Answer:
A) 3.367 × 10^(-6)
B) 2.97 × 10^(7) N/C
C) Upwards
Explanation:
We are given;
Mass of bee; m = 100 mg = 100 × 10^(-6) kg
Charge on bee;q=33 pC = 33 × 10^(-12)C
Electric field strength; E = 100 N/C
A) Formula for weight of bee; W = mg = 100 × 10^(-6) × 9.8 = 9.8 × 10^(-4) N
Electric force on Bee; F = qE = 33 × 10^(-12) × 100 = 33 × 10^(-10) N
ratio of the electric force on the bee to the bee's weight; F/W = (33 × 10^(-10))/(9.8 × 10^(-4)) = 3.367 × 10^(-6)
B) For the bee to be suspended in the air, it means the weight of the bee must be equal to the electric force. Thus;
mg = qE
100 × 10^(-6) × 9.8 = 33 × 10^(-12) × E
E = (100 × 10^(-6) × 9.8)/(33 × 10^(-12))
E = 2.97 × 10^(7) N/C
C) From Newton's law, sum of forces = 0.
Thus;
F_n + F + W = 0
Where F is the normal force.
Thus;
F_n = -(F + W)
F_n = - ((33 × 10^(-10)) + (9.8 × 10^(-4)))
F_n = -9.8 × 10^(-4) N
Thus, applied electric field is;
E_a = F_n/q = (-9.8 × 10^(-4))/(33 × 10^(-12)) = -2.97 × 10^(7) N/C
This is negative and so it means the direction will be opposite the Earth's electric filed which is upwards.
here did you walk? What did you find most enjoyable while walking: listening to music, listening to an audio book, or nothing? How did your body react to this introductory amount of exercise? Was it more exercise or less exercise than you are used to? If you did not walk, what other type of physical movement did you do?
HELP ME PLS
Chlorine (chemical symbol Cl) is located in Group 17, Period 3. Which is
chlorine most likely to be?
A. A metalloid with properties of both metals and nonmetals
B. A gaseous, highly reactive nonmetal
C. A soft, shiny, highly reactive nonmetal
D. A soft, shiny, highly reactive metal
Answer:
Option B
Explanation:
A gaseous, highly reactive non-metal
Answer:B
Explanation:I just took the test
If the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change
Answer:
A neutral atom has no net charge and should not be affected by a point charge, and the distance of separation is not a factor.
The force will remain the same and is equal to zero.
We have a point charge and a neutral atom.
We have to investigate if the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change.
State Coulomb's Law of Electrostatic force.The Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically -
[tex]$F=\frac{q _{1}\cdot q _{2}}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
According to question, we have -
A point charge and a neutral atom.
If initially the distance between the point charge and neutral atom is r meters, then -
q(1) = Q (say)
q(2) = 0 ( Neutral atom has zero charge)
Using Coulomb's law -
[tex]$F=\frac{Q\times 0}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
F = 0 Newtons.
Now, if you will increase the distance by the factor of six, still the force will remain zero as there is only one point charge and other is neutral atom. There is no electrostatic force between a charged and uncharged particle.
Hence, the force will remain the same and is equal to zero.
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An object 1.00cm high is placed 18.0cm from a converging lens, forms a real Image 2.00cm high Calculate the forcal length of to the lens
Answer:
focal length=12cm
Explanation:
object size is equal to 1.00cm
object distance = 18cm
heigh of image = 2.00cm
image distance = ??
but magnification is given by;
M = 2.00/1.00 = 2
but u/v = M
u/18 = 2
u = 36
1/f = 1/u+1/v
1/f = 1/18+ 1/36
1/f = 1/12
f = 12cm
The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3.25 × 10-9 mm.
Answer:
Option B. 3.25×10¯¹⁰ mm.
Explanation:
Measurement (cm) = 0.00325×10⁻⁸ cm
Measurement (mm) =?
The measurement in mm can be obtained as follow:
1 cm = 10 mm
Therefore,
0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm
0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm
Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.
The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm
The number given is in standard form and can be written as 3.25*10^-11 cm.
To convert this from centimeter to millimeter, we have to multiply this value by 10.
Conversion Units1 cm - 10mm100cm = 1m1000m = 1kmSo, let's 3.25*10^-11 by 10 and get our value in mm
[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]
From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]
Learn more about conversion of units here;
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In addition to acceleration, what else will be a maximum at the amplitude for SHM?
