A perfect electric dipole doesn't feel a net force in a uniform electric field.
To determine whether this statement is true or false, we need to know about the behavior of perfect dipole in an electric field.
How does a perfect dipole behave in uniform electric field?When a perfect dipole is kept in uniform electric field, both the charges of the dipole experience same amount of force.But these forces are not linear, so they collectively rotate the dipole.So, in a uniform electric field, the perfect dipole only experiences torque about its center of axis.How does a perfect dipole behave in non-uniform electric field?When non-uniform electric field is applied, both charges of the perfect dipole experience different amounts of force.So the dipole experiences a torque as well as a net linear force.Thus, we can conclude that a perfect electric dipole doesn't feel a net force in a uniform electric field.
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1. What do you need to change the momentum of a system?
2. What are the features of a typical modern running shoe? How does this change the ground reaction force during a heel strike when running with shoes compared to a heel strike when running without shoes?
3. When running with shoes how does the ground reaction force change from a heel strike run to a forefoot strike run?
A hot air balloon is ascending at a rate of 12 m/s. when it is 80m above the ground, a package is dropped over the side of he passenger basket. What is the speed of the package just before it hits the ground?
Answer:
41.4 m/s
Explanation:
Consider downward direction of motion as negative
v₀ = initial velocity of the package as it is dropped over the side = 12 m/s
v = final velocity of the package just before it hits the ground
y = vertical displacement of the package = - 80 m
a = acceleration = - 9.8 m/s²
Using the kinematics equation
v² = v₀² + 2 a y
v² = 12² + 2 (- 9.8) (- 80)
v² = 144 + 1568
v = - 41.4 m/s
The negative sign indicates the downward direction of motion.
Hence the speed of package comes out to be 41.4 m/s
You are driving directly behind a pickup truck, going at the same speed as the truck. A crate falls from the bed of the truck to the road. (a) Will your car hit the crate before the crate hits the road if you neither brake nor swerve? (b) During the fall, is the horizontal speed of the crate more than, less than, or the same as that of the truck?
Final answer:
The crate will hit the road before your car hits the crate. The horizontal speed of the crate is the same as that of the truck.
Explanation:
(a) If you are driving directly behind a pickup truck at the same speed and neither brake nor swerve, the crate will hit the road before your car hits the crate. This is because the crate and your car are both traveling at the same horizontal speed, and the crate will have a shorter distance to fall than your car would have to travel to reach the crate.
(b) During the fall, the horizontal speed of the crate is the same as that of the truck. This is because both the truck and the crate are moving at the same speed horizontally, and gravity acts only vertically on the falling crate.
A car moves a distance of 50.0 km West, followed by a distance of 64.9 km North What was the magnitude of the displacement of the car, in units of kilometers?
Answer:
Displacement of the car, XY = 81.92 km
Explanation:
From the attached figure,
OX = 64.9 km (in north direction)
OY = 50 km (in west direction)
We have to find the displacement of the car. The shortest path covered by a car is called displacement of the car. Here, XY shows the displacement of the car :
Using Pythagoras equation as :
[tex]XY^2=OX^2+OY^2[/tex]
[tex]XY^2=(64.9\ km)^2+(50\ km)^2[/tex]
XY = 81.92 km
Hence, the displacement of the car is 81.92 km.
The magnitude of the car's displacement is calculated using the Pythagorean theorem by treating the westward and northward movements as the sides of a right-angled triangle. The displacement, representing the hypotenuse of that triangle, is found to be approximately 81.93 km.
The student has asked to find the magnitude of displacement of a car that moves 50.0 km West and then 64.9 km North.
Calculating Displacement
To calculate the resultant displacement, we treat the distances as vector quantities and use the Pythagorean theorem. The car's westward and northward movements are at right angles to each other, so we can draw this as a right-angled triangle where the westward distance is one side (50.0 km), the northward distance is the other side (64.9 km), and the hypotenuse will be the displacement.
Using the Pythagorean theorem:
Displacement = \/(westward distance)^2 + (northward distance)^2Displacement = \/(50.0 km)^2 + (64.9 km)^2Displacement = \/(2500 + 4212.01) km^2Displacement = \/(6712.01) km^2Displacement = 81.93 kmThe magnitude of the car's displacement, therefore, is approximately 81.93 km.
A ball is thrown straight up with an initial speed of 16.9 m/s. At what height above its initial position will the ball have one‑half its initial speed?
Answer:
10.9 m
Explanation:
We can solve the problem by using the law of conservation of energy.
The initial mechanical energy is just the kinetic energy of the ball:
[tex]E = K_i = \frac{1}{2}mu^2[/tex]
where m is the mass of the ball and u = 16.9 m/s the initial speed.
At a height of h, the total mechanical energy is sum of kinetic energy and gravitational potential energy:
[tex]E=K_f + U_f = \frac{1}{2}mv^2 + mgh[/tex]
where v is the new speed, g is the gravitational acceleration, h is the height of the ball.
Due to the conservation of energy,
[tex]\frac{1}{2}mu^2 = \frac{1}{2}mv^2 +mgh\\u^2 = v^2 + 2gh[/tex] (1)
Here, at a height of h we want the speed to be 1/2 of the initial speed, so
[tex]v=\frac{1}{2}u[/tex]
So (1) becomes
[tex]u^2 = (\frac{u}{2})^2+2gh\\\frac{3}{4}u^2 = 2gh[/tex]
So we can find h:
[tex]h=\frac{3u^2}{8g}=\frac{3(16.9 m/s)^2}{8(9.8 m/s^2)}=10.9 m[/tex]
Final answer:
To find the height where the ball has one-half its initial speed, we can use the equations vf = v0 + gt and d = v0t - 0.5gt2.
Explanation:
To find the height above its initial position where the ball has one-half its initial speed, we need to use the fact that the initial velocity (v0) of the ball is 16.9 m/s. At the highest point of the ball's trajectory, the velocity will be zero. We can use the formula vf = v0 + gt, where vf is the final velocity, g is the acceleration due to gravity, and t is the time it takes for the ball to reach its highest point.
By substituting vf = 0 and v0 = 16.9 m/s, we can solve for t. Once we have the value of t, we can use the equation d = v0t - 0.5gt2 to calculate the height (d) above the initial position where the ball will have one-half its initial speed.
By substituting the calculated value of t into the equation, we can find the value of d.
When during new product development is Design For Manufacture and Assembly (DFMA) most effective? a) At all times b) During production c) During process design and development d) During product design and development e) Before design
Answer:most likely E
Explanation:
Why would anybody do something after design there done with there work after that
A test charge is A a very small negative charge with little miee B a point charge of q 100 C C a spbere of charge D. a very amall positive charge with little s
Answer:
D. a very small positive charge with little s
Explanation:
A test charge is a very small charge with positive value which do not disturb the electric field exist in the region
So test charge is to find out the strength of electric field that exist in the region.
If the magnitude of test charge is large then it will change the strength of the existing electric field and due to this the value of the force will be altered.
So here in this case the test charge must be small as well as it must be positive nature
A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. What is the magnitude of the ball's velocity just before it hits the ground?
The ball's position vector has components
[tex]x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ t[/tex]
[tex]y=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2[/tex]
where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. The ball hits the ground when [tex]y=0[/tex]:
[tex]0=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2\implies t=1.22\,\mathrm s[/tex]
The ball's velocity vector has components
[tex]v_x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ[/tex]
[tex]v_y=\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ-gt[/tex]
so that after 1.22 s, the velocity vector is
[tex]\vec v=(6.13\,\vec\imath-6.79\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]
and the magnitude is
[tex]\|\vec v\|=\sqrt{6.13^2+(-6.79)^2}\,\dfrac{\rm m}{\rm s}=\boxed{9.14\dfrac{\rm m}{\rm s}}[/tex]
A disk with a radius of R is oriented with its normal unit vector at an angle\Theta with respect to a uniform electric field. Which of the following represent the electric flux through the disk? A: E(πR^2)cosϕ B: E(πR^2)sinΘ C: E(πR^2)cosΘ D: E(2πR)sinΘ E: E(2πR)cosΘ F: E(πR^2)sinϕ
Answer:
option (A)
Explanation:
electric flux is defined as the number of electric field lines which crosses through any area.
It is given by
Ф = E . A (It is the dot product of electric field vector and area vector)
According to the question, the angle between electric filed vector and area vector is θ.
