Answer:
the charge of the bees must be of the same sign
Explanation:
The electric field is given by the relation
E = k q / r²
This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.
The force generated by this field on a test charge is
F = q E
Since the charge is a scalar, the direction of the force is the same as the electric field.
In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.
The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,
Consequently the charge of the bees must be of the same sign
PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)
A) 59 eV
B) 58 eV
C) 57 eV
D) 56 eV
Answer:
a. 59 ev. helpful answer
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
djfjci3jsjdjdjdjdjddndn
ds
A laser beam enters one of the sloping faces of the equilateral glass prism (n=1.42) and refracts through the prism. Within the prism the light travels horizontally. What is the angle between the direction of the incident ray and the direction of the outgoing ray?
Answer:
30.5°
Explanation:
Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from
n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).
Making D subject of the formula, we have
n = [sin(D + α)/2]/sin(α/2)
nsin(α/2) = [sin(D + α)/2]
(D + α)/2 = sin⁻¹[nsin(α/2)]
D + α = 2sin⁻¹[nsin(α/2)]
D = 2sin⁻¹[nsin(α/2)] - α
So, substituting the values of the variables into the equation, we have
D = 2sin⁻¹[nsin(α/2)] - α
D = 2sin⁻¹[1.42sin(60°/2)] - 60°
D = 2sin⁻¹[1.42sin(30°)] - 60°
D = 2sin⁻¹[1.42 × 0.5] - 60°
D = 2sin⁻¹[0.71] - 60°
D = 2(45.23°) - 60°
D = 90.46° - 60°
D = 30.46°
D ≅ 30.5°
A toddler weighs 10 kg and raises herself onto tiptoe (on both feet). Her feet are 8 cm long with each ankle joint being located 4.5 cm from the point at which her feet contact the floor. While standing on tip toe:
(a) what is the upward normal force exerted by the floor at the point at which one of the toddler's feet contacts the floor?
(b) what is the tension force in one of her Achilles tendons? (c) what is the downward force exerted on one of the toddler's
ankle joints?
Answer:
a.49 n
b. 63 n
c. 112 n
Explanation:
a.10 times 9.8 from gravity/2 = 49 n
b. 49n times 4.5/8-4.5 = 63 n
c 49n + 63 n = 112 n
Một vô lăng sau khi bắt đầu quay được một phút thì thu được vận tốc 700
vòng/phút. Tính gia tốc góc của vô lăng
which of the following is the correct description of momentum?
-the product of mass and acceleration -the product of mass and velocity
-velocity divided by mass
-acceleration divided by mass
Answer:
The product of mass and velocity is the correct answer
Explanation:
Momentum is defined as mass × velocity
p = mv
Answer:
The product of mass and velocity
Explanation:
I just did it and got it right with a 100%
Posted 1/3/23
Need in hurry important please
Answer:
I don't see anything on your question?
how can the starch be removed from the leaves of potted plants
Answer:
Explanation:
There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.
I NEED THE ANSWER QUICK PLEASEE
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 mm, and a baseball has a mass of 145 g.
Required:
a. Draw a free-body diagram of the ball during the pitch.
b. What force did the pitcher exert on the ball during this record-setting pitch?
c. Estimate the force in part b as a fraction of the pitcher's weight.
Answer:
Following are the solution to the given points:
Explanation:
For point a:
Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:
The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.
For point b:
[tex]160.2\ N[/tex]
First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that
where
[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]
Calculating the mass:
[tex]m = 145 g = 0.145 kg[/tex]
Calculating the force:
[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]
For point c:
0.195 times the pitcher's weight
[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]
Solving for W:
[tex]W=84 \times 9.8= 823.2 \ N[/tex]
Now the force of Part B could be defined as the fraction of the mass of the pitcher:
[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
following:
blue chip
b. yellow chip
c. yellow chip
Answer:
Explanation:
Blue: 10/30
Red: 5/30
Yellow: 15/30
The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
What is the probability?
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
Given that
Blue chips =10
Red chips = 5
Yellow chips = 15
Number of the total samples =10+15+5=30
Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]
Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]
Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]
Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
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2. Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how much work has she done?
Answer:
W = 354.9 J
Explanation:
Given that,
The mass of a bale of hay, m = 23 kg
The displacement, d = 3.9 m
The horizontal force exerted on the hay, F = 91 N
We need to find the work done. We know that,
We know that,
Work done, W = Fd
So,
W = 91 N × 3.9 m
W = 354.9 J
So, the required work done is 354.9 J.