A. Potential energy
B. Kinetic energy
C. Nuclear energy
D. Chemical energy
It is Potential energy's
Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric potential difference depends on current and resistance according to this function V(I,R) = IR. Your measured current and resistance have the following values and uncertainties I = 5.9 Amps, delta I space equals space 0.4 Amps and R = 42.7 Ohms and delta R space equals space 0.6 Ohms. What is the uncertainty in the , delta V ? Units are not needed in your answer.
Answer:
ΔV = 2 10¹ V
Explanation:
The calculation of the uncertainty or error in an expression is given by
ΔV = [tex]\frac{dV}{di}[/tex] |Δi| + [tex]\frac{dV}{dR}[/tex] |ΔR |
V = i R
let's make the derivatives
[tex]\frac{dV}{di}[/tex] = R
[tex]\frac{dV}{dR}[/tex] = i
we substitute
ΔV = R | Δi | + i | ΔR |
in the exercise give the values
i = (5.9 ± 0.4) A
R = (42.7 ± 0.6) Ω
we calculate
ΔV = 42.7 0.4 + 5.9 0.6
ΔV = 20.6 V
ΔV = 2 10¹ V
the voltage is
V = i R
V = 5.9 42.7
V = 251.9 V
the result is
V = (25 ± 2) 10¹ V
The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20 times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
Answer:
the largest distance we can measure is 10¹⁴ km
Explanation:
Given the data in the question;
Threshold hearing = 10⁻²⁰
smallest distance measured = 1 mm
Largest distance measured will be;
⇒ ( threshold hearing )⁻¹ × smallest distance
= ( 1 / 10⁻²⁰ ) × 1 mm
= 10²⁰ × 1mm
= 10²⁰ mm
we know that; 1000 mm = 10⁶ km
Largest distance = ( 10²⁰ / 10⁶ ) km
= 10¹⁴ km
Therefore, the largest distance we can measure is 10¹⁴ km
Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 10s. 2. What force is on a 10kg mass that accelerates at 3m/s/s. 3. What is the potential energy of a 7kg object 4m off the ground *
show all your work please
Explanation:
1. Distance, d = 5 m
Time, t = 10 s
Speed = distance/time
[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]
2. Mass, m = 10 kg
Acceleration, a = 3 m/s³
Force, F = mass (m) × acceleration (a)
F = 10 × 3
= 20 N
3. Mass, m = 7 kg
Height, h = 4 m
Potential energy, E = mgh
E = 7 × 9.8 × 4
E = 274.4 J
Hence, this is the required solution.
The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specific heat of the solid is 390 J/kgK and its heat of fusion is 4000 J/kg.
Answer and I will give you brainiliest
Hey again!
Ok..
Now... The melting Point of this solid is 90°C.
Meaning That as soon as it gets to this temp... It STARTS Melting.
So at that temp... It still has some solid parts in it.
You can say its a Solid Liquid Mixture.
Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.
After Fusion...It'd then Be a Pure Liquid with no solids in it.
So
Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c
Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.
So
Q= Q' + Q"
Q= mc∆0 + ml
∆0 = 90°c - 30°c = 60°c
Q= 2.5(390)(60) + (2.5)(4000)
Q=6.9 x 10⁴Joules
The heat required to change 2.5 kg of the solid at 30.0C to a liquid is 6.9 x 10⁴J.
What is specific heat?The specific heat is the amount of heat energy required to change the temperature of 1kg of object by 1°C.
The heat needed to change the solid's temperature from 30°C - 90°C is
Q' = mC∆T
The heat used to change the phase solid-liquid phase .i.e.
Q'' =mL where L =latent heat of fusion
The total heat required is
Q= Q' + Q"
Q= mc∆T + ml
Q= 2.5(390)(90 - 30) + (2.5)(4000)
Q=6.9 x 10⁴Joules
Thus, the heat required to change the solid to liquid is 6.9 x 10⁴J.
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dòng điện là gì ?/???????
Answer:
Dòng điện là một dòng các hạt mang điện, chẳng hạn như electron hoặc ion, di chuyển qua vật dẫn điện hoặc không gian. Nó được đo bằng tốc độ thực của dòng điện tích qua một bề mặt hoặc vào một thể tích điều khiển.
Xin lưu ý rằng tôi đã sử dụng một trình dịch để nhập nội dung này, vì vậy có thể có một số từ không hợp lý.
Explanation:
A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)
Answer:
Distance = speed * time
55*5
275 meters.
The train would have covered a distance of 275 m
What is distance ?
We can define distance as to how much ground an object has covered despite its starting or ending point.
Distance = speed * time
given
speed= 55 m/s
time = 5 sec
Distance = 55 * 5 = 275 m
The train would have covered a distance of 275 m
learn more about distance
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two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
Part A
Calculate the work done when a force of 6.0 N moves a book 2.0 m
Express your answer with the appropriate units.
Answer:
Work done applied = 12 newton-meter
Explanation:
Given examples:
Force applied = 6 newton
Distance of book = 2 meter
Find from the given data:
Work done
Computation:
The equation can be used to compute work.
Work done applied = Force applied x Distance of book
Work done applied = Force x Distance
Work done applied = 6 x 2
Work done applied = 12 newton-meter
a system absorb 500 J of heat and the same time 400J of work is done one the system find change in internal enery ?
Answer:
+ 900 J
Explanation:
Since the total energy change ΔE = internal energy change ΔU since there is no change in kinetic and potential energy,
ΔE = ΔU
ΔE = Q - W where Q = heat absorbed by system and W = work done by system
Now since the system absorbs 500 J of heat, Q = + 500 J and work of 400 J is done on the system, W = -400 J
So, the values of the variables into the equation, we have
ΔE = Q - W
ΔE = + 500 J - (-400 J)
ΔE = + 500 J + 400 J
ΔE = + 900 J
So, the internal energy change, ΔE = + 900 J
The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 87.9 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?
Answer:
[tex]d=79.9m[/tex]
Explanation:
From the question we are told that:
coefficient of static friction [tex]\mu=0.38[/tex]
Velocity [tex]v=87.9=>24.41667m/s[/tex]
Generally the equation for Conservation of energy is mathematically given by
[tex]\mu*mgd = 0.5 m v^2[/tex]
[tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]
[tex]d=79.9m[/tex]
A computer monitor uses 200 W of power. How much energy does it use in
10 seconds?
Answer: see below explanation, should be straight forward from there? ;)
Explanation: 1 watt = 1 joule per second
Watt is a measure of energy over time
So 10 seconds... u got this :)
Two resistors of 10 and 15 n are connected. What is their combined resistance if they are connected: a) in series b) in parallel?
Explanation:
Given that,
Two resistors of 10 ohms and 15 ohms are connected.
In series combination, the equivalent resistance is given by :
[tex]R_s=R_1+R_2\\\\R_s=10+15\\\\R_s=25\ \Omega[/tex]
In parallel combination, the equivalent resistance is given by :
[tex]\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_p}=\dfrac{1}{10}+\dfrac{1}{15}\\\\R_p=6\ \Omega[/tex]
Hence, this is the required solution.
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
What is the magnitude of a vector that has the following components: x = 32 m y = -59 m
Answer:
Explanation:
Since the x and y components are given
The vectors Magnitude = √32²+(-59)²
=67.12m
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Force required to lift the wheelbarrow
Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
1. A skydiver carries a buzzer which emits a steady 1800 Hz tone. A friend on the ground (directly below the skydiver) listens to the doppler shifted frequency. Assume the air is calm and that the speed of sound is independent of altitude. While the skydiver is falling at terminal velocity (the constant speed reached due to the balance of the downward gravitational force and the upward drag force due to air resistance), her friend on the ground hears waves of frequency 2150 Hz. a. How fast is the skydiver falling
Answer:
vs = 55.84 m/s
Explanation:
In this case, the source (skydiver with buzzer) is moving towards the observer (friend). The formula for this case will be:
[tex]f' = \frac{v}{v-v_s} f[/tex]
where,
f' = shifted frequency = 2150 Hz
f = actual frequency = 1800 Hz
v = speed of sound = 343 m/s
vs = speed of skydiver = ?