So, electric flux
Ф = E x π R^2 Cosθ
The electric flux through a disk in a uniform electric field is represented by E(πR^2)cosΘ, so the correct answer is C: E(πR²)cosΘ.
The question is asking about the electric flux through a disk when the disk's normal is oriented at an angle Θ with respect to a uniform electric field. Electric flux is given by the product of the electric field strength, the area through which the field is passing, and the cosine of the angle between the field and the normal to the surface. The formula for the electric flux through a surface is Φ = E * A * cos(Θ), where E is the electric field strength, A is the area of the surface, and Θ is the angle between the electric field and the normal to the surface. For a disk with radius R, the area is πR². Thus, the correct answer for the electric flux through the disk is C: E(πR²)cosΘ.
a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. the shot hits the ground 2.08 s later. you can ignore air resistance. how far did she throw the shot?
Answer:
15.7 m
Explanation:
The range (horizontal distance) of the projectile is determined only by its horizontal motion.
The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:
[tex]v_x = v cos \theta[/tex]
where
v = 12.0 m/s is the initial velocity
[tex]\theta=51.0^{\circ}[/tex] is the angle between the direction of v and the horizontal
Substituting,
[tex]v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s[/tex]
We know that the projectile hits the ground in a time of
t = 2.08 s
so the horizontal distance covered is
[tex]d = v_x t = (7.55 m/s)(2.08 s)=15.7 m[/tex]
Final answer:
The shot putter threw the shot approximately 15.71 meters by using the horizontal component of the initial velocity and the time of flight.
Explanation:
To calculate the distance the shot putter threw the shot, we need to break down the velocity into its horizontal and vertical components. Given that the shot was released with a velocity of 12.0 m/s at an angle of 51.0 degrees above the horizontal, we can use trigonometry to find these components.
The horizontal velocity (vx) is given by vx = v * cos(θ) = 12.0 m/s * cos(51.0) = 7.55 m/s. The vertical velocity (vy) is vy = v * sin(θ) = 12.0 m/s * sin(51.0) = 9.35 m/s.
Since there's no acceleration in the horizontal direction (ignoring air resistance), this component of the velocity will remain constant until the shot hits the ground. Thus, the horizontal distance (range) the shot travels is simply the product of the horizontal velocity and the time it's in the air, given by Range = vx * t = 7.55 m/s * 2.08 s = 15.71 meters. So, the shot putter threw the shot approximately 15.71 meters.
g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force acting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 0.900 N. what is the maximum height reached by the stone?
Answer:
Height reached will be 28.35 m
Explanation:
Here we can use the work energy theorem to find the maximum height
As we know by work energy theorem
Work done by gravity + work done by friction = change in kinetic energy
[tex]-mgh - F_f h = 0 - \frac{1}{2}mv_i^2[/tex]
now we will have
[tex]-1.60(9.8)(h) - 0.900(h) = - 470[/tex]
[tex]-16.58 h = -470[/tex]
[tex]h = 28.35 m[/tex]
so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 81.1 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy U that flows through the area of 0.0253 m^2 over an interval of 19.9 s, assuming that the area is perpendicular to the direction of wave propagation.
Answer:
U = 4.39 J
Explanation:
Electric field energy stored in the medium or vacuum is given as
[tex]U = \frac{1}{2}\epsilon_0 E^2 V[/tex]
here we know that
[tex]\epsilon_0 = 8.85 \times 10^{-12} [/tex]
E = 81.1 V/m
V = volume
[tex]V = (0.0253)(speed \times time)[/tex]
[tex]V = (0.0253)(3\times 10^8 \times 19.9)[/tex]
[tex]V = 1.51 \times 10^8 m^3[/tex]
now from above formula we have
[tex]U = \frac{1}{2}(8.85 \times 10^{-12})(81.1)^2(1.51 \times 10^8)[/tex]
[tex]U = 4.39 J[/tex]
A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?
Answer:
The angle the wire make with respect to the magnetic field is 30°
Explanation:
It is given that,
Length of straight wire, L = 0.6 m
Current carrying by the wire, I = 2 A
Magnetic field, B = 0.3 T
Force experienced by the wire, F = 0.18 N
Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :
[tex]F=ILB\ sin\theta[/tex]
[tex]\theta=sin^{-1}(\dfrac{F}{ILB})[/tex]
[tex]\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\times 0.3\ T})[/tex]
[tex]\theta=30^{\circ}[/tex]
So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.
A chain link fence should be cut quickly with a
Answer: it should be cut with a chainsaw
Explanation:
A bolt cutter is usually the preferred tool to use to cut through a chain link fence quickly, taking into account the thickness and hardness of the chain link fence. Safety precautions should be taken while using such tools.
Explanation:To cut through a chain link fence quickly without undue strain, the preferred tool is typically a bolt cutter. Bolt cutters possess the strength and design needed to snip through metal links easily. They come in various sizes, and the size needed would depend on the thickness and hardness of the chain link fence. Ideally, a medium-sized bolt cutter would be used for a standard fence. However, it's advisable to wear protective gear while using such tools, as the cut metal links might be razor-sharp and could cause injuries.
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A rope attached to a load of 175 kg bricks Ilifts the bricks with a steady acceleration of 0.12.m/s^2 straight up. What is the tension in the rope? (a)2028N (b)1645 N (c) 1894 N (d) 1976 N (e) 1736 N (f) 1792 N
Answer:
Tension, T = 1736 N
Explanation:
It is given that,
Mass of bricks, m = 175 kg
A rope is attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12 m/s² in vertically upwards direction. let T is the tension in the rope. Using second equation of motion as :
T - mg = ma
T = ma + mg
T = m(a + g)
T = 175 kg ( 0.12 m/s² + 9.8 m/s² )
T = 1736 N
Hence, the tension in the wire is 1736 N.
Answer:
The tension in the rope is 1736 N.
(e) is correct option.
Explanation:
Given that,
Mass of bricks = 175 kg
Acceleration = 0.12 m/s²
Let T is the tension in the rope.
A rope attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12.m/s^2 in vertically upward direction.
Using equation of balance
[tex]T-mg=ma[/tex]
[tex]T=mg+ma[/tex]
[tex]T=175(9.8+0.12)[/tex]
[tex]T= 1736\ N[/tex]
Hence, The tension in the rope is 1736 N.
If a converging lens forms a real, inverted image 24.0 cm to the right of the lens when the object is placed 48.0 cm to the left of a lens, determine the focal length of the lens
Answer:
Focal length, f = 16 cm
Explanation:
Image distance, v = 24 cm
Object distance, u = -48 cm
We need to find the focal length of the lens. It can be determined using the lens formula as :
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex]
[tex]\dfrac{1}{24\ cm}-\dfrac{1}{(-48\ cm)}=\dfrac{1}{f}[/tex]
f = 16 cm
So, the focal length of the converging lens is 16 cm. Hence, this is the required solution.
Answer:
f = 16 cm
Explanation:
If a converging lens forms a real, inverted image 24.0 cm to the right of the lens when the object is placed 48.0 cm to the left of a lens, the focal length of the lens is 16 cm.
A simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.2 m/s in 2.14 s, what will be his total time?
Final answer:
The sprinter's total time in the 100 m dash is 10 seconds.
Explanation:
In this problem, the sprinter reaches his top speed of 11.2 m/s in 2.14 seconds.
To find the total time, we need to consider two parts: the time it takes to reach the top speed and the time it takes to cover the remaining distance at that speed.
First, calculate the time it takes to reach the top speed using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values: 11.2 = 0 + a(2.14), we can solve for a to find a = 5.23 m/s².
Now, to find the time it takes to cover the remaining distance at the top speed, we can use the equation:
t = d / v
where d is the distance and v is the velocity.
Since the total distance is 100 m, subtracting the distance covered during the acceleration phase, we get the remaining distance as 100 - (0.5)(5.23)(2.14)² = 88.2 m.
Dividing this distance by the top speed of 11.2 m/s, we find t = 88.2 / 11.2 = 7.86 s.
To find the total time, we add the time it takes to reach the top speed and the time it takes to cover the remaining distance, so the sprinter's total time is 2.14 s + 7.86 s = 10 s.
A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume free fall : Part A) In how many seconds after being thrown does the rock strike the ground? Part B) What is the speed of the rock just before it strikes the ground?
