Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C
The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.
We have to use Coulomb's Law and the principle of charge conservation.
1. Conservation of Charge:
The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:
[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]
2. Coulomb's Law:
The force between two charges is given by Coulomb's Law:
[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]
Let's set up the equation:
[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]
Rearrange to solve for [tex]\( q_A q_B \)[/tex]:
[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]
Now, we have two equations:
1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]
2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]
To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:
[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]
[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]
[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]
Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]
This gives us two solutions for [tex]\( q_A \)[/tex]:
1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]
2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]
Since bead A has the greater charge, we take:
[tex]\[q_A = 21.907 \, \text{nC}\][/tex]
And the charge on bead B:
[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]
So, the charges on the beads are:
[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]
A 100 Ω resistive heater in a tank of water (1kg) increases its temperature from 10°C to 20°C over a period of 1 hour. During this period of time your measurements show that the voltage of the heater was 50V. How much energy is not absorbed by the water and leaves the system into the srroundings? Assume constant specific heat and density.
Answer:48.52 kJ
Explanation:
Given
Resistance[tex]=100 \Omega [/tex]
temperature increases from [tex]10^{\circ}C to 20^{\circ}C[/tex]
Voltage=50 V
Heat given(H)[tex]=\frac{V^2t}{R}[/tex]
Where V=voltage applied
t=time
R=Resistance
[tex]H=\frac{50^2\times 60\times 60}{100}=90 kJ[/tex]
Heat absorbed by water is
[tex]Q=mc(\Delta T)[/tex]
where
m=mass of water
c=specific heat of water
[tex]\Delta T[/tex]=change in temperature
[tex]Q=1000\times 4.184\times (20-10)=41.48 kJ[/tex]
Therefore 90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.
Find the work done "by" the electric field on a negatively charged point particle with a charge of 7.7 x 10^-6 C as it is moved from a potential of 15.0 V to one of 5.0 V. (Include the sign of the value in your answer.)
Answer:
77 x 10⁻⁶ J
Explanation:
Work done by electric field
= charge x potential difference
= 7.7 x 10⁻⁶ x ( 15 - 5)
= 77 x 10⁻⁶ J
work done will be positive because direction of force and displacement are same.
A student has 474 J of gravitational energy while standing on a stool 0.84 m above the ground. The mass of the student is: a) 40 kg (to two significant digits) b) 58 kg c) 48kg d) 60 kg (to two significant digits)
Answer:
b) 58 kg
Explanation:
Gravitational potential energy is the energy that an object has due to its state or position in which it rests.
Gravitational potential energy = U = mass x gravity x height = m g h = 474 J
Height of the stool = h = 0.84 m
rearranging m g h and solving for the mass m gives m =
474 / (9.8)(0.84) = 57.6 kg = 58 kg (rounded to 2 significant digits).
A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A is the mass number of the nucleus. Estimate the den- sity of the nucleus of 127I (which has a nuclear mass of 2.1 × 10222 g) in grams per cubic centimeter. Compare with the density of solid iodine, 4.93 g cm23.
Answer:
Density of 127 I = [tex]\rm 1.79\times 10^{14}\ g/cm^3.[/tex]
Also, [tex]\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]
Explanation:
Given, the radius of a nucleus is given as
[tex]\rm r=kA^{1/3}[/tex].
where,
[tex]\rm k = 1.3\times 10^{-13} cm.[/tex]A is the mass number of the nucleus.The density of the nucleus is defined as the mass of the nucleus M per unit volume V.
[tex]\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.[/tex]
For the nucleus 127 I,
Mass, M = [tex]\rm 2.1\times 10^{-22}\ g.[/tex]
Mass number, A = 127.
Therefore, the density of the 127 I nucleus is given by
[tex]\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.[/tex]
On comparing with the density of the solid iodine,
[tex]\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]
A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed vertically upward. After traveling in this field for 3.90×10^−7 s , the proton’s velocity is directed 45° above the +x-axis. What is the strength of the electric field?
Answer:
The strength of the electric field is [tex]1.35\times10^{4}\ N/C[/tex].
Explanation:
Given that,
Speed [tex]v= 5.05\times10^{5}\ m/s[/tex]
Time [tex]t= 3.90\times10^{-7}\ s[/tex]
Angle = 45°
We need to calculate the acceleration
Using equation of motion
[tex]v = u+at[/tex]
[tex]5.05\times10^{5}=0+a\times3.90\times10^{-7}[/tex]
[tex]a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}[/tex]
[tex]a=1.29\times10^{12}\ m/s^2[/tex]
We need to calculate the strength of the electric field
Using relation of newton's second law and electric force
[tex]F= ma=qE[/tex]
[tex]ma = qE[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
Put the value into the formula
[tex]E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}[/tex]
[tex]E=1.35\times10^{4}\ N/C[/tex]
Hence, The strength of the electric field is [tex]1.35\times10^{4}\ N/C[/tex].
A man loads 260.5 kg of black dirt into his pickup. 60,452 g blows out on the ride home. How much black dirt does the man have when he reaches home? (with the correct number of significant figures)
Answer:
200.048 kg
Explanation:
Total Black dirt loaded in the pickup (A) = 260.5 kg = 260.500 kg
Quantity of the dirt that blew away (B )= 60452 g = 60.452 kg
Remaining quantity of the black dirt is
[tex]= 260.500 - 60.452[/tex]
[tex]= 200.048[/tex]
Thus, the amount of black dirt the man has when he reaches home = 200.048 kg.
A 2200 kg truck is coming down a hill 50 m high towards a stop sign. What force will the brakes need to provide (in N) in order to stop the truck in the 300 m before the sign? (This is a conservation of energy problem). Use g = 10 m/s^2.
Answer:
The force required by the brakes is [tex]3.67\times 10^{3} N[/tex]
Solution:
As per the question:
Mass of the truck, M = 2200 kg
Height, h = 50 m
distance moved by the truck before stopping, x = 300 m
[tex]g = 10 m/s^{2}[/tex]
Force required by the brakes to stop the truck, [tex]F_{b}[/tex] can be calculated by using the law of conservation of energy.
Now,
Kinetic Energy(K.E) downhil, K.E = reduction in potential energy, [tex]\Delta PE[/tex]
[tex]\Delta PE = Mgh = 2200\times 10\times 50 = 1100 kJ[/tex]
Work done is provided by the decrease in K.E,i.e., change in potential energy.
W = [tex]F\times x = 300 F = 1100\times 10^{3}[/tex]
F = [tex]3.67\times 10^{3} N[/tex]
In a movie, a monster climbs to the top of a building 30 m above the ground and hurls a boulder downward with a speed of 25 m/s at an angle of 45° below the horizontal. How far from the base of the building does the boulder land?
Final answer:
The boulder will land approximately 26.34 meters from the base of the building.
Explanation:
To determine the horizontal displacement of the boulder, we need to analyze the horizontal and vertical components of its motion separately.
First, we can determine the time it takes for the boulder to hit the ground using the vertical component of its motion. The initial vertical velocity can be found by multiplying the initial velocity (25 m/s) by the sin of the launch angle (45°). We can then use the equation h = v0yt + 0.5gt2 to solve for the time, where h is the vertical displacement (30 m - 0 m = 30 m), v0y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s2), and t is the time. Solving for t, we get t ≈ 1.87 s.
Next, we can determine the horizontal displacement by multiplying the horizontal component of the initial velocity (25 m/s cos 45°) by the time of flight (1.87 s). Multiplying the values, we get approximately 26.34 m. Therefore, the boulder lands approximately 26.34 m from the base of the building.
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding the speed limit?
Answer:
yes driver exceed the car speed
Explanation:
given data
speed of car = 38 m/s
speed limit = 75.0 mi/h
to find out
Is the driver exceeding the speed limit
solution
we know car speed is 38 m/s and limit is 75 mi/hr
so for compare the speed limit we convert limit and make them same
as we know
1 m/s = 2.236 mi/hr
so
car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr
as this car speed is exceed the speed limit that is 75 mi/hr
yes driver exceed the car speed
Two parallel plates have equal but opposite charges on their surface. The plates are separated by a finite distance. A fast moving proton enters the space between the two plates through a tiny hole in the left plate A. The electric potential energy of the proton increases as it moves toward plate B. (a) How is the speed of the proton affected as it moves from plate A to plate B
Answer:
Explanation:
The plates A and B are charged by opposite charges but which plate is positively charged and which is negatively charged is not clear.
Now a proton which is positively charged is moving from plate A to B . If it is attracted by plate B then its kinetic energy will be increased and potential energy will be decreased due to conservation of energy . In that case B will be negatively charged .
But in the given case it is stated that
potential energy of the proton increases . That means its kinetic energy decreases . In other words its speed decreases . It points to the fact that plate B is also positively charged.
So proton will be repelled and its speed will be decreased.
You are standing on a log and a friend is trying to knock
youoff.
He throws the ball at you. You can catch it, or you can let
itbounce off of you.
Which is more likely to topple you, catching the ball or letting
itbounce off?
Briefly explain what physics you used to reach
yourconclusion.
Answer: catching the ball is a better choice.
Explanation:
The collision of 2 objects involves involves large impact force since the force is inversely proportional to the time in which the momentum of the object changes.
Mathematically
[tex]F=\frac{\Delta p}{\Delta t}[/tex]
If we catch the ball we increase the time in which the momentum of the ball is decreased thus the impact force that acts on us is lower as larger time is allowed for the ball to decrease it's momentum.
If we allow the ball to hit us the momentum of the ball changes in a short period of time thus applying a large impact force on our body thus increasing the chances of toppling.
Letting the ball bounce off of you is more likely to topple you.
What is Collision?This happens when two bodies hit each other with force. In this scenario, the ball has the force and we know that force is inversely proportional to the time in which the momentum of the object changes.
Letting the ball bounce off of you will decrease the time in which the momentum of the ball is increased thus the force that acts on us is higher which ius why it will most likely topple us.
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An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.99 x 10^-7 C/m^2, and the plate separation is 1.69 x 10^-2 m. How fast is the electron moving just before it reaches the positive plate?
Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area[tex] = \sigma = 1.99*10^{-7} C/m^2[/tex]
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is[tex] E = \frac{\sigma}{\epsilon}[/tex]
force acting on charge is F = q E
Work done by charge q id[tex] \Delta X =\frac{ q\sigma \Delta x}{\epsilon}[/tex]
this work done is converted into kinectic enerrgy
[tex]\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}[/tex]
solving for v
[tex]v = \sqrt{\frac{2q\Delta x}{\epsilon m}[/tex]
[tex]\epsilon = 8.85*10^{-12} Nm2/C2[/tex]
[tex]v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}[/tex]
v = 1.15*10^{7} m/s
Raindrops fall 1.59 x 10^3 m from a cloud to the ground. If they were not slowed by air resistance, how fast would the drops be moving when they struck the ground?
Answer:
18 seconds
Explanation:
s = Displacement = 1.59×10³ m
a = Acceleration due to gravity = 9.81 m/s²
u = Initial velocity = 0
v = Final velocity
t = Time taken
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1590=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1590\times 2}{9.81}}\\\Rightarrow t=18\ s[/tex]
Time taken by the raindrop to reach the ground neglecting air resistance is 18 seconds
A car is designed to get its energy from a rotating
flywheelwith a radius of 1.50 m and a mass of475 kg. Before a trip,
the flywheel isattached to an electric motor, which brings the
flywheel'srotational speed up to 4000 rev/min.
(a) Find the kinetic energy stored in
theflywheel.
b) If the flywheel is to supply energy to the car as would a15.0-hp
motor, find the length of timethe car could run before the flywheel
would have to be brought backup to speed.
Answer:
(a). The kinetic energy stored in the fly wheel is 46.88 MJ.
(b). The time is 1.163 hours.
Explanation:
Given that,
Radius = 1.50 m
Mass = 475 kg
Power [tex]P= 15.0 hp = 15.0\times746=11190 watt[/tex]
Rotational speed = 4000 rev/min
We need to calculate the moment of inertia
Using formula of moment of inertia
[tex]I=\dfrac{1}{2}mr^2[/tex]
Put the value into the formula
[tex]I=\dfrac{1}{2}\times475\times(1.50)^2[/tex]
[tex]I=534.375\ kg m^2[/tex]
(a). We need to calculate the kinetic energy stored in the fly wheel
Using formula of K.E
[tex]K.E=\dfrac{1}{2}I\omega^2[/tex]
Put the value into the formula
[tex]K.E = \dfrac{1}{2}\times534.375\times(4000\times\dfrac{2\pi}{60})^2[/tex]
[tex]K.E=46880620.9\ J[/tex]
[tex]K.E =46.88\times10^{6}\ J[/tex]
[tex]K.E =46.88\ MJ[/tex]
(b). We need to calculate the length of time the car could run before the flywheel would have to be brought backup to speed
Using formula of time
[tex]t=\dfrac{46.88\times10^{6}}{11190}[/tex]
[tex]t=4189.45\ sec[/tex]
[tex]t=1.163\ hours [/tex]
Hence, (a). The kinetic energy stored in the fly wheel is 46.88 MJ.
(b). The time is 1.163 hours.
The kinetic energy stored in the flywheel is calculated using its moment of inertia and angular velocity. The time the car could run driven by the flywheel's energy can be determined by dividing the total energy by the rate at which the energy is supplied.
Explanation:The kinetic energy stored in the flywheel (part a) can be obtained using the formula for rotational kinetic energy: K.E. = 0.5 * I * ω^2, where I is the moment of inertia of the flywheel and ω is its angular velocity. The moment of inertia for a solid disc (like our flywheel) is given by I = 0.5 * m * r^2, and ω can be found by converting the given rate of 4000 rev/min to rad/s. Inputting the given values, we can calculate the kinetic energy of the flywheel.
For part b, we first convert the horsepower of the motor to Watts (1 hp = 746 Watts). This gives us the rate at which the flywheel is supplying energy. We then divide the total energy obtained in part a by this rate to find the time the car could run.
Learn more about Kinetic Energy here:https://brainly.com/question/26472013
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Calculate the minimum wavelength (λ0) for the continuous spectrum of X-rays emitted when.. a) 30-kV electrons strike a cobalt (Co) target
b) 60-kV electrons strike a copper (Cu) target
Explanation:
The relationship between the wavelength and the potential difference V is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2me}}\times \dfrac{1}{\sqrt{V}}[/tex]
Putting the values of known parameters,
[tex]\lambda=\dfrac{12.28}{\sqrt{V} }\times 10^{-10}\ m[/tex]
(a) [tex]V=30\ kV=30000\ V[/tex]
[tex]\lambda=\dfrac{12.28}{\sqrt{30000} }\times 10^{-10}\ m[/tex]
[tex]\lambda=7.08\times 10^{-12}\ m[/tex]
(b) [tex]V=60\ kV=60000\ V[/tex]
[tex]\lambda=\dfrac{12.28}{\sqrt{60000} }\times 10^{-10}\ m[/tex]
[tex]\lambda=5.01\times 10^{-12}\ m[/tex]
Hence, this is the required solution.
Three point charges are on the x axis: q1 = -6.0 µC is at x = -3.0 m, q2 = 1.0 µC is at the origin, and q3 = -1.0 µC is at x = 3.0 m. Find the electric force on q1.
Answer:
The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].
Explanation:
Given that,
Charge on [tex]q_{1}=-6.0\ \mu C[/tex]
Distance [tex]x= -3.0\ m[/tex]
Charge on [tex]q_{2}=1.0\ \mu C[/tex] at origin
Distance [tex]x= 3.0\ m[/tex]
Charge on [tex]q_{3}=-1.0\ \mu C[/tex]
We need to calculate the electric force on q₁
Using formula of electric force
[tex]F_{12}=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F_{12}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times1.0\times10^{-6}}{(3.0)^2}[/tex]
[tex]F_{12}=-0.006\ N[/tex]
Negative sign shows the attraction force.
We need to calculate the electric force F₁₃
[tex]F_{13}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times(-1.0\times10^{-6})}{(6.0)^2}[/tex]
[tex]F_{13}=0.0015\ N[/tex]
Positive sign shows the repulsive force.
We need to calculate the net electric force
[tex]F=F_{12}+F_{13}[/tex]
[tex]F=-0.006+0.0015[/tex]
[tex]F=0.0045\ N[/tex]
[tex]F=4.5\times10^{-3}\ N[/tex]
Hence, The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].
The correct answer is that the electric force on q1 is 2.55 x 10^5 N, directed towards the origin.
To find the electric force on q1, we need to calculate the net electric force due to q2 and q3. According to Coulomb's law, the force between two point charges is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
[tex]where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \) N m^2/C^2), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.[/tex]
First, we calculate the force between q1 and q2:
[tex]\[ F_{12} = k \frac{|q_1 q_2|}{r_{12}^2} \][/tex]
[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})|}{(-3.0 \text{ m})^2} \][/tex]
[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})}{(9.0 \text{ m}^2)} \][/tex]
[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{6.0 \times 10^{-12} \text{ C}^2}{9.0 \text{ m}^2} \][/tex]
[tex]\[ F_{12} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) (6.67 \times 10^{-13} \text{ C}^2/\text{m}^2) \][/tex]
[tex]\[ F_{12} = 5.99 \times 10^{-3} \text{ N} \][/tex]
Since q1 and q2 have opposite signs, the force [tex]\( F_{12} \)[/tex] is attractive, meaning it is directed towards q2, which is towards the origin.
Next, we calculate the force between q1 and q3:
[tex]\[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} \][/tex]
[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(-6.0 \times 10^{-6} \text{ C})(-1.0 \times 10^{-6} \text{ C})|}{(-3.0 \text{ m} - 3.0 \text{ m})^2} \][/tex]
[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.0 \times 10^{-6} \text{ C})(1.0 \times 10^{-6} \text{ C})}{(6.0 \text{ m})^2} \][/tex]
[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{6.0 \times 10^{-12} \text{ C}^2}{36.0 \text{ m}^2} \][/tex]
[tex]\[ F_{13} = (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) (1.67 \times 10^{-12} \text{ C}^2/\text{m}^2) \][/tex]
[tex]\[ F_{13} = 1.50 \times 10^{-2} \text{ N} \][/tex]
Since q1 and q3 have the same sign, the force [tex]\( F_{13} \)[/tex] is repulsive, meaning it is directed away from q3, which is also towards the origin in this case.
Now, we add the forces vectorially. Since both forces are directed towards the origin, we can simply add their magnitudes:
[tex]\[ F_{net} = F_{12} + F_{13} \][/tex]
[tex]\[ F_{net} = 5.99 \times 10^{-3} \text{ N} + 1.50 \times 10^{-2} \text{ N} \][/tex]
[tex]\[ F_{net} = 2.099 \times 10^{-2} \text{ N} \][/tex]
Converting this to a more standard scientific notation:
[tex]\[ F_{net} = 2.55 \times 10^5 \text{ N} \][/tex]
Therefore, the electric force on q1 is [tex]\( 2.55 \times 10^5 \)[/tex] N, directed towards the origin.
Two fixed charges, +1.0 x 10^-6 C and -3.0 x 10^-6 C, are 10 cm apart. (a) Where may a third charge be located so that the net force acting on this charge is zero? (b) Is the equilibrium stable or not?
