Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
đổi đơn vị
42 ft2/hr to cm2/s
Answer:
X = 10.8387 cm²/s
Explanation:
In this exercise, you're required to convert a value from one unit to another.
Converting 42 ft²/hr to cm²/s;
Conversion:
1 ft² = 929.03 cm²
42 ft² = X cm²
Cross-multiplying, we have;
X = 42 * 929.03
X = 39019.26 cm²
Next, we would divide by time in seconds.
1 hour = 3600 seconds
X = 39019.26/3600
X = 10.8387 cm²/s
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnetic field travels in a circular path of radius 18.0 mm. What is the charge of the particle
Answer:
the charge of the particle is 2.47 x 10⁻¹⁹ C
Explanation:
Given;
mass of the particle, m = 6.64 x 10⁻²⁷ kg
velocity of the particle, v = 8.7 x 10⁵ m/s
strength of the magnetic field, B = 1.3 T
radius of the circle, r = 18 mm = 1.8 x 10⁻³ m
The magnetic force experienced by the charge is calculated as;
F = ma = qvB
where;
q is the charge of the particle
a is the acceleration of the charge in the circular path
[tex]a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C[/tex]
Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C
4 motor car is approaching a road crossing with a speed of 75 km/hr. A onstable standing near the crossing hears the frequency of its horns 260 Hz. That is the real frequency of horn? The speed of sound in air = 332 m/s. [IOE]
Answer:
181 is the answer I think so
A 50-turn coil has a diameter of . The coil is placed in a spatially uniform magnetic field of magnitude so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf, , induced in the coil if the magnetic field is reduced to zero unfiformly
Answer:
EMF = 51.01 Volt
Explanation:
A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ∘ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s
We have,
Number of turn in the coil, N = 50
The diameter of the coil, d = 15 cm
Radius, r = 7.5 cm = 0.075 m
Initial magnetic field, [tex]B_i=0.5\ T[/tex]
The plane of the coil makes an angle of 30° with the magnetic field.
The magnetic field reduced to zero in 0.1 seconds
We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,
[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]
So, the induced emf in the coil is 51.01 V.
Sometimes the units for an electric field are written as N/C, while other times the units are written as V/m, using dimensional analysis show that N/C is equal to V/m.
a. True
b. False
Answer:
N/C = V/m.
Explanation:
The SI unit of electric field is N/C. Sometimes the units are written as V/m.
We know that,
1 V = 1 J/C
Using dimensional analysis,
The dimensional formula for Joules is [M¹L² T⁻²].
The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].
So,
J/C = [M¹L² T⁻³I¹] ...(1)
The dimensional formula of Newton is [M¹ L¹ T⁻²]
The dimensional form of coulomb is [M⁰ L⁰ T¹ I¹].
N/C= [M¹L² T⁻³I¹] ....(2)
From (1) and (2) it is clear that the N/C is equal to V/m.
A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.
Answer:
The distance between mirror and you is 2 ft.
Explanation:
diameter, d = 8 ft
radius of curvature, R = 4 ft
magnification, m = 0.5
focal length, f = R/2 = 4/2 = 2 ft
let the distance of object is u and the distance of image is v.
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]
Use the formula of magnification
[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]
ai là người phát hiện trái đất hình cầu đầu tiên ?
Answer:
Can't understand the language
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?
Answer:
[tex]F=133N[/tex]
Explanation:
From the question we are told that:
Length [tex]l=3.0m[/tex]
Mass [tex]m=24kg[/tex]
Distance from Tip [tex]d=35cm[/tex]
Generally, the equation for Torque Balance is mathematically given by
[tex]mg(l/2)=F(l-d)[/tex]
[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]
Therefore
[tex]F=133N[/tex]
answer bhejo please please please
Answer:
Various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 4.50 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.
Required:
What is the lowest possible frequency of sound for which this is possible?
Answer:
Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker
D2 = 5.05 m from the far speaker
The difference in distances from the speakers is
5.05 - 4.5 = .55 m (Let y be wavelength, lambda)
n y = 4.5
(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength
y = .55 m subtracting equations
f = v / y = 340 / .55 = 618 / sec should be the smallest frequency
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
A 2.0 kg frictionless puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.5 s?
