Binary code is the answer
Answer:
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the magnitude of the electrical force acting between a +2.4x10-8c charge and 1+1.8x10-6 charge that are separated by 1.008m is
Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
sometimes balance point may not be obtained on the potentiometer wire why
Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
I’ve been stuck please help !!
Answer:
The slope of the position time graph gives the velocity.
Explanation:
The slope of the position time graph gives the value of velocity.
In first graph,
The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts. and more than the second part, so the initial velocity is more than the final velocity.
In the second graph,
The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.
1. What are Earth's natural climate cycles?
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.
Required:
How much thermal energy is added to the air by the drag force?
Answer:
by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Explanation:
Given the data in the question;
drag coefficient of Cd = 1.4
speed v = 6.0 m/s
One model expands to a square 1.8 mm on a side
Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²
distance travelled s = 240 mm = 0.24 m
we know that; density of air e = 1.225 kg/m³
Now,
Dragging force F[tex]_D[/tex] = ( Cd × e × v² × A ) / 2
thermal energy = F[tex]_D[/tex] × s
so
thermal energy = ( 1.4 × 1.225 × (6)² × (3.24 × 10⁻⁶) × 0.24 ) / 2
thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2
thermal energy = 2.4004512 × 10⁻⁵ J
Therefore, by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Help please help please
Answer:
No. D is the right answer
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 man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s. the stone clears the cliff edge on the way down and falls all the way to the ground. what is the maximum height of the stone above the ground
Answer:
h = 90.10 m
Explanation:
Given that,
A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m
The initial speed of the stone, u = 10 m/s
The path followed by the projectile is given by :
[tex]h(t)=-4.9t^2+10t+85[/tex] ....(1)
For maximum height,
Put dh/dt = 0
So,
[tex]\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s[/tex]
Put the value of t in equation (1).
[tex]h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m[/tex]
So, the maximum height of the stone is equal to 90.10 m.
2 Lights slows down when it enters water from air.
a What happens to its speed?
b What happens to its wavelength?
c What happens to its frequency?
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
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Question 2 of 32
A water-skier with a mass of 68 kg is pulled with a constant force of 980 N by
a speedboat. A wave launches him in such a way that he is temporarily
airbome while still being pulled by the boat, as shown in the image below.
Assuming that air resistance can be ignored, what is the vertical acceleration
that the water-skier experiences on his return to the water surface? (Recall
that g = 9.8 m/s2)
Rope Force
ODON
Weight
O A. - 18.1 m/s2
OB. - 15.6 m/s2
O C. -11.2 m/s2
OD. -9.8 m/s2
Answer:
OD. -9.8 m/s2
Explanation:
The only force vertical force that is acting on the skier is gravity and since its pulling him back it's a negative force down the y axis.
An object is in free fall Group of answer choices Anytime it has an acceleration equal to 9.8 m/s/s Anytime it feels the force of gravity Only if the weight is the only force acting on the object Only when it's moving vertically down
Answer:
Anytime it feels the force of gravity.
Explanation:
Gravity is considered to be a universal force of attraction which acts between all objects that has both mass, energy and can occupy space. Therefore, it acts in such a way as to bring objects together i.e causing the objects to fall down towards the Earth.
This ultimately implies that, an object is in free fall anytime it feels the force of gravity i.e regardless of how fast the object moves or the direction it moves, the only force acting on the object is the force of gravity (g).
For example, when you throw any object up, it will naturally fall down due to the gravitational force between the Earth and the object.
Additionally, the gravity of earth makes it possible for all physical objects to possess weight.
On planet Earth, the acceleration due to gravity that all physical objects experience is 9.8 meters per seconds square.
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
Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?
a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.
Answer:
Option (a).
Explanation:
Let the angular velocity is w.
The centripetal acceleration is given by
[tex]a = r w^2[/tex]
where, r is the distance between the axle and the spoke.
So, more is the distance more is the centripetal acceleration.
(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.
The statement is true.
(b) The direction of centripetal acceleration is always towards the center, so the statement is false.
(c) It is false.
(d) It is false.
Option (a) is correct.
The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be 3.7 m/s. An archerfish spots an insect sitting 18 cm above the waterline and a horizontal distance of 28 cm away. The fish aims its stream at an angle of 39° from the waterline.
Required:
Determine the height above the waterline that the stream reaches at the horizontal position of the insect.
Answer:
The fish gobbles the mosquito at height 18 cm.
Explanation:
Initial velocity, u = 3.7 m/s
horizontal distance, d = 28 cm
Angle, A = 39 degree
Let the time is t.
Horizontal distance = horizontal velocity x time
d = u cos A x t
0.28 = 3.7 cos 39 x t
t = 0.097 s
Let the height is h.