The rock strikes the ground after approximately 3.67 seconds. The speed of the rock just before it strikes the ground is approximately 17.0 m/s.
Explanation:To find the time it takes for the rock to strike the ground, we can use the equation for vertical motion. Assuming negligible air resistance, the initial velocity is 18.0 m/s and the vertical displacement is 50.0 m.
Using the equation y = yo + vyo*t - 1/2*g*t^2, where y is the final position, yo is the initial position, vyo is the initial velocity, g is the acceleration due to gravity, and t is the time, we can solve for t. Plugging in the values, we get:
50.0 = 0 + 18.0*t - 1/2*9.8*t^2
After rearranging the equation and solving the quadratic equation, we find that t = 3.67 seconds.
To find the speed of the rock just before it strikes the ground, we can use the equation for vertical motion. The final velocity v is equal to the initial velocity vyo - g*t. Plugging in the values, we get:
v = 18.0 - 9.8*3.67
Calculating the value, we find that the speed of the rock just before it strikes the ground is approximately 17.0 m/s.
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Describe Lenz's law.
Answer:
Explanation:
Lenz law is used to find the direction of induced emf in the coil.
It state taht the direction of induced emf in the coil is such that it always opposes the change due to which it is produced.
Suppose there is a coil and a north pole of the magnet comes nearer to the coil. Due to changing magnetic flux an induced emf is developed in the coil whose direction is such that the north pole moves away. That means this face of the coil behaves like a north pole and the current flows at this face is in anticlockwise wise direction.
Final answer:
Lenz's law is a manifestation of the conservation of energy in physics. It states that the direction of the induced electromotive force (emf) drives current around a wire loop to always oppose the change in magnetic flux that causes the emf. Lenz's law ensures that the induced current produces a magnetic field that tries to cancel out the change in flux caused by a changing magnetic field.
Explanation:
Lenz's law is a manifestation of the conservation of energy in physics. It states that the direction of the induced electromotive force (emf) drives current around a wire loop to always oppose the change in magnetic flux that causes the emf. This means that when there is a change in the magnetic field through a circuit, the induced current will create a magnetic field that acts against the change.
For example, if a magnet is brought near a wire loop, the induced current will flow in such a way that it creates a magnetic field that opposes the motion of the magnet. This is because the changing magnetic field induces an emf in the wire loop, and Lenz's law ensures that the induced current produces a magnetic field that tries to cancel out the change in flux caused by the magnet.
The nuclear potential that binds protons and neutrons in the nucleus of an atom is often approximated by a square well. Imagine a proton conned in an innite square well of length 105 nm, a typical nuclear diameter. Calculate the wavelength and energy associated with the photon that is emitted when the proton undergoes a transition from the rst excited state (n 2) to the ground state (n 1). In what region of the electromagnetic spectrum does this wavelength belong?
3. The nuclear potential that binds protons and neutrons in the nucleus of an atom
is often approximated by a square well. Imagine a proton confined in an infinite
square well of length 10−5 nm, a typical nuclear diameter. Calculate the wavelength
and energy associated with the photon that is emitted when the proton undergoes a
transition from the first excited state (n = 2) to the ground state (n = 1). In what
region of the electromagnetic spectrum does this wavelength belong?
Answer 3
We are given that,
Length of square well = L = 10−5
nm = 10−14 m.
Energy of proton in state n is given by,
En =
π
2n
2~
2
2mpL2
,
where L is the width of the square well.
⇒ E1 =
π
2~
2
2mpL2
E2 =
4π
2~
2
2mpL2
·
A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^?
Answer:
The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]
Explanation:
Given that,
Mass [tex]m = 1.81\times10^{-3}\ kg[/tex]
Velocity [tex]v = (3.00\times10^{4}\ m/s)j[/tex]
Charge [tex]q = 1.22\times10^{-8}\ C[/tex]
Magnetic field [tex] B= (1.63\hat{i}+0.980\hat{j})\ T[/tex]
We need to calculate the acceleration of the particle
Formula of the acceleration is defined as
[tex]F = ma=q(v\times B)[/tex]
[tex]a =\dfrac{q(v\times B)}{m}[/tex]
We need to calculate the value of [tex]v\times B[/tex]
[tex]v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})[/tex]
[tex]v\times B=4.89\times10^{4}[/tex]
Now, put the all values into the acceleration 's formula
[tex]a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}[/tex]
[tex]a= -0.3296\ \hat{k}\ m/s^2[/tex]
Negative sign shows the opposite direction.