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
A race car starts from rest on a circular track of radius 378 m. The car's speed increases at the constant rate of 0.580 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
a. The speed of the race car
b. The distance traveled
c. the elapsed time
Answer:
a) [tex]V=14.904m/s[/tex]
b) [tex]d = 191.49 m[/tex]
c) [tex]t= 25.696 s[/tex]
Explanation:
From the question we are told that:
Radius [tex]r =378m[/tex]
Acceleration [tex]a=0.580[/tex]
a)
Generally the equation for speed of the car is mathematically given by
[tex]a=\frac{v^2}{r}[/tex]
[tex]V=\sqrt{a*r}[/tex]
[tex]V=\sqrt{0.58*383}[/tex]
[tex]V=14.904m/s[/tex]
b)
Generally the equation for distance traveled of the car is mathematically given by
[tex]V^2=u^2+2ad[/tex]
[tex]d=\frac{V^2}{2a}[/tex]
[tex]d=\frac{14.904^2}{2*0.58}[/tex]
[tex]d = 191.49 m[/tex]
c)
Generally the equation for time of the car is mathematically given by
[tex]V=u+at[/tex]
[tex]t=\frac{V}{a}[/tex]
[tex]t=\frac{14.904}{0.58}[/tex]
[tex]t= 25.696 s[/tex]
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
Need an answer in hurry u can make the pic big
2 coplas o pregones inventadas por ti relacionadas con la región caribe.
A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute
Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
What happens in the gray zone between solid and liquid?-,-
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
what does Newton's third law ? Describe.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.
Hope it helps
Answer:
Newton's third law state that every action there is equal but opposite reactions
hope this will help you more
Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
P5. A bullet with an initial velocity of 280 m/s in the x-direction penetrates an initially stationary block of mass 11 kg and emerges on the other side with a final velocity of 70 m/s in the x-direction. The velocity of the block after the collision is 0.2 m/s, also in the x-direction. Assume the block slides on a horizontal frictionless surface. What is the mass of the bullet
Answer:
the mass of the bullet is 10.5 g
Explanation:
Given;
initial velocity, u₁ = 280 m/s
final velocity of the bullet, v₁ = 70 m/s
final velocity of the block, v₂ = 0.2 m/s
mass of the block, m₂ = 11 kg
initial velocity of the block, u₂ = 0
let the mass of the bullet = m₁
Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
280m₁ + 11(0) = 70m₁ + 11 x 0.2
280m₁ = 70m₁ + 2.2
280m₁ - 70m₁ = 2.2
210m₁ = 2.2
m₁ = 2.2/210
m₁ = 0.0105 kg
m₁ = 10.5 g
Therefore, the mass of the bullet is 10.5 g
You should extending your throwing hand straight up to the sky to follow-through.
O True
O False
False
It's not straight up
A circular ice rink is 20 m in diameter and is to be temporarily enclosed in a hemispherical dome of the same diameter. The ice is maintained at 270 K. On a particular day the inner surface of the dome is measured to be 290 K. Estimate the radiant heat transfer from the dome to the rink if both surfaces can be taken as blackbody.
570k or 1050k
270k+290k= 570k
270k•290k = 1050k
A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.
Answer:
Explanation:
Given that:
Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm
For objective lens of focal length f₁ = 20 mm
s₁' = 120 mm + 20 mm = 140 mm
∴
Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{140}{20}[/tex]
[tex]m_1 = 7 \ m[/tex]
For objective lens of focal length f₁ = 4 mm
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{124}{4}[/tex]
[tex]m_1 = 31 \ m[/tex]
For objective lens of focal length f₁ = 1.4 mm
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{121.4}{1.4}[/tex]
[tex]m_1 = 86.71 \ m[/tex]
The magnification of the eyepiece is given as:
[tex]m_e = 5X \ and \ m_e = 15X[/tex]
Thus, the largest angular magnification when [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]
[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]
= 86.71 × 15
= 1300.65
The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]
[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]
= 7 × 5
= 35
The largest magnification will be 1300.65 and the smallest magnification will be 35.
What is magnification?Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.
Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.
It is given in the question:
Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm
now we will calculate the magnification for all three focal lengths of the objective lens.
Also, each objective forms an image 120 mm beyond its second focal point.
(1) For an objective lens of focal length [tex]f_1=20 \ mm[/tex]
[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]
Magnification will be calculated as
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]
(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]
(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]
Now the magnification of the eyepiece is given as:
[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]
Thus, the largest angular magnification when
[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]
[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]
[tex]m_{large}=86.71\times 15=1300.65[/tex]
The smallest angular magnification derived when
[tex]m_1=7\ \ \ \ m_e=5[/tex]
[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]
[tex]m_{small}=7\times 5=35[/tex]
Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.
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PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Answer:
[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]
Explanation:
Given that,
The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]
A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.
We need to find the decay constant for this decay. The formula for half life is given by :
[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]
Put all the values,
[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]
So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].
What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced