If an electron moves in a direction perpendicular to the same magnetic field with this same linear speed,
what is the radius of its circular orbit?
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z
The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.
Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar
The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;
= 1000 MB - 976 MB
= 24 MB
Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
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A football is kicked with an initial velocity of 50.0 m/s, 60° above the horizontal line. Find the following: The time it takes to reach the maximum height; The maximum height reached by the projectile; The time of flight; and The range of projectile.
Answer:
Explanation:
Initial vertical velocity
vy₀ = 50.0sin60 = 43.3 m/s
This initial velocity is reduced to zero by gravity in a time of
t = v/a = 43.3/9.81 = 4.41 s
h(max) = ½gt² = ½(9.81)4.41² = 95.6 m
The ball will return to earth in the same amount of time
t(max) = 2(4.41) = 8.82 s
The horizontal velocity is
vx = 50.0cos60 = 25.0 m/s
d = vt = 25.0(8.82) = 221 m
That 's one heck of a kick! No air resistance of course.
A race car, starting from rest, travels around a circular turn of radius 22.5 m. At a certain instant the car is still speeding up, and its angular speed is 0.541 rad/s. At this time, the car’s total acceleration vector (centripetal plus tangential) makes an angle of 39.0 with respect to the car’s velocity. What is the magnitude of the car’s total acceleration
Answer:
Explanation:
The answer:
https://www.chegg.com/homework-help/questions-and-answers/race-car-starting-rest-travels-around-circular-turn-radius-247-m-certain-instant-car-still-q402991
Total acceleration of car is 148.31 m/s².
What is centripetal acceleration?Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Given parameters:
Radius of the circular path: r = 22.5 m.
Angular speed: ω = 0.541 rad/s.
So, centripetal acceleration; α = ω²r = (0.541)² × 22.5 m/s² = 6.58 m/s².
Tangential acceleration: [tex]\alpha_t[/tex] = αr = 6.58 × 22.5 = 148.16 m/s².
Hence, total acceleration = √(α² + [tex]\alpha_t[/tex]²) = √(6.58² +148.16²) = 148.31 m/s².
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A straight wire in a magnetic field experiences a force of 0.011 N when the current in the wire is 2.3 A. The current in the wire is changed, and the wire experiences a force of 0.051 N as a result. What is the new current
Answer:
A straight wire in a magnetic field experiences a force of 0.022 N when the current in the wire is 1.1 A. The current in the wire is changed ...
1 answer
·
Top answer:
The magnetic force that acts on the wire in the first case can be written as
Explanation:
A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).
We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"
Answer:
Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]
From the question we are told
This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
A) Resistivity is given as,
[tex]p = \frac{RA}{l}[/tex]
where,
[tex]R = \frac{V}{I}[/tex]
Therefore,
[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]
B) Conductivity is given as,
[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]
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How long will it take a car, starting from rest, accelerating at 2 meters per second square to travel the same distance that another car traveling at a constant rate of 20m/s will travel?
20 seconds
Explanation:
Let [tex]x_a[/tex] be the distance traveled by the accelerating car and [tex]x_c[/tex] be the distance traveled by the car moving with a constant velocity. When they cover the same distance, we can write
[tex]x_a = x_c \Rightarrow v_{0a}t + \frac{1}{2}at^2 = v_ct[/tex]
where [tex]v_c[/tex] is the velocity of car moving at a constant rate and a is the acceleration of the accelerating car. Since the accelerating car started from rest, then [tex]v_{0a}[/tex] is zero so our equation above simplifies to
[tex]\frac{1}{2}at^2 = v_ct[/tex]
Note that the variable t cancels out so solving for t, we get
[tex]\frac{1}{2}at = v_c \Rightarrow t = \dfrac{2v_c}{a}[/tex]
Plugging in the given values,
[tex]t = \dfrac{2(20\:\text{m/s})}{2\:\text{m/s}^2} = 20\:\text{s}[/tex]
EXAM ENDS IN 30 MINS
PLSSS HELPPP ILL MAKE U BRAINLIEST
Explanation:
F = Icurrent×length×Bfieldstrength×sin(angle field to wire)
in our case
Icurrent = 10 A
length = 0.02km = 20 meters
B = 10^-6 T
angle = 30 degrees.
F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =
= 200 × 10^-6 = 2 × 10^‐4 N
Which statement best describes how light behaves with liquids, gases, and solids?
A. Light is unable to travel through liquids but travels easily through solids and some gases.
B. Light is unable to travel through gases but does travel through liquids and solids.
C. Light travels easily through liquids and gases, as well as through some solids like
glass.
D. Light travels easily through solids but is unable to travel through liquids and gases.
(20 points!)
Answer:
C number is write i think
What is the difference between real and apparent weightlessness?
Answer:
In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.
Explanation:
Hope this helps:)
The amount of work done in example B is:
Answer:
Explanation:
20 n is an unknown amount
If that is supposed to be 20 N(ewtons)
then W = Fd = 20(15) = 300 J
Answer: it will be 300 newton meters
Explanation:
Two trucks leave at different times (from the same place) headed for the same city. Both trucks arrive at the same time. Based on this information, which of the following sentences is true? Select one:
a. The trucks travelled the same distance in the same amount of time.
b. The trucks were traveling at the same average speed.
c. The trucks travelled different distances.
d. The truck that left later was travelling faster.
Answer:
d. The truck that left later was travelling faster
Explanation:
Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;
They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;
The only variable that can account for this difference is speed;
The one that left later, therefore, must have been going faster.
Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when it is subjected to tension of 50.0 N.
Hi there!
We can use the following equation to find the frequency of each harmonic:
[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]
n = nth harmonic
L = length of string (m)
T = Tension of string (N)
λ = linear density (kg/m)
Begin by converting the linear mass density to kg:
2.00g /m · 1 kg / 1000g = 0.002 kg/m
Now, we can use the equation to find the first three harmonics.
First harmonic:
[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]
Second harmonic:
[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]
Third harmonic:
[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Can someone solve by showing the steps?
This question involves the concepts of work done and the frictional force.
a. Work done by the person is "692.82 N".
b. Work done by the frictional force is "490.5 N".
a.
Work done by the person can be given by the following formula:
[tex]W=FdCos\theta[/tex]
where,
W = work done by the person = ?
F = Force applied by the person = 80 N
d = distance traveled = 10 m
θ = angle between force and motion = 30°
Therefore,
[tex]W=(80\ N)(10\ m)Cos30^o[/tex]
W = 692.82 N
b.
Work done by the frictional force is given by the following formula:
[tex]W_f=fd\\W_f=\mu mgd[/tex]
where,
[tex]W_f[/tex] = work done by the frictional force = ?
μ = coefficient of friction = 0.5
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]W_f=(0.5)(10\ kg)(9.81\ m/s^2)(10\ m)[/tex]
[tex]W_f=490.5\ N[/tex]
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A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is
W = 25 J
Explanation:
Work done on an object is defined as
[tex]W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}[/tex]
5 kg block of iron is heated to 800°C. It is placed in the tub containing 2 L of water at 15°C. Assuming all the water is brought to the boil rapidly, calculate the mass of water which boils off. (The specific heat capacity of iron 800°C is 220 J kg-1 K-1)
Answer:
Heat Loss = 5 kg * 700 deg K * 220 J / (kg*deg K) = 7.70E5 J
Since there are 4.186 J/cal
Heat Loss = 7.70E5J / 4.186 J/cal = 1.84E5 cal
Heat Gain = 2000 g * 85 deg K / cal / (deg K g) + M * 540 cal/g
Heat Gain = 1.70E5 cal + M * 540 cal/g
M = (1.84 - 1.70) E5 g / 540
25.9 g
25.9 g or 25.9 cm^3 or .0259 L of water will boil away
Which statement describes electromagnetic waves with wavelengths grater than 700 nanometers
Answer:
They take the form of heat, I think thats it but I cant see if you put down any answers
Explanation:
Answer:a
Explanation:
they form hear
A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the ball to reach the ground?