Use the second equation of motion
[tex]h =u t-0.5 gt^2\\\\h= u sin A t - 0.5 gt^2\\\\h= 3.7 sin 39 \times 0.097 - 0.5\times 9.8\times 0.097\times0.097\\\\h =0.226 -0.046 \\\\h=0.18 m=18 cm[/tex]
A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes 0.9 s for the wave to travel the 26 m to the opposite tower. If one meter of the rope has a mass of 0.28 kg, find the tension in the tightrope.
Answer:
the tension in the tightrope is 233.68 N
Explanation:
Given the data in the question;
Time taken to reach the opposite tower t = 0.9 s
Distance between the two towers S = 26 m
mass per one meter length = 0.28 kg
First we calculate the velocity;
Velocity V = Distance / time
we substitute
Velocity V = 26 m / 0.9 s
Velocity V = 28.889 m/s
We know that Velocity V can also be expressed as;
V = √( T / m )
we make T the subject of formula
V² = T / m
T = mV²
we substitute
T = 0.28 × ( 28.889 )²
T = 233.68 N
Therefore, the tension in the tightrope is 233.68 N
6. A transverse periodic wave on a string with a linear density of 0.200 kg/m is described by the following equation: y = 0.08 sin(469t – 28.0x), where x and y are in meters and t is in seconds. What is the tension in the string? A) 3.99 N B) 32.5 N C) 56.1 N D) 65.8 N
Answer:
T = 56.11 N
Explanation:
Given that,
The equation of a wave is :
y = 0.08 sin(469t – 28.0x),
where x and y are in meters and t is in seconds
The linear mass density of the wave = 0.2 kg/m
The speed of wave is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Also,
[tex]v=\dfrac{\omega}{k}[/tex]
We have,
[tex]k=469\ and\ \omega=28[/tex]
Put all the values,
[tex]\dfrac{\omega}{k}=\sqrt{\dfrac{T}{\mu}}\\\\(\dfrac{\omega}{k})^2=\dfrac{T}{\mu}\\\\T=(\dfrac{\omega}{k})^2\times \mu[/tex]
Put all the values,
[tex]T=(\dfrac{469}{28})^2\times 0.2\\\\T=56.11\ N[/tex]
So, the tension in the string is 56.11 N.
44.7
When Xavier places his hands near a light bulb, he notices that certain areas around the light bulb are warmer than
others. Which best explains this?
The areas to the sides of the light bulb are warmest because of conduction,
O The areas to the sides of the light bulb are warmest because of convection,
The area directly above the light bulb is warmest because of conduction,
The area directly above the light bulb is warmest because of convection.
Save and Exit
Submit
Mark this and retum
Nex
Answer:
The area directly above the light bulb is warmest because of convection.
Explanation:
if all the sides of the bulb are equally close to the light source inside the bulb, all area of the bulb would be equally heated by conduction. however, convection heating mainly heats up the surface above the light source. in convection heating, the air above the surface of the light source get heated by the light source and expands, casuing it to be less dense and rise to the top of the bulb. colder denser air at the top of the bulb sink to the light source adn gain heat and expands, becoming less dense. this process repeats and the surface above the light source becomes the warmest due to convection heating
A train is moving with a speed of 100 m/s. If the train is traveling south, at what position will it be 3 minutes after passing the +1,000-meter position marker ?
Remember, south is the negative direction and when you use the time, it must be in units of seconds. You will be applying one of the equations from above to solve this problem. And you must include a + sign if the final position is positive or a - sign if the final position is negative.
Answer:
The position after 3 minutes is - 800 m.
Explanation:
speed, v = 100 m/s
time, t = 3min = 180 s
initial position, x = 1000 m
let the distance traveled in 3 minutes is d
d = 100 x 180 = - 18000 m
So, the position is
= - 18000 + 1000 = - 800 m
You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s.
1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.
Answer:
Explanation:
Given that:
the putty initial speed (u) = 9.10 m/s
distance (s) between hand and the ceiling = 3.50 m
the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.
v² - u² = 2as
Since the putty is moving in a vertical motion(i.e. in an upward direction)
v² - u² = -2gs
v² = u² - 2gs
[tex]v = \sqrt{u^2 - 2gs}[/tex]
[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]
[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]
[tex]v = \sqrt{82.81 -68.6}[/tex]
[tex]v = \sqrt{14.21}[/tex]
v = 3.77 m/s
2.
The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:
v = u + at
In an upward direction
v = u - gt
making time t the subject;
v - u = -gt
[tex]t = \dfrac{v-u}{-g}[/tex]
[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]
[tex]t = \dfrac{-5.33}{-9.8}[/tex]
t = 0.54s
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
Answer: 1 cal is 4.186 J, 1 kcal = 4186 J A : 1014 m , B 200 m
Explanation: A) Work done by climber is change in potential energy.