Hence, The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]
The magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]
Explanation:A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^?
A charged particle is a particle with an electric charge. Whereas electric charge is the matter physical property that causes to experience a force when placed in an electromagnetic field. Uniform magnetic field is the condition when magnetic field lines are parallel then magnetic force experienced by an object is same at all points in that field
From Newton's second law, the force is given by:
[tex]F=ma[/tex]
Magnetic force is
[tex]F= qv \times B[/tex]
[tex]ma = qv \times B[/tex]
[tex]a = \frac{qv \times B}{m}[/tex]
Subsituting with the givens above we get
[tex]a = \frac{(1.22 \times 10^{-8} C) (3 \times 10^{4} m/s) (1.63 T ) (\hat{j} \times \hat{i})}{1.81 \times 10^{-3} kg} = -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]
Therefore the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]
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The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7300 L of blood. Assume that the work done by the heart is equal to the work required to lift that amount of blood a height equal to that of the average citizen of Atlantic Falls, approximately 1.6 m. The density of blood is 1050 kg/m3. What is the heart's power output in watts?
Answer:
1.39 W
Explanation:
The volume of blood is
[tex]V=7300 L = 7.3 m^3[/tex]
the density is
[tex]\rho = 1050 kg/m^3[/tex]
So the total mass of blood lifted in one day is
[tex]m=\rho V=(1050 kg/m^3)(7.3 m^3)=7665 kg[/tex]
So the total work done is:
[tex]W=mgh=(7665 kg)(9.8 m/s^2)(1.6 m)=1.2\cdot 10^5 J[/tex]
The total time taken is one day, so
[tex]t=24 h = 86400 s[/tex]
So the power output is
[tex]P=\frac{W}{t}=\frac{1.2\cdot 10^5 J}{86400 s}=1.39 W[/tex]
A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below the horizontal, what is the magnetic flux through the desk surface?
Answer:
The magnetic flux through the desk surface is [tex]6.6\times10^{-4}\ T-m^2[/tex].
Explanation:
Given that,
Magnetic field B = 0.42 T
Angle =68°
We need to calculate the magnetic flux
[tex]\phi=BA\costheta[/tex]
Where, B = magnetic field
A = area
Put the value into the formula
[tex]\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}[/tex]
[tex]\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927[/tex]
[tex]\phi=6.6\times10^{-4}\ T-m^2[/tex]
Hence, The magnetic flux through the desk surface is [tex]6.6\times10^{-4}\ T-m^2[/tex].
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.6 m/s. (a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.) kg · m/s (b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist. N
(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:
[tex]I= \Delta p = m \Delta v = m (v-u)[/tex]
where
m is the mass of the object
[tex]\Delta v[/tex] is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is
[tex]I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s[/tex]
(b) 155 N
The impulse can also be rewritten as
[tex]I=F \Delta t[/tex]
where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
[tex]\Delta t[/tex] is the duration of the collision
In this situation, we have
[tex]\Delta t = 0.06 s[/tex]
So we can re-arrange the equation to find the magnitude of the average force:
[tex]F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N[/tex]
If a ball is thrown vertically upward with a velocity of 144 ft/s, then its height after t seconds is s = 144t − 16t2. (a) What is the maximum height reached by the ball? ft (b) What is the velocity of the ball when it is 320 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 320 ft above the ground on its way down? ft/s
(a) 168.2 ft/s
The vertical position of the ball is given by
[tex]s = 144t - 16t^2[/tex]
where t is the time.