I need the Formula
Hi there!
We can use the equation:
d = v₀t + 1/2at²
Where:
v₀ = initial velocity downward
a = acceleration due to gravity
t = time
Plug in given values:
d = 4(12) + 1/2(9.8)(12²)
d = 48 + 705.6 = 753.6 m
One of the great challenges of cosmology today is to ---
A
determine the amount of matter in the Universe
B
find intelligent signals emanating from outer space
С
look backward in time to before the Big Bang
D
locate wormholes to help define the structure of the Universe
Answer:
Steady-state theory, in cosmology, a view that the universe is always expanding but maintaining a constant average density, with matter being continuously created to form new stars and galaxies at the same rate that old ones become unobservable as a consequence of their increasing distance and velocity of recession.
Cosmic inflation is a faster-than-light expansion of the universe that spawned many others. ... Cosmic inflation solves these problems at a stroke. In its earliest instants, the universe expanded faster than light (light's speed limit only applies to things within the universe).
A 20-N falling apple encounters a 4-N of air resistance. The magnitude of the net force on the apple is?
A. 0n
B.4n
C.16n
D. 20n
E. None of the above
Answer:
C
Explanation:
Pls brainliest
What is the cause of plate tectonics on Earth?
The magnetic field
Tides in the ocean
Convection in the mantle
Volcanic eruptions
Answer:
im pretty sure the answer is C
PLEASE HELP ME WITH THISSSS
Answer:
she will move in the same direction at the same speed forever.
Explanation:
If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady
If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, what is the direction of rotation and the sign of the angular acceleration
From the diagram, the angular speed will increase clockwise, the sign of the angular acceleration will be negative and the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative. The correct answer is option B
Given that two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Counterclockwise is considered the positive rotational direction.
If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, The moment of object m1 will be equal to the moment of object m2 without the Mbar
Let assume that the length L of the seesaw is 9 cm.
Anticlockwise moment = 24 x 9/3 = 72Nm
Clockwise moment = 12 x 2(9/3) = 72 Nm
With the consideration of mass of the bar Mbar, this will add to clockwise moment of the seesaw.
Therefore, the direction of rotation will be clockwise direction.
Angular acceleration is positive when object is speeding up and negative when slowing down. Also, angular acceleration is positive when speed increases in an anticlockwise direction and negative when speed increases in the clockwise direction.
From the diagram, since the angular speed increase clockwise, the sign of the angular acceleration will be negative.
We can conclude that the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative.
The correct answer is option B
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Disk A, with a mass of 2.0 kg and a radius of 40 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 20 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together.
After the collision, what is magnitude of their common angular velocity (in rev/s)?
Hi there!
For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:
I₁w₁ + I₂w₂ = (I₁ + I₂)wf
We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:
Disk 1: 1/2(2)(0.40²) = .16 kgm²/s
Disk 2: 1/2(2)(0.20²) = .04 kgm²/s
Plug in the values. Let counterclockwise be positive.
.16(-50) + .04(50) = (.16 + .04)wf
Solve:
wf = -30 rev/s
The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight W = 684 N, which is uniformly distributed. The left tire has a contact area with the ground of AL = 6.20 × 10-4 m2, whereas the right tire is underinflated and has a contact area of AR = 9.20 × 10-4 m2. Find (a) the force from the left tire, (b) the pressure from the left tire, (c) the force from the right tire, (d) the pressure from the right tire that each tire applies to the ground.
Answer:
Explanation:
Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N
Pressure on the left
PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa
Pressure on the right
PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa
A) The force from the left tire is; FL = 342 N
B) The pressure from the left tire is; PL = 551613 N/m²
C) The force from the right tire is; FR = 342 N
D) The pressure from the right tire is; PR = 371739 N/m²
We see that;
FL and FR are upward forces
W is the downward force.