W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.
Solve h = 160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m
B Energy is only 20 % : Then h = 0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.
Actually, muscles also produce heat from most of the energy provided by food.
A two-slit interference experiment in which the slits are 0.200 mm apart and the screen is 1.00 m from the slits. The m = 1 bright fringe in the figure is 9.49 mm from the central bright fringe. Find the wavelength of the ligh
Answer:
λ = 1.90 10⁻⁶ m
Explanation:
The interference pattern for the two-slit case is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
interference experiments angles are small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
λ = [tex]\frac{ d \ y}{m \ L}[/tex]
we calculate
λ = 0.2000 10⁻³ 9.49 10⁻³ / (1 1.00)
λ = 1.898 10⁻⁶ m
λ = 1.90 10⁻⁶ m
The wavelength of the light after calculation is find out to be λ = 1.90 *10⁻⁶ m
What is wavelength?
The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.
The interference pattern for the two-slit case is
d sin θ = m λ
let's use trigonometry
[tex]tan\theta=\dfrac{y}{L}[/tex]
interference experiments angles are small
[tex]sin\theta=\dfrac{y}{L}[/tex]
we substitute
[tex]\dfrac{dy}{L}=m\lambda[/tex]
[tex]\lambda=\dfrac{dy}{mL}[/tex]
we calculate
[tex]\lambda=\dfrac{0.2\times 10^{-3}\times 9.49\times 10^{-3}}{1\times 1}[/tex]
[tex]\lambda=1.90\times 10^{-6}\ m[/tex]
Hence the wavelength of the light after calculation is find out to be λ = 1.90 *10⁻⁶ m
To know more about wavelength follow
https://brainly.com/question/10728818
20 points and brainliest‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
parallel (at 0°) to a magnetic field of
0.0579 T. What is the magnetic force
on the charge?
Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
Find the weight of a man whose mass is 40 kg on earth.
(also
write complete data plus proper formula).
I am sure it help you with that much ☺️
Explanation:
pleasae give me some thanks please good morning sister
A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the south bank pointed in a direction 26o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground
Answer:
(a) The speed is 7.96 m/s
(b) The direction is 76 degree from positive X axis in counter clockwise direction.
Explanation:
Width of river = 280 m
speed of river, vR = 4.7 m/s towards east
speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north
[tex]vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j[/tex]
(a) The velocity of boat with respect to ground is
[tex]\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s[/tex]
(b) The direction is given by
[tex]tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o[/tex]
A compact car has a maximum acceleration of 3.0 m/s2 when it carries only the driver and has a total mass of 1300 kg. What is its maximum acceleration after picking up four passengers and their luggage, adding an additional 400 kg of mass?
Answer:
[tex]a_2=3.88m/s^2[/tex]
Explanation:
From the Question we are told that:
Initial Mass [tex]m_1=1300kg[/tex]
Final mass [tex]m_2=1300+400=>1700kg[/tex]
[tex]a_1=3.0m/s^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=m_1a_1[/tex]
[tex]F=1300*5[/tex]
[tex]F=6500N[/tex]
Generally the equation for Final acceleration is mathematically given by
[tex]F'=m_2*a_2[/tex]
[tex]a_2=\frac{6500}{1700}[/tex]
[tex]a_2=3.88m/s^2[/tex]
Jacie made a model to show the water cycle. The model she made is shown
below.
Which process in the model represents condensation?
A. As water vapor transfers heat to ice cubes, it forms clear droplets outside the
plastic wrap.
B. As water vapor gains heat from ice cubes, it forms clear droplets outside the
plastic wrap.
C. As water vapor transfers heat to ice cubes, it forms colored droplets inside the
plastic wrap.
D. As water vapor gains heat from ice cubes, it forms colored droplets inside the
plastic wrap
Answer:
option C
Explanation:
as water vapor transfer heat, colored drops are seen inside the wrap.
A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surface of the water directly below the bird. Does the bird appear to the diver to be closer to or farther from the surface than the actual bird
Answer:
The bird appears farther
Explanation:
This is because as the light from the bird travels into the water which has a higher refractive index than air, light rays from the kingfisher bird bend towards the normal at the water surface and thus enter the eye of the scuba diver. Now, if we project the light rays from the eyes of the scuba diver into the air, we see that they appear to come from a point farther than that of the actual kingfisher bird perched on the branch.
So, the bird appears to the diver to be farther from the surface than the actual bird
semiconductor have negative temperature coefficient of resistance why
Answer:
As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.
Explanation:
