You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.681 m apart. At what distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source
Answers
Complete Question
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.681 m apart. At what distance, in kilometres, are you marginally able to discern that there are two headlights rather than a single light source?Take the wavelength of the light to be 549 nm and your pupil diameter as 4.63 mm.
Answer:
The distance is \(z = 4707.6 \ m\)
Explanation:
From the question we are told that
The is distance between the headlight is \(d = 0.681 \ m\)
The wavelength is \(\lambda = 549 \ nm = 549 *10^{-9} \ m\)
The pupil diameter is \(D = 4.63 \ mm = 0.00463 \ m\)
Generally, we can mathematically evaluate the resolution of the eye as
\(\theta = \frac{1.22 * \lambda }{D}\)
substituting values
\(\theta = \frac{1.22 * 549 *10^{-9} }{0.00463}\)
\(\theta = (1.45 *10^{-4} )^o\)
Now according to SOHCAHTOA rule
\(sin \theta = \frac{ d}{z}\)
Where z is the distance at which the eye can discern the two head light
given that the angle is very small \(sin \theta = \theta\)
=> \(\theta = \frac{ d}{z}\)
substituting values
\(1.45*10^{-4} = \frac{ 0.681}{z}\)
=> \(z = \frac{0.681}{1.45 *10^{-4}}\)
=> \(z = 4707.6 \ m\)
Related Questions
The mass of a satellite orbiting Earth is 15000 kg.
Answers
Newton's universal law of gravity
\(F=G\frac{m1\cdot m2}{r^2^{}}\)Where:
F= force between objects
m1= mass 1 = 15,000 kg
m2= mass 2 = 6x10^24
r = distance = 34,000,000 m
G= universal contant of gravitation = 6.67 x10^-11
Replacing:
\(F=6.67\cdot10^{-11}\cdot\frac{15,000\cdot6x10^{24}}{(34,000,000)^2}\)F= 5,193 N
All of the following steps are included in a self-modification program except __________. A. specifying target behavior B. designing a program C. focusing on one goal at a time D. gathering data about target behavior
(Psychology)
Answers
Answer:
C)focusing on one goal at a time
Explanation:
Self-modification programs could be regarded as a program that help individual in managing unwanted as well as dysfunctional behavioral responses whenever they are going through a problem and try to deal with it. For instance the dysfunctional behavioral response for someone with a panic attack is avoidance. Some of the the steps that are involved in in a self-modification program are;
✓ specifying target behavior ✓designing a program
✓ gathering data about target behavior
Answer:
c
Explanation:
on edge 2020
Planet Earth exerts a force of 175N on a boy named Bobby. How much force will the
Earth exert on Bobby if his mass increases by 4 times?
Answers
Answer:
175(4) = 700 N
Explanation:
Car P travels due East along a straight highway at a constant speed of 30 m/s. At 9:00
a.m., P passes Exit 17. At precisely the same moment, car Q passes Exit 16, traveling due
West at a constant 26 m/s. Slightly later, car P and car Q pass the same point. Knowing
the exits are exactly 7 km apart, determine how many minutes past 9:00 a.m. the cars pass
each other.
Answers
Knowing the exits are exactly 7 km apart, the cars pass each other at 9:29 and 15 seconds a.m.
How to calculate time?The relative velocity of the cars is 30 m/s - 26 m/s = 4 m/s.
The distance between the cars is 7 km = 7000 m.
The time it takes for the cars to pass each other is 7000 m / 4 m/s = 1750 seconds.
1750 seconds is 29 minutes and 15 seconds.
To calculate the time in minutes;
Let:
v_p = the speed of car P (m/s)
v_q = the speed of car Q (m/s)
d = the distance between the cars (m)
t = the time it takes for the cars to pass each other (s)
Given that:
v_p = 30 m/s
v_q = 26 m/s
d = 7000 m
Use the equation for relative velocity to find the velocity of the cars relative to each other:
v_r = v_p - v_q
v_r = 30 m/s - 26 m/s = 4 m/s
Use the equation for distance to find the time it takes for the cars to pass each other:
d = v_r × t
7000 m = 4 m/s × t
t = 7000 m / 4 m/s = 1750 s
Convert 1750 seconds to minutes and seconds:
1750 s = 29 minutes and 15 seconds
Therefore, the cars pass each other at 9:29 and 15 seconds a.m.
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How much power does motor provides an elevator in a building when the total mass when loaded is 500 kg to a height of 3m in 25 seconds?
