free fall is true?
a. the effect of air resistance is factored in the equation of motion in the idealized model called free fall.
b. free fall only models motion for objects that do not have an initial velocity in the upward direction.
c. the idealized model of the motion called free fall applies in cases where distance of the fall is large compared with the radius of the astronomical body on which the fall occurs.
d. a freely falling object has a constant acceleration due to gravity.
A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 2.50 s after the bird lets go, the fish lands in the ocean. (a) Just before reaching the ocean, what is the horizontal component of the fish's velocity in m/s
Answer:
Horizontal Component of Fish's Velocity = 2.6 m/s
Explanation:
In this scenario, we will neglect the effects of the air resistance on the small fish. Since there is no resisting force available in the horizontal direction. Therefore, the horizontal component of the velocity of the fish will remain equal to the horizontal component of the velocity of the seagull and it will remain the same throughout the whole motion.
Horizontal Component of Fish's Velocity = Constant Horizontal Speed of Seagull
Horizontal Component of Fish's Velocity = 2.6 m/s
A child sleds down a snowy hillside starting from rest. The hill has a 15 degree slope, with a long stretch of level field at the foot. The child starts 50 ft up the slope and continues for 100 ft on the level field before coming to a complete stop. Find the coefficient of friction between the sled and the snow, assuming that it is constant throughout the ride. Neglect air resistance.
Answer:
0.0872
Explanation:
the solution to the problem can be found in the attachment section. Please go through and feel free to ask your doubts.
What do the spheres in this model represent?
A. Molecules
B. electrons
C. Planets and the sun
D. Atoms
Answer:
Explanation:
the spheres cant be electrons as they will repel each other.
so the ans is D. Atoms
Answer:
D. Atoms
Explanation:
The spheres in this model represents the atoms. So, option (D) is correct answer.
A circular loop of wire with radius 10.0 cm is located in the xy-plane in a region of uniform magnetic field. A field of 2 T is directed in the z-direction, which is upward. (a) What is the magnetic flux through the loop
Answer:
[tex]\phi=628.3[/tex]
Explanation:
From the question we are told that
Radius [tex]r=10.0 cm[/tex]
Magnetic field[tex]B=2T[/tex]
Generally the equation for area of circular path is mathematically given by
[tex]Area=\pi r^2[/tex]
[tex]A=\pi 10^2[/tex]
[tex]A=314.15m^2[/tex]
Generally the equation for Magnetic flux is mathematically given by
[tex]\phi=BA[/tex]
[tex]\phi=2*314.15[/tex]
[tex]\phi=628.3[/tex]
What is the resistance a circult when 9V batter is connected the circult to generate 0.002A
Answer:
v= IR then you can R= v/I 9v÷0.002 = 450 Ohms
Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his car, mcartmcart, is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest.
Chuch then picks up a ball of mass mballmball and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is vc. The speed of the thrown ball relative to the ground is vb. Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie's speed relative to the ground after she catches the ball is vj.
When answering the questions in this problem, keep the following in mind:
1. The original mass mcartmcart of Chuck and his cart does not include the mass of the ball.
2. The speed of an object is the magnitude of its velocity. An object's speed will always be a non-negative quantity.
Required:
a. Find the relative speed u between Chuck and the ball after Chuck has thrown the ball.
b. What is the speed vb of the ball (relative to the ground) while it is in the air?
c. What is Chuck's speed vc (relative to the ground) after he throws the ball?
d. Find Jackie's speed vj (relative to the ground) after she catches the ball in terms of vb.
e. Find Jackie's speed vj (relative to the ground) after she catches the ball in terms of u.
Explanation:
a.)
We find the relative speed
u = vb + vc
b.)
chuck and the cart are at a rest position
mcartvc = mballvb
from part a above,
vc = u - vb
mcartu = vb(mcart + mball)
make vb the subject of the equation
[tex]vb=\frac{mcartu}{mcart +mball}[/tex]
c.)
from anser a,
vb = u - vc
then mcart vc = mball(u-vc)
[tex]vc= \frac{Mballu}{Mcart+Mball}[/tex]
d.
Mballvb = (Mcart + Mball)vj
we make vj the subject to get her relative speed
[tex]vj=\frac{MballVb}{Mcart+Mball}[/tex]
e.
given the solution in part d above,
we have
[tex]vj=\frac{MballVb}{Mcart+Mball}[/tex]
remember,
vb = u - vc
such that
[tex]Vj = \frac{Mball(u-v)}{Mball +Mcart}[/tex]
thank you!
One coulomb represents how many electrons?
a. 1 electron
b. 100 electrons
C. 6.25 quintillion electrons
d. 6.25 million-million electrons
e, none of the above
Answer:
6.24 x 1018 electrons.
Explanation:
So I think C
What is the momentum of a lab cart with a mass of 0.60 [kg] and a speed of 2.2 [m/s]?
-3.67 [kg m/s)
-0.27 [kg m/s]
-1.32 [kg m/s)
-0.82 [kg m/s)
Answer:
1.32kgm/s
Explanation:
use the formula: p=mv
The momentum of a lab cart = -1.32 kg m/s
What is momentum?It is measure of the inertia of a body/ object .It can be calculated by multiplying mass with velocity .
General formula for momentum = M = m * v
given
mass = 0.60 kg
speed = 2.2 m/s
velocity = - 2.2 m/s ( answer is in negative , since mass is a scaler quantity but velocity is a vector quantity hence , velocity can be negative )
momentum = mass * velocity
= 0.60 * (-2.2 ) = -1.32 kg m/s
The momentum of a lab cart =c) -1.32 kg m/s
learn more about momentum
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pls can anyone solve this
Answer:
3 pls give me brainliest
Explanation:
A car sitting at rest begins
accelerating at 2.40 m/s2 for
15.0 seconds. How far has the
car gone?
The distance covered by the car for the speed of 2.40 m/s² for 15 seconds is 36 meters
To find the distance the given datas are:
Speed = 2.40 m/s²
Time = 15 seconds.
What is distance?Distance is the total movement of an object without any regard to direction.Distance can be evaluated how much an object moves from starting point to the end point.The distance completely depends upon the speed and time, i,e., the object covering some area with some particular time interval with the particular speed.Formula of distance,Distance = Speed × Time.
Distance will be measured in meter, kilometer, etc..Distance is a Scalar quantity.Substituting the given datas in the formula,
Distance = 2.40 × 15
= 36 m
The Car went at the distance of about 36 meters.
Learn more about Distance,
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1./ The upward net force on the space shuttle at launch is 10,000,000 N. What is the least amount of charge you could move from its nose to the launch pad, 60 m below, and thereby prevent it from lifting off
Answer:
[tex]Q=2C[/tex]
Explanation:
From the question we are told that:
Force [tex]F=10000000N[/tex]
Distance [tex]d=60m[/tex]
Where
[tex]Q_1=Q_2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{KQ^2}{r^2}[/tex]
Where
[tex]K=9*10^9[/tex]
Therefore
[tex]Q^2=\frac{Fr^2}{K}[/tex]
[tex]Q^2=\frac{10000000*60^2}{98*10^9}[/tex]
[tex]Q=\sqrt{\frac{10000000*60^2}{9*10^9}}[/tex]
[tex]Q=2C[/tex]
Your tutor says atomic nuclei are converted to nuclear energy in a nuclear reaction. why should you seek a new tutor?
Answer:
The answer is "Momentum and net Energy(that contains mass)".
Explanation:
No matter which type of nuclear reaction, the momentum always is maintained. According to the formula, E = mc^2 of Einstein, energy conservation on the atomic level requires the change in mass of its core which may have been shared throughout the reaction as mass and energy, that's why we can say that the nuclei aren't fully converted into electrical energy rather rearranged and the mass could be changed and dispensed to the nucleon as energy.
A bullet of m=30g reaches 900m/s in a 550mm long gun barrel. What total energy does the bullet have upon exiting the gun? *
A)12150000J
B)810000J
C)12150J
D)14850000J
show your work please
Answer:
C)12150J
Explanation:
KE = (1/2)(m)(v²)
KE = .5*30g(1kg/1000g)*(900m/s)²
KE = 12,150J
what type of material is good at transferring heat?
A. A thermal insulator, such as air
B. A thermal conductor, such as water
C. A thermal insulator, such as water
D, A thermal conductor, such as air
A thermal conductor, such as water is the type of material that is good at transferring heat. Option B is correct.
What is a heat conductor?Heat conductors are substances that effectively conduct thermal energy.
The kind of substance that is effective in transferring heat is a thermal conductor, such as water.