By differentiating this expression, we find the velocity:
[tex]v = 144-32 t[/tex]
The maximum height is reached when the velocity is zero, so:
[tex]0 = 144 - 32 t[/tex]
From which we find
[tex]t = \frac{144}{32}=4.5 s[/tex]
And substituting this value into the equation for s, we find the maximum height:
[tex]s = 144(4.5 s)-16(4.5 s)^2=324 ft[/tex]
(b) 16 ft/s
We want to find the velocity of the ball when the position of the ball is
s = +320 ft
Substituting into the equation for the position,
[tex]320 = 144t-16t^2[/tex]
[tex]16t^2 -144t +320 = 0[/tex]
Solving for t, we find two solutions:
t = 4 s
t = 5 s
The first one corresponds to the instant in which the ball is still on its way up: Substituting into the equation for the velocity, we find the velocity of the ball at that time
[tex]v = 144 - 32 t=144- 32(4 s)=16 ft/s[/tex]
(c) -16 ft/s
Now we want to find the velocity of the ball when the position of the ball is
s = +320 ft
but on its way down. In the previous part, we found
t = 4 s
t = 5 s
So the second time corresponds to the instant in which the ball is at s = 320 ft but on the way down.
Substituting t = 5 s into the equation for the velocity, we find:
[tex]v = 144 - 32 t=144- 32(5 s)=-16 ft/s[/tex]
And the negative sign means the direction is downward.
The answers for the ball thrown vertically upward with a velocity of 144 ft/s and with a height after t seconds of s = 144t - 16t² are:
a) The maximum height reached by the ball is 324 ft.
b) The velocity of the ball when it is 320 ft above the ground on its way up is 16 ft/s.
c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s.
a) The maximum height reached by the ball can be calculated with the given equation:
[tex] s = 144t - 16t^{2} [/tex] (1)
Where:
s: is the height
t: is the time
We can find the time with the following equation:
[tex] v_{f} = v_{i} - gt [/tex] (2)
Where:
[tex] v_{f} [/tex]: is the final velocity = 0 (at the maximum height)
[tex] v_{i} [/tex]: is the initial velocity = 144 ft/s
g: is the acceleration due to gravity = 32 ft/s²
Solving equation (2) for t and entering into equation (1), we can find the maximum height:
[tex]s = 144t - 16t^{2} = 144(\frac{v_{i}}{g}) - 16(\frac{v_{i}}{g})^{2} = 144(\frac{144 ft/s}{32 ft/s^{2}}) - 16(\frac{144 ft/s}{32 ft/s{2}})^{2} = 324 ft[/tex]
Hence, the maximum height is 324 ft.
b) To find the velocity of the ball when it is 320 ft above, we can use the following equation:
[tex] v_{f}^{2} = v_{i}^{2} - 2gs [/tex]
[tex]v_{f}^{2} = (144 ft/s)^{2} - 2*32 ft/s^{2}*320 ft[/tex]
The above equation has two solutions:
[tex]v_{f_{1}} = 16 ft/s[/tex]
[tex]v_{f_{2}} = -16 ft/s[/tex]
Since the question is for the velocity of the ball on its way up and considering the way up as the positive direction, the answer is the positive value [tex]v_{f_{1}} = 16 ft/s[/tex].
c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s (we take the negative value calculated above, [tex] v_{f_{2}}[/tex]). We consider the way down as the negative direction.
Find more about vertical motion here https://brainly.com/question/13665920?referrer=searchResults
I hope it helps you!
When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 936 N and the drag force has a magnitude of 1032 N. The mass of the sky diver is 95.5 kg. Take upward to be the positive direction. What is his acceleration, including sign?
Answer: [tex]1.0052m/s^{2}[/tex]
Explanation:
Assuming there is only force in the y-component, the total net force [tex]F_{y}[/tex] acting on the parachute and the sky diver is:
[tex]F_{y}=F_{D}-W[/tex] (1)
Where:
[tex]F_{D}=1032N[/tex] is the drag force acting upwards
[tex]W=936N[/tex] is the weight of the sky diver acting downwards, hence with negative sign
Then:
[tex]F_{y}=1032N-936N=96N[/tex] (2) This is the total net force excerted on the system parachute-sky diver, and the fact it is positive means is upwards
Now, according Newton's 2nd Law of Motion the force is directly proportional to the mass [tex]m[/tex] and to the acceleration [tex]a[/tex] of a body:
[tex]F_{y}=m.a[/tex] (3)
Where [tex]m=95.5kg[/tex] is the mass of the diver.
Substituting the known values and finding [tex]a[/tex]:
[tex]a=\frac{F_{y}}{m}[/tex] (4)
[tex]a=\frac{96N}{95.5kg}[/tex] (5)
Finally:
[tex]a=1.0052m/{s^{2}}\approx 1m/s^{2}[/tex] This is the acceleration of the sky diver. Note it has a positive sign, which means its direction is upwards.
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).
(a) -2451 N
We can start by calculating the acceleration of the car. We have:
[tex]u=85.0 km/h = 23.6 m/s[/tex] is the initial velocity
v = 0 is the final velocity of the car
d = 125 m is the stopping distance
So we can use the following equation
[tex]v^2 - u^2 = 2ad[/tex]
To find the acceleration of the car, a:
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2[/tex]
Now we can use Newton's second Law:
F = ma
where m = 1100 kg to find the force exerted on the car in order to stop it; we find:
[tex]F=(1100 kg)(-2.23 m/s^2)=-2451 N[/tex]
and the negative sign means the force is in the opposite direction to the motion of the car.
(b) [tex]-1.53\cdot 10^5 N[/tex]
We can use again the equation
[tex]v^2 - u^2 = 2ad[/tex]
To find the acceleration of the car. This time we have
[tex]u=85.0 km/h = 23.6 m/s[/tex] is the initial velocity
v = 0 is the final velocity of the car
d = 2.0 m is the stopping distance
Substituting and solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2[/tex]
So now we can find the force exerted on the car by using again Newton's second law:
[tex]F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N[/tex]
As we can see, the force is much stronger than the force exerted in part a).
he electric fux through a surface is zero. Thereloee there are no chargm inside the ace A. True, if there in no fus theee ran be no chargs B. Fase, flux hae nothing to do with nclosed charge C. True, thst is how Ewoks control the world banks D. Fabe, the sum of all thargrs inside can be aro
Answer:
Option (D)
Explanation:
According to the Gauss theorem in electrostatics, the electric flux passing through any surface is equal to the one divided by epsilon note ties the total cahrge enclosed in the surface.
As the flux is zero it means the enclosed charge is zero. It means the sum of the total cahrge inside is zero.
If the speed of light in a vaccum is c, the speed of light in a medium like glass with an index of refraction of 1.5 is : (a) 3c/2 (b) 3c (c) 2c/3 (d) 9c/4 (e) 4c/9 Please explain in detail why it is the answer you have chosen
Answer:
The speed of light in the medium is [tex]\dfrac{2c}{3}[/tex]
(c) correct option.
Explanation:
Given that,
Speed of light in vacuum = c
Refraction index = 1.5
We need to calculate the value of speed of light in the medium
The refractive index is equal to the speed of light in vacuum divide by the speed of light in medium.
Using formula of refractive index
[tex]\mu = \dfrac{c}{v}[/tex]
[tex]v=\dfrac{c}{\mu}[/tex]
Where, c = speed of light in vacuum
v = speed of light in medium
Put the value into the formula
[tex]v=\dfrac{c}{1.5}[/tex]
[tex]v=\dfrac{2c}{3}[/tex]
Hence, The speed of light in the medium is [tex]\dfrac{2c}{3}[/tex]
A ball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. Neglect any effects due to air resistance. How much time elapses between the throwing of the ball and its return to the original launch point?
Answer:
4 s
Explanation:
u = 19.6 m/s, g = 9.8 m /s^2
Let the time taken to reach the maximum height is t.
Use first equation of motion.
v = u + at
At maximum height, final velocity v is zero.
0 = 19.6 - 9.8 x t
t = 19.6 / 9.8 = 2 s
As the air resistance be negligible, is time taken to reach the ground is also 2 sec.
So, total time taken be the ball to reach at original point = 2 + 2 = 4 s
The ball takes 4 seconds to return to its original launch point.
Explanation:To find the time it takes for the ball to return to its original launch point, we need to determine the total time it takes for the ball to reach its maximum height and come back down. The ball is shot straight up, which means its initial velocity is positive and its final velocity is negative (when it returns to the launch point). We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed. In this case, the final velocity is -19.6 m/s (negative because it's going down), the initial velocity is 19.6 m/s, and the acceleration is -9.8 m/s^2 (due to gravity). Plugging the values into the equation and solving for t:
v = u + at
-19.6 = 19.6 - 9.8t
-39.2 = -9.8t
t = -39.2 / -9.8
t = 4 seconds
It takes 4 seconds for the ball to return to its original launch point.