Answers
What is radioactivity????
Answers
Radioactivity is a nuclear phenomenon. It is the process of spontaneous emission of α (alpha particles) or β (beta particles) and γ (gamma particles ) radiations from the nucleus of atoms during their decay.
Uranium and radium are the two radioactive substances.Explanation:
Hope it helps you!!
Answer:
radioactivity is the act of emitting radiation spontaneously. This is done by an atomic nucleus that, for some reason, is unstable; it "wants" to give up some energy in order to shift to a more stable configuration.
Explanation:
Can someone explain how to do the algebra for this question? I know everything else, I just don’t know how to rearrange the question to solve for v.
Answers
Answer:
Refer to the step-by-step Explanation.
Step-by-step Explanation:
Simplify the equation with given substitutions,
Given Equation:
\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)
Given Substitutions:
\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)
Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)
Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the left-hand side of the equation:
\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
Simplifying the third term.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)
"R²'s" cancel, we are left with:
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)
We have like terms, combine them.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)
Each term has an "m" in common, factor it out.
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)
Now we have the following equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the right-hand side of the equation:
\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)
\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)
Now we have the equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
\(\hrulefill\)
Now solving the equation for the variable "v":
\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
Dividing each side by "m," this will cancel the "m" variable on each side.
\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)
Subtract the term "gh" from either side of the equation.
\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)
Multiply each side of the equation by "10/7."
\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)
Now squaring both sides.
\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)
Thus, the simplified equation above matches the simplified equation that was given.
Each of the following figures shows a person (not to scale) located on Earth at either 40°N or 40°S latitude. Rank the figures based on how much time the person spends in daylight during each 24-hour period, from most to least. To rank items as equivalent, overlap them.
Answers
The ranking is based on the tilt of the Earth's axis and its orbit around the Sun. The figure at 40°N in June receives the most daylight because it is located at a high latitude during the summer solstice in the Northern Hemisphere. The Earth's axis tilts towards the Sun, resulting in longer days and shorter nights. The figure at 40°S in December receives a moderate amount of daylight as it is located at a lower latitude during the summer solstice in the Southern Hemisphere.
The figure at 40°N in December experiences less daylight because it is located at a high latitude during the winter solstice in the Northern Hemisphere, with shorter days and longer nights. Lastly, the figure at 40°S in June receives the least amount of daylight as it is located at a lower latitude during the winter solstice in the Southern Hemisphere, where the days are shortest and the nights are longest. Based on the information given, the ranking of figures based on the amount of daylight they experience in a 24-hour period, from most to least.
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You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
of 0.50 m/s. What is the work done by gravity on the book?
1) +10 J
2) -10
3) +5.0
4) -5.0
Answers
The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).
The force exerted on the book by gravity has magnitude
F = mg = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N
You raise the book 1.0 m in the opposite direction, so the work done is
W = (10 N) (-1.0 m) = -10 J
The work done by gravity on the book is -10 Joules. The correct option is 2.
What is gravity?The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.
Given are the three positions. The Earth has the acceleration due to gravity, g = 9.81 m/s²,
The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).
The force exerted on the book by gravity has magnitude
F = mg = 10 N x( -9.80 m/s² )= -9.8 N
F ≈ -10 N
The book is raised by 1.0 m in the opposite direction, so the work done is
W =- 10 N x 1.0 m = -10 J
Thus, the correct option is 2.
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If energy is conserved in a pendulum then the potential energy at the top does what?
Answers
At the greatest point of the pendulum's swing (the top of the arc), the kinetic energy is at its least since the velocity is momentarily zero.
What is Mechanical Energy?Mechanical energy is the product of potential and kinetic energy. The principle of conservation of mechanical energy states that if an isolated system is subject only to conservative forces, then the mechanical energy is constant.
There are two types of mechanical energy – motion (kinetic energy) and stored (potential energy)
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Question 7 of 10Which of the following is a correct statement of Ohm's law?V=O A.RRDVIO B. R=O c. R = 1R=OOD. I=RV
Answers
According to Ohm's law,
\(\begin{gathered} V=IR \\ I=\frac{V}{R} \\ R=\frac{V}{I} \end{gathered}\)Thus, the correct option is B.
A sine bar is used to determine the angle of a part feature. The length of the sine
bar=8.000in. The rolls have a diameter of 1.000 in. All inspection is performed on a surface plate. In
order for the sine bar to match the angle of the part, the following gage blocks must be stacked: 2.0000,
0.5000, 0.2500, and 0.0050. Determine the angle of the part feature.
Answers
A sine bar is used to get the angular measurement of a part feature. The angle of the part feature is 18.24°
From the question
Given that,
Length of the sine bar, L = 8.000 in
Diameter of the rolls = 1.000 in
Height under the roll, H = 2.0000 + 0.5000 + 0.0050
= 2.505 in
From sine bar formula,
we know that,
H = sin A x L
sin A = H ÷ L
where,
A ⇒ angle of part feature
H ⇒ height under the roll
L ⇒ length of the sine bar
Substituting values in the above equation,
sin A = H / L
A = sin⁻¹ ( 2.505 ÷ 8 )
A = sin⁻¹ (0.3131)
A = 18.24⁰
Hence the angle of the part feature = 18.24°
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Ann is driving down a street at 56 km/h. Suddenly a child runs into the street. If it takes Ann 0.749 s to react and apply the brakes, how far will she have moved before she begins to slow down? Answer in units of m.
Answers
The distance travelled by Ann before she begins to slow down is 11.65 m
How do I determine the distance travelled by Ann?First, we shall enlighten ourselves on what speed is. This is given below.
Speed is the distance an object travelled per unit time. It can be expressed as:
Speed = distance / time
Finally, we can obtain the distance Ann travelled as illustrated below.
From the question given above, the following data were obtained:
Speed = 56 Km/h = 56 / 3.6 = 15.56 m/sTime = 0.749 sDistance =?Speed = distance / time
15.56 = distance / 0.749
Cross multiply
Distance = 15.56 × 0.749
Distance = 11.65 m
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If an object that floats on the surface displaces 10 cm³ of water, how much
does that object weigh? Explain how you know.
Answers
The weight of the body is 0.0979N
What do you mean by the weight of a body?
The force of gravity acting on an object is known as its weight, which may be computed using the formula
w = mg,
or mass times gravity acceleration.
The SI unit for weight is the newton since it is a force.
In order for an object to be buoyant, its weight must be equal to its own or, if it is floating, equal to water volume times a specific water weight equals W.
W=FB
W=γVd
where
γ = specific weight of the fluid [kN/m^3]
Vd = volume of displaced fluid
The specific weight of water at 20∘C is 9.79 kN/m^3
∴ W = 9.79 (kN/m^3) (10cm^3)(10^−6 m^3/cm^3)
= 97.9 x 10^−6 kN
W = 0.0979N
Hence, The weight of the body is 0.0979N
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(b) Figure 11.1 shows a uniform meter rule balanced horizontally on a knife-edge placed at the 58cm mark when a mass of 20g is suspended from the end. 0cm 58cm Figure 11.1 (i) Find the mass of the rule. (ii) What is the weight of the rule. (taking g = 10m/s²)? 100cm 20g [2] [2]
Answers
The mass of the rule is 3.36 kg and the weight of the rule is 33.6 N.
A knife should be balanced somewhere, right?The butt of the blade is designed to balance the majority of fine chef's knives. This is due to the fact that a pinch grip is used in both Western and Eastern cutting styles for improved control. Therefore it makes sense that you'd want your knife to be balanced where you'll be holding it.
To balance the meter rule, Assume the mass of the rule "M".
It is possible to determine the rule's magnitude and weight, which act downward:
weight of rule = M * g
where g = acceleration due to gravity.
weight of rule * (50 cm) = (20 g) * (100 cm - 58 cm)
M * g * (50 cm) = (20 g) * (42 cm)
M = (20 g * 42 cm) / (50 cm * g)
M = 33.6 g / g
M = 33.6 g / 10 m/s^{2}
M = 3.36 kg
(ii) The weight of the rule:
weight of rule = M * g
weight of rule = 3.36 kg * 10 m/s^{2}
weight of rule = 33.6 N
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The half-life of a radioactive isotope is 210 d. How many days would it take for the decay rate of a sample of this isotope to fall to 0.58 of its initial rate?
Answers
It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.
1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.
2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.
3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.
4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.
5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.
6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".
7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).
8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).
9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.
10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.
11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).
12. Evaluating the expression, we find: t ≈ 546.
13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.
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A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Answers
Answer:
2.1x10^6m/s
Explanation:
One electron has a charge of –1.602e-19 C
mass of electron is 9.1e-31 kg
mass of proton is 1.6726e−27 kg
mass ratio is 1.6726e−27 / 9.1e-31 = 1838
The force is constant, F
distance is constant, d
a = F/m
a increases by a factor 1838, as m decreases by that factor
a = a₀1838
v₀² = 2a₀d
v² = 2a₀d1838
v²/v₀² = 2a₀d1838 / 2a₀d = 1838
v² = 1838v₀² = 1838(45000)²
v = 45000√1838 = 2.1e6 m/s
11.) If figure skater spins with a circular acceleration of 275 m/s^2 in a circle 5 poin with a radius of 0.55 meters, calculate the velocity of the skater."
Answers
Answer:
we don't care what ure saying thats absurd
A uniform solid cylindrical flywheel has a mass of 50 kg and a radius of 40 cm. The flywheel begins to rotate faster with an acceleration of 1.5 rad/s2. The kinetic energy of the flywheel after 1 minute of rotation is:
A. 16.2 KJ
B. 180 KJ
C. 40.5 KJ
D. 32.4 KJ
Answers
The kinetic energy of the flywheel after 1 minute of rotation, given that it has a mass of 50 and radius of 40 cm is 32.4 KJ (Option D)
How do I determine the kinetic energy?We'll begin by obtaining the velocity of the flywheel. This is shown below:
Radius (r) = 40 cm = 40 / 100 = 0.4 mAcceleration (a) = 1.5 rad/s² = 1.5 × 0.4 = 0.6 m/s²Time (t) = 1 minute = 1 × 60 = 60 sVelocity (v) = ?v = at
v = 0.6 × 60
v = 36 m/s
Finally, we shall determine the kinetic energy of the flywheel. Details below:
Mass (m) = 50 KgVelocity (v) = 36 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 50 × 36²
KE = 25 × 1296
KE = 32400 J
Divide by 1000 to express in KJ
KE = 32400 / 1000
KE = 32.4 KJ
Thus, the kinetic energy is 32.4 KJ (Option D)
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Two equal and opposite charges are placed 40mm apart,if the force between them is found to be 0.5N Calculate the magnitude of the charge
Answers
F = k * (q1 * q2) / r^2
where k is Coulomb's constant (9 x 10^9 N m^2 C^-2).
In this problem, we have two equal and opposite charges, so we can assume that q1 = -q2 = q. The distance between them is 40 mm, which is 0.04 m. We are given that the force between them is 0.5 N. Therefore, we can set up an equation:
0.5 N = k * (q * q) / (0.04 m)^2
Simplifying and solving for q:
q^2 = (0.5 N * (0.04 m)^2) / k
q^2 = (0.5 N * (0.04 m)^2) / (9 x 10^9 N m^2 C^-2)
q^2 = 8.88 x 10^-12 C^2
q = sqrt(8.88 x 10^-12 C^2)
q = 2.98 x 10^-6 C
Therefore, the magnitude of each charge is 2.98 x 10^-6 C.
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answers
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
\(\Phi_B=S\cdot B=SBcos\alpha\) (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
\(B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT\)
The strength of the magntetic field is 4.1mT
Two tiny objects with equal charges of 7.25 μC are placed at the two lower corners of a square with sides of 0.201 m, as shown.
Find the electric field at point B, midway between the upper left and right corners. If the direction of the electric field is upward, enter a positive value. If the direction of the electric field is downward, enter a negative value. answer in N/C
Answers
The electric field at point B, located at the midpoint between the upper left and right corners of the square, can be approximated as 3.244 x \(10^6\) N/C in the upward direction.
To find the electric field at point B, we can consider the contributions from the two charges placed at the lower corners of the square. Since the charges are the same and the distance to point B is the same for both charges, the magnitudes of the electric fields produced by each charge will be equal.
First, let's calculate the magnitude of the electric field produced by one of the charges at point B using Coulomb's Law:
Electric field due to a point charge (E) = (k * q) / \(r^2\)
Where:
- k is the electrostatic constant, approximately equal to 8.99 x 10^9 \(10^9\)N \(m^2/C^2\)
- q is the charge of the object, 7.25 μC (7.25 x \(10^-^6\) C)
- r is the distance from the charge to point B, which is half the length of the square's side, 0.201 m / 2 = 0.1005 m
Plugging in the values, we have:
E = (8.99 x \(10^9 N m^2/C^2\) * 7.25 x \(10^-^6\) C) / (0.1005 \(m)^2\)
E ≈ 1.622 x \(10^6\) N/C
Since the electric fields from the two charges at the lower corners have equal magnitudes and point in the same direction (upward), the total electric field at point B is twice the magnitude of the individual electric field:
Electric field at point B = 2 * E ≈ 2 * 1.622 x \(10^6\) N/C
Electric field at point B ≈ 3.244 x \(10^6\) N/C
Therefore, the electric field at point B, midway between the upper left and right corners of the square, is approximately 3.244 x \(10^6\)N/C upward.
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What is the total charge of barium oxide?
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I need help with this question
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Explanation:
a boy walks at 2m/s for 30sec and run at 5m/s for 20sec. what is his average speed
Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the source, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.
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Answer:
d. unchanged.
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other
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Answer:
The force remains the same.
Explanation:
Let the magnitude of the forces in each case be F1 and F2 respectively.
The charges are Q1 and Q2.
The distance of separation between them is R.
Hence, for F1;
F1 = KQ1Q2/R^2
For F2:
F2 = K * 3Q1 * 3Q2/(3R)^2
F2 = 9KQ1Q2/9R^2
F2 =KQ1Q2/R^2
Hence F1=F2
The force is the same in both cases!
Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride.Part AIf a boat and its riders have a mass of 800 kg and the boat drifts in at 1.4 m/s how much work does Sam do to stop it?
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Given data
*The given mass is m = 800 kg
*The given speed is v = 1.4 m/s
The formula for the work done by the Sam to stop it is given as
\(W=\frac{1}{2}mv^2\)Substitute the known values in the above expression as
\(\begin{gathered} W=\frac{1}{2}(800)(1.4)^2 \\ =784\text{ J} \end{gathered}\)Hence, the work done by the Sam to stop it is W = 724 J
areaAgNO3 + ___Response area_Li →_____Response areaLiNO3 +____Response areaAg
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Answer:
1 is 3 -1
Explanation:
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a pound of body fat stores in the amount of chemical energy equivalent to 3500 calories. When sleeping the average adult burns on and expends about 0.45 Cal/h How many calories would a 150 lb person burn during 8 hours of sleep?How how long would the personhave to sleep continuously to lose one pound of body fat
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We know that
• The average is 0.45 calories per hour.
,• The energy stored is 3500 calories per pound of body fat.
,• The mass is 150 pounds.
,• The hours of sleep are 8.
First, we have to multiply the mass 150 lb by the average rate
\(150lb\times0.45=67.5\text{cal}/hr\)This means that this person burns 67.5 calories per hour. Now, we have to multiply this rate by 8 hours.
\(67.5\times8=540\text{cal}\)Therefore, a 150 lb person would burn 540 calories during 8 hours of sleep.Now, we know that 1 pound of body fat is equivalent to 3500 calories. So, let's use a rule of three to find the period of time would take this person to lose one pound of body fat (3500 cal).
\(\frac{67.5\text{cal}}{1hr}=\frac{3500\text{cal}}{x}\)Let's solve for x
\(\begin{gathered} x=\frac{3500}{67.5}hr \\ x\approx51.9hr \end{gathered}\)Therefore, this person would have to sleep continuously for 51.9 hours to lose one pound of body fat.34. JAnswer this: Pure (24 carat) gold has a density of 19 g/mL. If you bought"gold" ring and found it had a volume of 0.3 ml and that it weighed 5.7 grams, is it pure gold? Show your work to support your answer.
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Explanation:
Density = mass / volume
ρ = 5.7 g / 0.3 mL
ρ = 19 g/mL
Yes, it's pure gold.
Yes, it's pure gold.
How do you calculate gold density?The mass of an object divided by its volume is the formula for determining density. This is expressed mathematically as d = m/v, where d denotes density, m denotes mass, and v denotes object volume. The common measurements are kg/m³.What is the formula of volume?The basic formula for volume is length, breadth, and height, as opposed to length, width, and height for the area of a rectangular shape.What is the volume of pure gold?Gold has a specific gravity of 19.3. In light of this, 19.3 grams take up 1 cubic centimeter. 31.1 grams make up a Troy ounce. As a result, we need 31.1/19.3 = 1.61 cubic centimeters to equal 1 Troy Oz.According to the question:
Density = mass / volume.
ρ = 5.7 g / 0.3 mL.
ρ = 19 g/mL.
Hence, Yes it's pure gold.
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