If a spaceship has a momentum of 30,000 kg-m/s to the right and a mass of
400 kg, what is the magnitude of its velocity?
A. 12,000,000 kg-m/s
B. 75 m/s
C. 1,200 kg-m/s
D. 300 m/s
Answer:
75m/s
Explanation:
...................
A 64.7 cm long straight section of wire is located entirely inside a uniform magnetic field of |B| = 0.370 T. The wire is perpendicular to the direction of the magnetic field. When a current runs through the wire a magnetic force of |F| = 0.110 N acts on that section of wire. Calculate the size of the current.
Answer:
The current is [tex]I = 0.4595 \ A[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 64.7 \ cm = 0.647 \ m[/tex]
The magnetic field is [tex]B = 0.370 \ T[/tex]
The magnetic force is [tex]F = 0.110 \ N[/tex]
Generally the magnetic force exerted by the field is mathematically represented as
[tex]F = IL B sin \theta[/tex]
Making I the subject we have
[tex]I = \frac{F}{LB \ sin\theta}[/tex]
The angle here is 90° since the wire is perpendicular to the direction of the magnetic field
Substituting values
[tex]I = \frac{0.110}{ (0.370) * 0.647 \ sin(90)}[/tex]
[tex]I = 0.4595 \ A[/tex]
A woman on a bicycle travels at 9 m/s relative to the ground as she passes a little boy on a tricycle going in the opposite direction. If the boy is traveling at 1 m/s relative to the ground, how fast does the boy appear to be moving relative to the woman
Answer:
Useless
Explanation:
Question
A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, with the plane of the coil parallel to the field. In 0.230 ss , it is rotated so that the plane of the coil is perpendicular to the field.
(a) What is the change in the magnetic flux through the coil due to the rotation?
(b) What is the magnitude of the average emf induced in the coil during the rotation?
(c) What is the average current induced in the coil during the rotation?
Answer:
(a) ΔФ = -0.109W
(b) emf = 28.43V
(c) Iin = emf/R
Explanation:
(a) In order to calculate the magnetic flux you use the following formula:
[tex]\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)[/tex] (1)
B: magnitude of the magnetic field = 1.40T
A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2
Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.
Then, you obtain:
[tex]\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W[/tex]
The change in the magnetic flux is -0.109 W
(b) During the rotation of the coil the emf induced is given by:
[tex]emf=-N\frac{\Delta \Phi}{\Delta t}[/tex] (2)
N: turns of the coil = 60
ΔФ: change in the magnetic flux = 0.109W
Δt: lapse time of the rotation = 0.230s
You replace the values of the parameters in the equation (2):
[tex]emf=-(60)(\frac{-0.109W}{0.230s})=28.43V[/tex]
The induced emf is 28.43V
(c) The induced current in the coil is given by:
[tex]I_{in}=\frac{emf}{R}[/tex] (3)
R: resistance of the coil (it is necessary to have this value)
emf :induced emf = 28.43V
Which of the following selections completes the given radioactive decay? A). Alpha particle B). Gamma radiation. C). Positron. D). Neutron. E). Beta particle.
Answer:
Beta Particle
Explanation:
The beta particle decay completes the following reaction,
¹⁴C₆ ⇒ ¹⁴N₇ + Beta particle
Therefore the answer is option D.
What is radioactive decay?Radioactive decay is a type of nuclear reaction in which the unstable nucleus of a radioactive element releases energy in the form of nuclear radiation. Alpha decay, decay, and beta decay are a few examples of radioactive decay.
The radioactive deacy is acompained by the release of the enormous amount of energy,the nuclear power generation with the help of the nuclear reaction is one of the most significant application of the radioactive decay.
¹⁴C₆ ⇒ ¹⁴N₇ + Beta particle
Beta particles are the negatively charged electron emitted from radioactive decay.
For the given reaction the unstable carbon atom losses a beta particle to produce a stable nitrogen atom.
Thus the given radioactive decay can is completed by beta particle emission.
Learn more about radioactive decay from here
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1. Who is Nancy Lieberman?
2. What were some of her accomplishments throughout her life?
3. How could her accomplishments motivate other women in their lives?
4. Describe how you think she felt at the peak of her success?
5. How did her success contribute to women’s sports?
who can help me with that 5 questions please, I need help
Answer:
Nancy Elizabeth Lieberman (nacido el 1 de julio de, 1958), apodado "Señora mágica", ] es un ex profesional jugador de baloncesto y entrenador de la WNBA (WNBA) que es actualmente un organismo de radiodifusión para los New Orleans Pelicans de la Nacional La Asociación de Baloncesto (NBA) y la entrenadora principal de Power , un equipo en el BIG3 que lideró en su Campeonato 0.]bLieerman es considerada como una de las figuras más importantes del baloncesto femenino estadounidense.
Explanation:
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east. If the boat (including its crew) has a mass of 260 kg, what are the magnitude and direction of its acceleration? magnitude m/s2 direction ° north of east
Answer:
The magnitude of the acceleration is [tex]a_r = 1.50 \ m/s^2[/tex]
The direction is [tex]\theta = 32.5 6^o[/tex] north of east
Explanation:
From the question we are told that
The force exerted by the wind is [tex]F_{sail} = (330 ) \ N \ north[/tex]
The force exerted by water is [tex]F_{keel} = (210 ) \ N \ east[/tex]
The mass of the boat(+ crew) is [tex]m_b = 260 \ kg[/tex]
Now Force is mathematically represented as
[tex]F = ma[/tex]
Now the acceleration towards the north is mathematically represented as
[tex]a_n = \frac{F_{sail}}{m_b}[/tex]
substituting values
[tex]a_n = \frac{330 }{260}[/tex]
[tex]a_n = 1.269 \ m/s^2[/tex]
Now the acceleration towards the east is mathematically represented as
[tex]a_e = \frac{F_{keel}}{m_b }[/tex]
substituting values
[tex]a_e = \frac{210}{260}[/tex]
[tex]a_e =0.808 \ m/s^2[/tex]
The resultant acceleration is
[tex]a_r = \sqrt{a_e^2 + a_n^2}[/tex]
substituting values
[tex]a_r = \sqrt{(0.808)^2 + (1.269)^2}[/tex]
[tex]a_r = 1.50 \ m/s^2[/tex]
The direction with reference from the north is evaluated as
Apply SOHCAHTOA
[tex]tan \theta = \frac{a_e}{a_n}[/tex]
[tex]\theta = tan ^{-1} [\frac{a_e}{a_n } ][/tex]
substituting values
[tex]\theta = tan ^{-1} [\frac{0.808}{1.269 } ][/tex]
[tex]\theta = tan ^{-1} [0.636 ][/tex]
[tex]\theta = 32.5 6^o[/tex]
Three long parallel wires, each carrying 20 A in the same direction, are placed in the same plane with the spacing of 10 cm. What is the magnitude of net force per metre on central wire?
Answer:
F/L = 8*10^-4 N/m
Explanation:
To calculate the magnitude of the force per meter in the central wire, you take into account the contribution to the force of the others two wires:
[tex]F_N=F_{1,2}+F_{2,3}[/tex] (1)
F1,2 : force between first and second wire
F2,3 : force between second and third wire
The force per meter between two wires of the same length is given by:
[tex]\frac{F}{L}=\frac{\mu_oI_1I_2}{2\pi r}[/tex]
μo: magnetic permeability of vacuum = 4pi*10^-7 T/A
r: distance between wires
Then, you have in the equation (1):
[tex]\frac{F_N}{L}=\frac{\mu_oI_1I_2}{2\pi r}+\frac{\mu_oI_2I_3}{2\pi r}\\\\\frac{F_N}{L}=\frac{\mu_oI_1}{2\pi r}[I_2+I_3][/tex]
But
I1 = I2 = I3 = 10A
r = 10cm = 0.1m
You replace the values of the currents and the distance r and you obtain:
[tex]\frac{F_N}{L}=\frac{\mu_oI^2}{\pi r}\\\\\frac{F_N}{L}=\frac{(4\pi*10^{-7}T/A)(20A)^2}{2\pi (0.1m)}=8*10^{-4}\frac{N}{m}[/tex]
hence, the net force per meter is 8*10^-4 N/m
A person's resting heart rate typically lessens
with age.
Answer:
Ok
Explanation:
Please mark brainliest
After t hours a freight train is s(t) = 18t2 − 2t3 miles due north of its starting point (for 0 ≤ t ≤ 9). (a) Find its velocity at time t = 3 hours. 108 Incorrect: Your answer is incorrect. mi/hr (b) Find its velocity at time t = 7 hours. 546 Incorrect: Your answer is incorrect. mi/hr (c) Find its acceleration at time t = 1 hour. 42 Incorrect: Your answer is incorrect. mi/hr2
Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
A loop of wire carrying a steady current I is initially at rest perpendicular to a uniform magnetic field of magnitude B, as shown above. The loop is then rotated about a diameter at a constant rate. The torque on the loop is maximum when the loop has rotated, with respect to its initial position, through an angle of:__________.
(A) 30°
(B) 45°
(C) 90°
(D) 180°
(E) 360°
your answer is b to be presided
Can someone please help me with this question thank you!
A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Explanation:
We have,
Mass of a baseball is 0.147 kg
Initial velocity of the baseball is 44.5 m/s
The ball is moved in the opposite direction with a velocity of 55.5 m/s
It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Change in momentum,
[tex]\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s[/tex]
Impulse = 14.7 kg-m/s
Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s
A cosmic ray electron moves at 7.6 x 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.02 x 10-5 T. What is the radius of the circular path the electron follows
Answer:
r = 4.23 m
Explanation:
To find the radius of the circular path of the electron you use the following formula:
[tex]r=\frac{mv}{qB}[/tex] (1)
This formula can be used because the motion of the electron is perpendicular to the direction of the magnetic field vector.
m: mass of the electron = 9.1*10^-31 kg
q: charge of the electron = 1.6*10^-19 C
B: magnitude of the magnetic field = 1.02*10^-5 T
v: velocity of the electron = 7.6*10^6 m/s
You replace the values of m, v, q and B in the equation (1):
[tex]r=\frac{(9.1*10^{-31}kg)(7.6*10^6 m/s)}{(1.6*10^{-19}C)(1.02*10^{-5}T)}\\\\r=4.23m[/tex]
hence, the raiuds of the orbit of the electron is 4.23m
Coulomb's law: A point charge Q is located a short distance from a point charge 3 Q, and no other charges are present. If the electrical force on Q is F, what is the electrical force on 3 Q
Answer:
The electrical force on 2nd charge (3 Q) will also be equal to F.
Explanation:
The electrical force between two electrically charged particles is given by the Coulomb's Law. The mathematical formula of Coulomb's Law is as follows:
F = kQ₁Q₂/r²
where,
F = Electrical Force exerted by both particles on each other
k = Coulomb's Constant
Q₁ = Magnitude of 1st Charge = Q
Q₂ = Magnitude of 2nd Charge = 3 Q
r = Distance between the charges
Therefore,
F = k(Q)(3 Q)/r²
F = 3 kQ²/r²
Since, it is clear from the formula that the magnitude of force applied by the first charge on 2nd charge is equal to the force applied by 2nd charge on 1st charge.
Therefore, the electrical force on 2nd charge (3 Q) will also be equal to F.
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 74.0 kg, and the height of the water slide is 11.3 m. If the kinetic frictional force does -5.42 × 103 J of work, how fast is the student going at the bottom of the slide?
Answer:
The student is going at the bottom of the slide with a velocity of 8.66 m/s
Explanation:
Given;
mass of the student, m = 74 kg
height of the water slide, h = 11.3 m
work done, W = -5.42 × 10³ J
Apply work energy theorem;
[tex]W = \frac{1}{2}mv_f^2+ mgh_f-(\frac{1}{2}mv_o^2 + mgh_o)\\\\ W + \frac{1}{2}mv_o^2 + mgh_o -mgh_f= \frac{1}{2}mv_f^2\\\\ W + \frac{1}{2}mv_o^2 +mg(h_o-h_f) = \frac{1}{2}mv_f^2\\\\\frac{W}{m} + \frac{1}{2} v_o^2 + g(h_o-h_f) = \frac{1}{2}v_f^2\\\\\frac{2W}{m}+ v_o^2 + 2g(h_o-h_f) = v_f^2\\\\v_f = \sqrt{\frac{2W}{m}+ v_o^2 - 2g(h_f-h_o)} \\\\v_f = \sqrt{\frac{2(-5.42*10^3)}{74}+ (0)^2 - 2*9.8(-11.3)}\\\\v_f=\sqrt{-146.4865+221.48} \\\\v_f = \sqrt{74.9935} \\\\v_f = 8.66 \ m/s[/tex]
Therefore, the student is going at the bottom of the slide with a velocity of 8.66 m/s
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a car is moving away from this stereo at a speed of 40.0 m/s. The frequency of sound waves that the car receives is ________. In addition, when the car is 70.0 meters away from the speaker, the car will hear sound waves with a sound intensity level of _________ .
Answer:
a) f' = 432 Hz
b) I = 8.12*10^-4 W/m^2
Explanation:
a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:
[tex]f'=f(\frac{v-v_o}{v+v_s})[/tex] (1)
where
f: frequency of the source = ?
v: speed of sound = 343 m/s
vo: speed of the observer = 40.0 m/s
vs: speed of the source = 0 m/s (stationary)
You replace the values of all parameters in the equation (1):
To calculate f' you first calculate the frequency of the sound wave, by using the following formula:
[tex]v=\lambda f\\\\[/tex]
v: speed of sound
λ: wavelength = 0.700 m
[tex]f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz[/tex]
Next, you replace the values of all parameters in the equation (1):
[tex]f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz[/tex]
hence, the frequency perceived by the car is 432 Hz
b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:
[tex]I=\frac{P}{4\pi r^2}[/tex]
P: power of the source = 50.0 W
r: distance to the source = 70.0 m
[tex]I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}[/tex]
hence, the intensity is 8.12*10^⁻4 W/m^2
With the wire back at its initial location, you connect a second identical battery in series with the first one. When you close the switch, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection
Answer:
It is larger
Explanation:
I believe the answer would be "It is larger" since v = 2v (since it's in series), which would increase the magnetic field and thus produced a bigger angle.
Choose the INCORRECT statement: A) Gauss' law can be derived from Coulomb's law B) Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface C) Coulomb's law can be derived from Gauss' law and symmetry D) Gauss' law applies to a closed surface of any shape E) If a closed surface encloses no charge, the electric field is zero everywhere on the surface
Answer:
A
Explanation:
A)
INCORRECT
No, Gauss's Law can not be derived from Coulomb's Law alone. Since, Coulomb's Law provides electric field produced by a single point charge. Therefore, it is important to assume principle of super position along with Coulomb's law to get the field due to all charges. So, Gauss's Law can be proved by the help of both Super Position Principle and Coulomb's law.
B)
CORRECT
Yes, Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface.
C)
CORRECT
Yes, Coulomb's law can be derived from Gauss' law and symmetry.
D)
CORRECT
Yes, Gauss's Law applies to a closed surface of any shape.
E)
CORRECT
Yes, if a closed surface encloses no charge, the electric field is zero everywhere on the surface.
Which explains what a velocity time graph would look like with no acceleration?
O curved line on a upward slope
O straight line on a downward slope
O straight line with no slope
O curved line on a downward slope
Answer:
Hey!
Your answer is C!
Explanation:
HOPE THIS HELPS!!
Straight line with no slope explains what a velocity time graph would look like with no acceleration. The correct option is 3.
When there is no acceleration, the item is travelling at a steady speed. The y-axis of a velocity-time graph represents velocity, while the x-axis represents time.
A straight line with no slope implies constant velocity, where the velocity remains constant over time. There is no change in velocity, which indicates that there is no acceleration.
Thus, on the velocity-time graph, a straight line with no slope depicts an object travelling with no acceleration. The correct option is 3.
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Your question seems incomplete, the probable complete question is:
Which explains what a velocity time graph would look like with no acceleration?
O curved line on a upward slopeO straight line on a downward slopeO straight line with no slopeO curved line on a downward slopeWILL VOTE MOST BRAINLIEST
Which of the following explains why the inner planets are different than the outer planets?
A) Due to the greater inertia of the dust particles, the dust remained close to the sun to form the inner planets leaving gases to form the outer planets.
B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.
C) Due to the greater momentum of the gas particles, the gas particles flew farther away from the sun than dust and metals to form the outer planets, leaving dust and metals to form the inner planets.
Answer:
B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.
Explanation:
Inner planets are smaller and rockier than outer gas planets,outer planets are larger,because of their lower gravity they don't attract extreme amounts of gas in their planets.The four outer planets were so far from the Sun that its winds could not blow away their ice and gases
17. A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?
Answer:
The monkey will get hit by the dart. This is because when the dart is fired the force of gravity makes it fall from its original position. The monkey also has the same fall from the branches which means they will both be falling at the same gravitational rate.
This will eventually make the monkey to get hit by the dart.
A gardener pushes a shovel into the ground with a force of 75 N. The angle of the shovel to the ground is 80 degrees. What is the downward force of the shovel? A. 13.0 N B. 73.9 N C. 75.2 N D. 80.0 N
Answer:
B
Explanation:
Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.
9. Santa brings a special baby a bouncy swing. The child bounces in a harness, with a spring constant k, and is suspended off of the ground. a. If the spring stretches x1 = 0.27 m from equilibrium while supporting an 7.15-kg child, what is its spring constant, in newtons per meter?
Answer:
259.52N
Explanation:
Force initiated on a spring that causes an extension is related by the expression below;
F= K×e
Where F is the Force
K is the spring constant
e is the extension caused by the spring.
By change of subject formula for K;
K = F/e
Now F is the same as weight and is given by mass× acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}
Hence W =7.15×9.8=70.07N
Hence K = 70.07/0.27 =259.5185N/m
=259.52N/m to 2 decimal place.
2.A 5 Ohm resistor is connected to a 9 Volt battery. How many Joules of thermal energy are produced in 7 minutes?
3.The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
4.If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
7.A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the circuit?
8.How many Joules of electric energy are delivered to a 60 Watt lightbulb if the bulb is left on for 2.5 hours?
Answer:
Explanation:
2 )
power of an electric device = V² / R where V is volts and R is resistance
putting given data
power = 9²/ 5
= 16.2 J/s
energy produced in 7 minutes
= 16.2 x 7 x 60
= 6804J .
3 ) Power of an electrical device
= V² / R
= V X I where I is current
= 4.5 x .5
= 2.25 W or J/s
4 )
energy used in 3 minutes with power of 2.25 W
= 2.25 x 3 x 60
= 405 J .
7 )
power of a electrical device
= V x I
IR x I where R is resistance .
= I²R
putting given data
power = .005² x 50
= 1.25 x 10⁻³ W .
8 )
Energy used up by a 60 W bulb in 2.5 hours
= 60 x 2.5 x 60 x 60
= 5.4 x 10⁵ J .
A gas at pressure 100 atm and volume 3.7 L has the pressure reduced to 32 atm what is the new volume of the gas
Answer:
The new volume is [tex]11.6L[/tex]
Explanation:
This problem is on the application of Boyle's law which states that" the volume of a given mass of gas at constant temperature is inversely proportional to the pressure"[tex]pressure \alpha \frac{1}{volume}[/tex]
Given data
initial pressure p1= 100atm
initial volume v1= 3.7L
final pressure p2= 32 atm
final volume v2= ?
Apply Boyle's law we have
[tex]p1v1= p2v2\\[/tex]
[tex]v2= \frac{p1v1}{p2}[/tex]
Substituting our data we have
[tex]v2= \frac{100*3.7}{32} \\v2=\frac{370}{32} \\v2= 11.6 L[/tex]
The new volume is [tex]11.6L[/tex]
What is a limitation of the electron cloud model theory that
a law about electrons would not have?
Answer:
Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.
An advertisement for a new fish food claims that lab studies show that fish grew three inches in three weeks while eating the food. Travis wondered if the food was worth the extra cost. To evaluate the claim in the advertisement, Travis should
Answer:
look at the data for the control group to see how much the fish grew without the new food.
Explanation:
i got it right
Particle A of charge 3.30 10-4 C is at the origin, particle B of charge -6.24 10-4 C is at (4.00 m, 0), and particle C of charge 1.06 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C?
Answer:
a) [tex]E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]
b) zero
Explanation:
a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:
[tex]E_T=E_1+E_2[/tex]
E1: electric field of charge 1
E2: electric field of charge 2
It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:
[tex]E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\[/tex]
r13: distance between charges 1 and 3
r12: charge between charges 2 and 3
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
Thus, you first calculate the distance r13 and r23, and also the angles:
[tex]r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°[/tex]
Next, you replace the values of all parameters in order to calculate E1 and E2:
[tex]E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}[/tex]
finally, you obtain for ET:
[tex]E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}[/tex]
b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the brakes than a sober driver's car? Assume that both are initially traveling at 50.0 mi/h and their cars have the same acceleration while slowing down, and that the sober driver takes 0.33 s to hit the brakes in a crisis, while the drunk driver takes 1.0 s to do so. (5280 ft = 1 mi)
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
ΔS = 49.13 ft