Therefore,
[tex]2150\ Hz = \frac{343\ m/s}{343\ m/s - v_s}(1800\ Hz)\\\\343\ m/s - v_s = \frac{343\ m/s}{2150\ Hz} (1800\ Hz)\\\\v_s = 343\ m/s - 287.16\ m/s\\[/tex]
vs = 55.84 m/s
The length of a cylindrical axon is 8 cm and its radius of 8μm,and the thickness of the membrane is 0.01μm,dielectric constant ( ε=24.78x10-12 F/m),the capacitance of nerve cell is
Answer:
9.965 nF
Explanation:
The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m
So, C = εA/d
C = ε2πrL/d
Substituting the of the values variables into the equation, we have
C = ε2πrL/d
C = 24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m
C = 9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m
C = 996463 × 10⁻¹⁴ F
C = 9.96463 × 10⁻⁹ F
C = 9.96463 nF
C ≅ 9.965 nF
a rock with the mass of 10 kg sits at the top of a hill 20 m high. what is the potential energy
Answer:
What is the potential energy? PE= mghPE= hwKE= 1/2mv2
Answer:1960J
Explanation:
Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical
region formed by two coaxial cylindrical surfaces of radii, 1.0 mm and 3.0 mm. Determine
the magnitude of the electric field at a point which is 4.0 mm from the symmetry axis.
Answer:
The electric field is given by 4.5 N/C.
Explanation:
Charge density = 80 nC/m3
inner radius, r' = 1 mm
outer radius, r'' = 3 mm
distance, r = 4 mm
The linear charge density is given by
[tex]\lambda =\rho \times\pi\times (r''^2 - r'^2)\\\\\lambda = 80\times 10^{-9}\times 3.14\times 10^{-6}\times(9-1)\\\\\lambda = 2\times 10^{-12}\\[/tex]
The electric field is given by
[tex]E = \frac{\lambda }{4\pi\varepsilon_or}\\E=\frac{9\times 10^9\times 2 \times 10^{-12}}{0.004}\\\\E=4.5 N/C[/tex]
A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam
Answer:
the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
Explanation:
Given the data in the question;
initial velocity; u = 0 m/s
height; h = 2.5 m
we find the velocity of the ball just before it touches the foam.
using the equation of motion;
v² = u² + 2gh
we know that acceleration due gravity g = 9.81 m/s²
so we substitute
v² = ( 0 )² + ( 2 × 9.81 × 2.5 )
v² = 49.05
v = √49.05
v = 7.00357 m/s
Now as the ball touches the foam
final velocity v₀ = 0 m/s
compresses S = 3 cm = 0.03 m
so
v₀² = v² + 2as
we substitute
( 0 )² = 49.05 + 0.06a
0.06a = -49.05
a = -49.05 / 0.06
a = -817.5 m/s²
Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
The number of current paths in a series circuit is:
a. one
b. two
C. three
d. four
Answer:
One
Explanation:
In series combination, the circuit follows one path whereas in parallel it follows two or more than two path
A 30g bullet at v=900m/s strikes a 1kg soft iron target stopping inside the iron [c=490J/kg°C. How much will the temperature of the iron increase? Ignore the heat that will be shared with the bullet
A) 25°C
B) 24795°C
C) 826°C
D) 82653°C
show your full work please
Answer:
ΔT = 25°C
Explanation:
Given that.
The mass of a bullet, m₁ = 30 g = 0.03 kg
The speed of the bullet, v = 900 m/s
Mass of soft iron, m₂ = 1 k
The specific heat of iron, c=490J/kg°C
We need to find the increase in temperature of iron. using the conservation of energy,
Kinetic energy = heat absorbed
[tex]\dfrac{1}{2}m_1v^2=m_2c\Delta T\\\\\Delta T=\dfrac{\dfrac{1}{2}m_1v^2}{m_2c}\\\\\Delta T=\dfrac{\dfrac{1}{2}\times 0.03\times 900^2}{1\times 490}\\\\=24.79^{\circ} C\\\\or\\\\\Delta T=25^{\circ} C[/tex]
So, the correct option is (A).
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
Điện tích trên một vật dẫn bất kỳ có giá trị bằng:
A. Tổng độ lớn các giá trị điện tích âm và điện tích dương có trên vật.
B. Tổng đại số các giá trị điện tích âm và điện tích dương có trên vật.
C. Không. Vì lúc nào số điện tích âm cũng bằng số điện tích dương.
D. Tất cả đều sai.
Answer:
A.
sửa cho tôi nếu tôi sai