Metals have excellent heat conductivity. Heat insulators are substances that poorly conduct thermal energy. Thermal insulators include gases like air and materials like plastic and wood.
Water conducts heat 24.17 times more quickly than air. The kind of substance that is effective in transferring heat is a thermal conductor, such as water.
Hence, option B is correct.
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How do solar systems, galaxies, and the universe show different frames of reference about space?
Answer:
Many people are not clear about the difference between our Solar System, our Milky Way Galaxy, and the Universe.
Let’s look at the basics.
Our Solar System consists of our star, the Sun, and its orbiting planets (including Earth), along with numerous moons, asteroids, comet material, rocks, and dust. Our Sun is just one star among the hundreds of billions of stars in our Milky Way Galaxy. If we shrink the Sun down to smaller than a grain of sand, we can imagine our Solar System to be small enough to fit onto the palm of your hand. Pluto would orbit about an inch from the middle of your palm.
Artist diagram of Milky Way galaxy
On that scale with our Solar System in your hand, the Milky Way Galaxy, with its 200 – 400 billion stars, would span North America (see the illustration on the right). Galaxies come in many sizes. The Milky Way is big, but some galaxies, like our Andromeda Galaxy neighbor, are much larger.
The universe is all of the galaxies – billions of them! NASA’s telescopes allow us to study galaxies beyond our own in exquisite detail, and to explore the most distant reaches of the observable universe. The Hubble Space Telescope made one of the deepest images of the universe, called the Hubble Extreme Deep Field (image at the top of this article). Soon the James Webb Space Telescope will be exploring galaxies forming at the very beginning of the universe.
You are one of the billions of people on our Earth. Our Earth orbits the Sun in our Solar System. Our Sun is one star among the billions in the Milky Way Galaxy. Our Milky Way Galaxy is one among the billions of galaxies in our Universe. You are unique in the Universe!
You can observe objects in our solar system and even see other galaxies at a star party near you-and rest assured that everything you are seeing is a part of the same universe as you!
Explanation:
A solar system is the system of celestial bodies built around a central star, the Sun. All of the system bodies, be they dwarf planets, small bodies and large planets, are held in a gravitational bond around the central star. Our solar system has eight large planets:
Four inner planets which are terrestrial, made entirely of rock and metal: Mars, Mercury, Earth and Venus;
Four outer planets which are gas and ice giants: Jupiter and Saturn (composed entirely of helium and hydrogen), Uranus and Neptune (composed of ices such as water, ammonia and methane).
The solar system also contains asteroid belts and the natural satellites of some of the planets. The trans-Neptunian region has the Kuiper belt, home to several dwarf planets, Pluto among them. Our solar system is located on the Orion Arm and is part of the Milky Way Galaxy. It was formed 4.6 billion years ago.
A galaxy is made out of billions of stars and their solar systems, held together by gravity, with a super- massive black hole at the center. Our Solar System is called the Milky Way; it is a spiral galaxy and the black hole in the center is called Sagittarius A*. Apart from the spiral shape, galaxies can also be elliptical or irregular in form. Galaxies gather in groups, clusters and super-clusters and there are billions of Galaxies in the Universe.
Some of these other galaxies are visible to the naked eye on a dark night and from places away from artificial light sources. The Andromeda Galaxy is the most recorded one throughout time and all over the world, its existence having been noted since the 10th century by Persian astronomer Al-Sufi, and having been the object of debate among other great thinkers up to the moment when the technology caught up to the discourse.
Solar System vs Galaxy
So what is the difference between a solar system and a galaxy?
A solar system represents the group of planets gravitationally bound to the central star. A galaxy has billions of stars and their solar systems. This difference in size is not only visible in the number of stars it is made out of, but also by how long it takes to cross it. It takes one light year to cross our solar system, and 100,000 light years to cross the galaxy.
While the biggest thing inside a solar system is the central sun, the biggest thing inside a galaxy is a massive black hole. The planets in a solar system orbit the sun, which is at the center, and the Sun, in turn, orbits the center of the Milky Way.
Comparison Chart
Solar system Galaxy
A group of planets orbiting the central sun A group of planetary systems whose central Suns are orbiting the center of the Galaxy
Gravitationally bound Gravitationally bound
Can be crossed in 1 light year Can be crossed in 100,000 light years
Most of the system mass is taken up by the central sun It hosts a super massive black hole, Sagittarius A*
More solar systems make up galaxies More galaxies make up the Universe
Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is ml2/3. Write your answer with one decimal place.
Answer:
2.2 s
Explanation:
Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point = mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)
So, T = 2π√(I/mgh)
T = 2π√(mL²/3 /mgL/2)
T = 2π√(2L/3g)
substituting the values of the variables into the equation, we have
T = 2π√(2L/3g)
T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))
T = 2π√(3.66 m/(29.4 m/s² ))
T = 2π√(0.1245 s² ))
T = 2π(0.353 s)
T = 2.22 s
T ≅ 2.2 s
So, the period of the man's leg is 2.2 s
a runner completed the 100 yard dash in 10 seconds. what was the runners average speed
Planet K2-116b has an Average orbital radius of 7.18x10^9 m around the star K2-116. It has a mass of about 0.257 times the mass of the earth and an orbital period of 2.7 days.
What is the orbital speed of the planet?
Determine the mass of the star.
a) v = 1.94 × 10^5 m/s
b) Ms = 2.09 × 10^24 kg
Explanation:
Given:
m = 0.257M (M = mass of earth = 5.972×10^24 kg)
= 1.535×10^24 kg
r = 7.18×10^9 m
T = 2.7 days × (24 hr/1 day) × (3600 s/1 hr)
= 2.3328×10^5 s
a) To find the orbital speed of the planet, we need to find the circumference of the planet's orbit first:
C = 2×(pi)×r
= 2(3.14)(7.18×10^9m)
= 4.51×10^10 m
The orbital speed v is then given by
v = C/T
= (4.51×10^10 m)/(2.33×10^5 s)
= 1.94 × 10^5 m/s
b) We know that centripetal force Fc is given by
Fc = mv^2/r
where v = orbital speed
r = average orbital radius
m = mass of planet
We also know that the gravitational force FG between the star K2-116 and the planet is given by
FG = GmMs/r^2
where m = mass of planet
Ms = mass of star K2-116
r. = average orbital radius
G = universal gravitational constant
= 6.67 × 10^-11 m^3/kg-s^2
Equating Fc and FG together, we get
Fc = FG
mv^2/r = GmMs/r^2
Note that m and one of the r's get cancelled out so we are left with
v^2 = GMs/r
Solving for the mass of the star Ms, we get
Ms = rv^2/G
=(7.18 × 10^9 m)(1.94 × 10^5 m/s)^2/(6.67 × 10^-11 m^3/kg-^2)
= 2.09 × 10^24 kg
If an EM wave has a wavelength of 595 nm, what is its frequency?
5.02 x 10^5 s
1.99 x 10^-6 Hz
5.02 x 10^14 Hz
1.99 x 10^-15 s
None of the above
Explanation:
1.9If an EM wave has a wavelength of 595 nm, what is its frequency9 x 10^-6 Hz
Explanation:
aah cmmon just come over to my meet
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
for lunch. If the lily pads are spaced 2.4 m apart, and Ferdinand jumps with a
speed of 5.4m/s, taking 0.60 s to go from lily pad to lily pad, at what angle
must Ferdinand make each of his jumps?
Answer:
θ = 33°
Explanation:
Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:
[tex]T = \frac{2\ u\ Sin\theta}{g} \\\\Sin\theta = \frac{Tg}{2u}[/tex]
where,
θ = launch angle = ?
T = Total time of flight = 0.6 s
g = acceleration due to gravity = 9.81 m/s²
u = launch speed = 5.4 m/s
Therefore,
[tex]Sin\theta = \frac{(0.6\ s)(9.81\ m/s^2)}{(2)(5.4\ m/s)}\\\\\theta = Sin^{-1}(0.545)[/tex]
θ = 33°
Which object has the most gravitational potential energy?
O A. A 5 kg book at a height of 2 m
B. A 8 kg book at a height of 2 m
C. An 8 kg book at a height of 3 m
D. An 5 kg book at a height of 3 m
help pls
Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.
Explanation:
Gravitational potential energy is the product of mass of object, height of object and gravitational field.
So, formula to calculate gravitational potential energy is as follows.
U = mgh
where,
m = mass of object
g = gravitational field = [tex]9.81 m/s^{2}[/tex]
h = height of object
(A) m = 5 kg and h = 2m
Therefore, its gravitational potential energy is calculated as follows.
[tex]U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J (1 J = kg m^{2}/s^{2})[/tex]
(B) m = 8 kg and h = 2 m
Therefore, its gravitational potential energy is calculated as follows.
[tex]U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J (1 J = kg m^{2}/s^{2})[/tex]
(C) m = 8 kg and h = 3 m
Therefore, its gravitational potential energy is calculated as follows.
[tex]U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J (1 J = kg m^{2}/s^{2})[/tex]
(D) m = 5 kg and h = 3 m
Therefore, its gravitational potential energy is calculated as follows.
[tex]U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J (1 J = kg m^{2}/s^{2})[/tex]
Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.
what are the laws of Physics
Answer:
thermodynamics
Explanation:
The laws of thermodynamics define a group of physical quantities, such as temperature, energy, and entropy, that characterize thermodynamic systems in thermodynamic equilibrium.
On the way home from school, Mr. X drives the first 10 miles at 55 mi/hr, the next 20 miles at 70 mi/hr, and the last 5 miles at 35 mi/hr. How far does Mr. X live from school?
Given :
On the way home from school, Mr. X drives the first 10 miles at 55 mi/hr, the next 20 miles at 70 mi/hr, and the last 5 miles at 35 mi/hr.
To Find :
How far does Mr. X live from school.
Solution :
To find the distance between Mr. X residence from school is simply given by summing all the distance he travelled .
So, distance = 10 + 20 + 5 miles
distance = 35 miles.
Hence, this is the required solution.
When you press on a circular table of radius 0.5 m, with a force of 20N, what will be exerted pressure?
50pa
25pa
40N/m
10N.m
Answer and I will give you brainiliest
Answer:
Note the word "circular".
Meaning the area would be found using that of a circle = πr²
Recall
Pressure = Force/Area
Area = π x 0.5² = 0.785m²
Pressure = 20/0.785
= 25pa ( to the nearest whole number)
Pressure = Force/Area
From the question,we have force ryt? so now let's find the area.As the table is circular, we have to use the formula of area of circle;
= πr²
= 3.14 × 0.5 × 0.5
= 3.14 × 0.25
= 0.785 m²
Put the values of force and area in formula;
= 20/0.785
= 20000/785
= 25.45 Pa
= 25 Pa (Approx)
Q5: An ice skater moving at 12 m/s coasts
to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?
u = 0.077
Explanation:
Work done by friction is
Wf = ∆KE + ∆PE
-umgx = ∆KE,. ∆PE =0 (level ice surface)
-umgx = KEf - KEi = -(1/2)mv^2
Solving for u,
u = v^2/2gx
= (12 m/s)^2/2(9.8 m/s^2)(95 m)
= 0.077
Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.
Given-
velocity of the ice skater is 12 m/ sec.
Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.
[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]
The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,
[tex]W_{f}=\bigtriangleup KE +0[/tex]
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,
[tex]W_{f} =u_{f} mgx[/tex]
Here, [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .
Equate the value of kinetic energy and work done of friction for further result, we get,
[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]
[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]
[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]
[tex]u_{f} =0.077[/tex]
Hence, the coefficient of kinetic friction between the ice and skates is 0.077.
For more about the friction, follow the link below-
https://brainly.com/question/13357196
If a cat can exert 2000N Of Force to move a trailer 50m is 20 seconds how much power did the car use ?
им putins брат, почему вы обманываете нашу систему образования, Это теперь запрещено в России.
The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 54 km/s . To the crew's great surprise, a Klingon ship is 110 km directly ahead, traveling in the same direction at a mere 22 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 3.4 s. The Enterprise's computers react instantly to brake the ship. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship?
Answer:
The acceleration will be "-5.12 km/s²".
Explanation:
The given values are:
This might be defined as a continuous inanimate object at either the speed between such organization as well as the Klingon boat (54 - 22) km/s.
The values given in the question are:
= 110 km
Final velocity will be:
= 0
By using to the third equation of motion, we get
⇒ [tex]v^2=u^2+2as[/tex]
[tex]0=(54-22)^2+2\times a\times 100[/tex]
[tex]0=(32)^2+2\times a\times 100[/tex]
[tex]0=1024+2\times a\times 100[/tex]
[tex]0=1024+200a[/tex]
[tex]-1024=200a[/tex]
[tex]a=-\frac{1024}{200}[/tex]
[tex]=-5.12 \ km/s^2[/tex]