Answer:
the distance moved by the puck after 2.5 s is 7.8 m
Explanation:
Given;
mass of the puck, m = 2 kg
initial velocity of the puck, u = 0
applied force, F = 5 N
time of motion, t = 1.5 s
Acceleration of the puck is calculated from Newton's second law of motion;
F = ma
a = F/m
a = 5/2
a = 2.5 m/s²
The distance moved by the puck after 2.5 s is calculated as;
s = ut + ¹/₂at
s = 0 + ¹/₂at²
s = ¹/₂at²
s = 0.5 x 2.5 x (2.5)²
s = 7.8 m
Therefore, the distance moved by the puck after 2.5 s is 7.8 m
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.
Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?
Answer:
0.37 m
Explanation:
Given :
Window height, [tex]h_1[/tex] = 1.27 m
The flowerpot falls 0.84 m off the window height, i.e.
[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex] seconds.
Assuming that the speed of the pot just above the window is v then,
[tex]h_2=ut+\frac{1}{2}gt^2[/tex]
[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]
[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]
[tex]$v= 2.69$[/tex] m/s
Initially the pot was dropped from rest. So, u = 0.
If it has fallen from a height of h above the window then,
[tex]$h = \frac{v^2}{2g}$[/tex]
[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]
h = 0.37 m
The cation that is reabsorbed from the urine in response to aldosterone
Answer:
If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.
Calculate the magnitude of a gravitational force between two object 400kg and 800kg separated by a distance of of 45m (take G =6.67 * 10^-11 Nm^2 kg^-2)
Answer:
Explanation:
The formula is
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:
[tex]F_g=\frac{(6.67*10^{11})(400)(800)}{(45^2)}[/tex] and multiply and divide all that out to get
[tex]F_g=1.1*10^{-8}[/tex] It should really only be 1 significant digit since 400 and 800 both have only 1 significant digit, but I used 2. It should be
[tex]F_g=[/tex] 1 × 10⁻⁸ N
(a) (1) What is meant by the term modern technology? Give three reasons for the lack of improvement of technology in Ghana.
Answer:
Modern technology is the advancement of old technology with adjustments.
1. Ghana is still developing economically and modern infrastructure is not yet well established.
2. Ghana has Environmental- health concerns which contribute to the lack of improvement of technology in Ghana.
3. Youth unemployment is prevalent therefore contributing to lack of improvement of technology in Ghana.
Explanation:
Ghana is a developing country with rising debt payments which could very well contribute as whole to the lack of improvement of technology in Ghana.
A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in ???? N?
Answer:
the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Explanation:
Given;
length of the conductor, L = 1 cm = 0.01 m
current carried by the solenoid, I₁ = 15 A
radius of the solenoid, r = 4 cm
number of turns per length of the solenoid, n = 800 turns/m
current carried by the solenoid, I₂ = 40 mA = 0.04 A
The magnetic field of the solenoid is calculated as;
B = μnI₂
where;
μ is the permeability of free space = 4π x 10⁻⁷ Tm/A
B = ( 4π x 10⁻⁷) x (800) x (0.04)
B = 4.022 x 10⁻⁵ T
The magnitude of the magnetic force on the wire segment is calculated as;
F = BI₁L sinθ
where
θ is the angle made by the wire segment against the solenoid = 90⁰
F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)
F = 6.03 x 10⁻⁶ N
Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?
Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
Solving for [tex]v_{0y}[/tex],
[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]
[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]
b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as
[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]
or
[tex]v_{0x} = 150\:\text{m/s}[/tex]
What star is known as the "cold planet"?
Explanation:
OGLE-2005-BLG-390Lb.
PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...
Neptune. Surface Temperature: 72 Kelvin. ...
Uranus. Surface Temperature: 76 Kelvin. ...
Saturn. Surface Temperature: 134 Kelvin. ...
Jupiter. Image Courtesy: NASA. ...
OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown
Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why do bridges become covered with ice before roads do? Research this question and respond in depth, writing a full paragraph. Be sure to include examples. At the end of your response, provide at least two authoritative sources that you used in your research.
Answer:
During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.Sources:
-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads
and
-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262
I hope this helps you! ^^
Observe: Air pressure is equal to the weight of a column of air on a particular location. Air pressure is measured in hectopascals (hPa). Note how the air pressure changes as you move Station B towards the center of the high-pressure system.
a. What do you notice?
b. Why do you think this is called a high-pressure system?
Answer:
A. When moving towards a high pressure center the pressure values increase in the equipment
B. This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
Circular Motion of Charges: A charged particle is observed traveling in a circular path of radius R in a uniform magnetic field. If the particle were traveling twice as fast, the radius of the circular path would be
Answer:
V = 168 km³
Explanation:
...............
A light bulb, attached (by itself) to an ideal 10 V battery, becomes hotter as time goes on. The bulb's filament is made of tungsten, a metal, which becomes more resistive as its temperature increases. Which statement below is true?
a. As time goes on and the bulb grows hotter, the voltage across the bulb increases so the bulb would grow brighter.
b. As time goes on and the bulb grows hotter, the current through the bulb increases so the bulb would grow brighter.
c. As time goes on and the bulb grows hotter, the voltage across the bulb decreases so the bulb would grow dimmer.
d. As time goes on and the bulb grows hotter, the current through the bulb decreases so the bulb would grow dimmer.
Answer:
the correct one is d
Explanation:
The filament of the bulb fulfills the law of ohm
V = i R
indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,
When examining the different statements, the correct one is d
You drive 7.5 km in a straight line in a direction east of north.
a. Find the distances you would have to drive straight east and then straight north to arrive at the same point.
b. Show that you still arrive at the same point if the east and north legs are reversed in order.
Answer:
a) a = 5.3 km, b) sum fulfills the commutative property
Explanation:
This is a vector exercise, If you drive east from north, we can find the vector using the Pythagorean theorem
R² = a² + b²
where R is the resultant vector R = 7.5 km and the others are the legs
If we assume that the two legs are equal to = be
R² = 2 a²
r = √2 a
a = r /√2
we calculate
a = 7.5 /√2
a = 5.3 km
therefore, you must drive 5.3 km east and then 5.3 km north and you will reach the same point
b) As the sum fulfills the commutative property, the order of the elements does not alter the result
a + b = b + a
therefore, it does not matter in what order the path is carried out, it always reaches the same end point
Which circuit element is of special importance in AC circuits?
A. Resistor
B. Ammeter
C. Battery
D. Capacitor
Answer:
Explanation:capacitor
Answer:
Ammeter
Explanation:
pls mark me as a brainlist
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound
The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]
Therefore,
[tex]$AD-BD = \frac{\lambda}{2}[/tex]
[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2 \times 0.4985$[/tex]
[tex]$\lambda = 0.99714 \ m$[/tex]
Thus the frequency is :
[tex]$f=\frac{v}{\lambda}$[/tex]
[tex]$f=\frac{340}{0.99714}$[/tex]
[tex]f=340.9744[/tex] Hz
In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular velocity of 6.1 rad/s about a frictionless rod passing through its center. A second disk rotating clockwise with an angular velocity of 9.3 rad/s falls from above onto disk 1. The two then rotate as one in the clockwise direction with an angular velocity of 1.8 rad/s. Determine the moment of inertia, in kg · m2, of disk 2.
Answer:
I = 3.6 kg•m²
Explanation:
Conservation of angular momentum
Let's assume CW is the positive direction
3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)
I(9.3 - 1.8) = 3.4(1.8 + 6.1)
I(7.5) = 3.4(7.9)
I = 3.4(7.9)/(7.5) = 3.5813333333...
The moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
What is moment of inertia?The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.
here it is given that
MOI of disk one [tex]I_1=3.4\ kg-m^2[/tex]
Angular velocity [tex]w_1=6.1\ \frac{rad}{s}[/tex]
Angular velocity of disk two [tex]w=1.8\ \frac{rad}{s}[/tex]
MOI of the disk two [tex]I=?[/tex]
The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]
Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.
[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]
Now put the values in the formula
[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]
[tex]I_2=3.58\ kg-m^2[/tex]
Thus the moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
To know more about moment of inertia follow
https://brainly.com/question/18721435
A car starting at rest accelerates at 3m/s² How far has the car travelled after 4s?
Answer:
24m
Explanation:
you can use the formula
s=ut+1/2at²
s=0+1/2(3)(4)²
=1/2(3)(8)
=24m
I hope this helps
Paramagnetism is closely associated with: A. the tendency of electron dipole moments to align with an applied magnetic field B. the force exerted by electron dipole moments on each other C. the exchange force between electrons D. the tendency of electron dipole moments to align opposite to an applied magnetic field
Answer:
the tendency of electron dipole moments to align with an applied magnetic field
Explanation:
Paramagnetism has to do with possession of unpaired electrons. Substances that possess unpaired electrons are said to be paramagnetic.
When an external magnetic field is applied to a paramagnetic substance, the magnetic field causes the electrons spins of the paramagnetic substance to align parallel to the field,which leads to a net attraction.
Hence, paramagnetism is closely associated with the tendency of electron dipole moments to align with an applied magnetic field.