We know that in equilibrium;
Sum of upward forces = sum of downward forces
Thus;
FL + FR = W
We are given W = 684 N
Since W is at the center, it means that FL = FR. Thus;
FL = FR = 684/2
FL = FR = 342 N
We are given;
Contact area of left tire; AL = 6.2 × 10⁻⁴ m²
Contact area of right tire; AR = 9.2 × 10⁻⁴ m²
Formula for pressure is;
Pressure = Force/Area
Pressure on the left tire;
PL = FL/AL
PL = 342/(6.2 × 10⁻⁴)
PL = 551613 N/m²
Pressure on the t right tire;
PR = FR/AR
PR = 342/(9.2 × 10⁻⁴)
PR = 371739 N/m²
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I need the answer fast for c pleaseeee
Answer:
F = - K x for spring (note that that F here is given in grams, F = m g is correct)
K here is 100 g / cm for the spring constant
x = -420 g / 100 g/cm = -4.2 cm
The spring would compress 4.2 cm for a total length of 20 - 4.2 = 15.8
d) to compress the spring 6.8 cm one can see that the load would be 680 g
How much energy is consumed by a 12 W night light left on for 10 hr?
Answer:
Energy consumed is 0.00033 Joules.
Explanation:
the formula of Energy is:
Energy = power/ time.
What is the intensity of the electromagnetic light waves coming from the Sun just outside of the atmosphere of Venus, Earth and Mars
The sun emits electromagnetic waves with a power of
4.0 ∗ 10 (26) W.
When 587.9 nm passes through a single slit 0.73 mm wide, it creates a diffraction pattern. (a) What distance away is the wall if the first minimum is 0.86 mm from the central maximum
From Young's single slit experiment, the distance away from the wall will be 1.068 m
Given that 587.9 nm of wavelength of light passes through a single slit 0.73 mm wide, it creates a diffraction pattern.
From the question, the following parameters are given:
The wavelength of the light λ = 587.9 nm
The width of the slit a = 0.73 mm
Fringe width X = 0.86 mm
The distance away from the wall D = ?
The fringe width is related to the wavelength of the light source by the equation:
X = Dλ ÷ a
Substitute all the parameters into the formula
0.83 × [tex]10^{-3}[/tex] = 587.9 × [tex]10^{-9}[/tex] D ÷ 0.73 ×
Cross multiply
587.9 × [tex]10^{-9}[/tex] D = 6.278 × [tex]10^{-7}[/tex]
make D the subject of the formula
D = 6.278 × [tex]10^{-7}[/tex] ÷ 587.9 × [tex]10^{-9}[/tex]
D = 1.068 m
Therefore, the distance away from the wall is 1.068 m
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If the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
Given the data in the question;
Wavelength; [tex]\lambda = 587.9nm = 5.879*10^{-7}m[/tex]Width of slit; [tex]a = 0.73mm = 0.00073m[/tex] First minimum; [tex]y = 0.86mm = 0.00086m[/tex]Since its first, order number; [tex]m = 1[/tex]Distance; [tex]L = \ ?[/tex]From Thomas Young's single slit experiment:
[tex]\frac{a*y}{L} = m * \lambda[/tex]
Where a is the width of the slit, y is first minimum, L is the distance, m is the order number and λ is the wavelength.
We substitute our values into the equation
[tex]\frac{0.00073m\ *\ 0.00086m}{L} = 1\ *\ ( 5.879*10^{-7}m)\\\\\frac{0.0000006278m^2}{L} = 5.879*10^{-7}m\\\\L = \frac{0.0000006278m^2}{5.879*10^{-7}m} \\\\L = 1.07m[/tex]
Therefore, if the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
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A small 1240-kg SUV has a wheelbase of 3.2 m. If 67% of its weight rests on the front wheels, how far behind the front wheels is the wagon's center of mass
Answer:
Explanation:
Let d be the distance to the center of mass from the front wheels
Sum moments about the front wheel contact point to zero
1240(9.8)[d] - 1240(9.8)(1 - 0.67)[3.2] = 0
1240(9.8)[d] = 1240(9.8)(1 - 0.67)[3.2]
d = (1 - 0.67)[3.2]
d = 1.056 m