Answer:
Part a)
the third charge will be placed at 13.66 cm on the other side of [tex]1.0 \times 10^{-6} C[/tex] charge
Part b)
If the charge is displaced by small distance towards left or right the force on it will move it away from its initial position
so this is Not stable equilibrium
Explanation:
Two charges are placed here on straight line are at 10 cm apart
here the two charges given are of opposite sign and hence the force on the third charge placed here on the same line will be zero where electric field is zero
Here electric field will be zero at the position near to the charge which is of small magnitude
so we will have
[tex]\frac{kq_1}{r^2} = \frac{kq_2}{(10 + r)^2}[/tex]
now we have
[tex]\frac{(1 \times 10^{-6})}{r^2} = \frac{(3\times 10^{-6})}{(10+ r)^2}[/tex]
[tex]10 + r = \sqrt3 r[/tex]
[tex]r = \frac{10}{\sqrt3 - 1} = 13.66 cm[/tex]
so the third charge will be placed at 13.66 cm on the other side of [tex]1.0 \times 10^{-6} C[/tex] charge
Part b)
If the charge is displaced by small distance towards left or right the force on it will move it away from its initial position
so this is Not stable equilibrium
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.0 m/s2 until he catches his friend. A.) How much time does it take until he catches his friend (after his friend passes him)? t=___s B.) How far has he traveled in this time? x= ____ m
C.) What is his speed when he catches up? v=____ m/s
Answer:
A) t = 7.0 s
B) x = 25 m
C) v = 10 m/s
Explanation:
The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
A)When both friends meet, their position is the same:
x bicyclist = x friend
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:
Position of the friend after 2 s:
x = v · t
x = 3.6 m/s · 2 s = 7.2 m
Then:
1/2 · a · t² = x0 + v · t v0 of the bicyclist is 0 because he starts from rest.
1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t
1 m/s² · t² - 3.6 m/s · t - 7.2 m = 0
Solving the quadratic equation:
t = 5.0 s
It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.
B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.
x = v · t
x = 3.6 m/s · 7.0 s = 25 m
(we would have obtained the same result if we would have used the equation for the position of the bicyclist)
C) Using the equation of velocity:
v = a · t
v = 2.0 m/s² · 5.0 s = 10 m/s
If a marathon runner averages 8.6 mi/h, how long does it take him or her to run a 26.22-mi marathon?
Answer:
Time, t = 3.04 hours
Explanation:
Given that,
Average speed of the marathon, v = 8.6 mi/h
Distance covered by the marathon runner, d = 26.22 mi
We need to find the time taken by the marathon runner to run that distance. Time taken is given by :
[tex]t=\dfrac{d}{v}[/tex]
[tex]t=\dfrac{26.22\ mi}{8.6\ mi/h}[/tex]
t = 3.04 hours
So, the time taken by the marathon runner is 3.04 hours. Hence, this is the required solution.
In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80 m/s^2 . Ignore air resistance. A flea jumps straight up to a maximum height of 0.380 m. What is its initial velocity v0 as it leaves the ground?
How long is the flea in the air from the time it jumps to the time it hits the ground?
Express your answer in seconds to three significant figures.
Answer:
Explanation:
Given
maximum height=0.380 m
initial velocity=[tex]v_0[/tex]
[tex]H_{max}=\frac{v_0^2}{2g}[/tex]
[tex]0.380=\frac{v_0^2}{2\times 9.81}[/tex]
[tex]v_0=2.73 m/s[/tex]
The time of flight will be [tex]t=\frac{2v_0}{g}[/tex]
time to reach top +time to reach bottom will be same
[tex]t=\frac{2\times 2.73}{9.81}[/tex]
t=0.556 s
The initial velocity of the flea as it leaves the ground is approximately 1.95 m/s. The flea is in the air for approximately 0.40 seconds.
Explanation:To find the initial velocity of the flea as it leaves the ground, we can use the equation for vertical motion: v^2 = v0^2 + 2aΔy. In this equation, v is the final velocity (which is 0 at the highest point), v0 is the initial velocity, a is the acceleration due to gravity (-9.80 m/s^2), and Δy is the change in height (0.380 m). Rearranging the equation gives us: v0^2 = 2aΔy. Plugging in the values for a and Δy and solving for v0, we get v0 = sqrt(2aΔy).
For the second part of the question, we can use the equation for the time of flight: Δt = 2v0/a. Plugging in the values for v0 and a, we get: Δt = 2 * v0 / a. Calculating the value of Δt will give us the time the flea is in the air.
Using the given values for a and Δy in the equation for v0 and evaluating the expression gives us the answer to the first part of the question: v0 = sqrt(2 * 9.80 * 0.380). To find the time of flight, we can plug in the values for v0 and a into the equation for Δt: Δt = 2 * 1.95 / 9.80. Evaluating this expression gives us the answer to the second part of the question: Δt = 0.40 s. Therefore, the initial velocity of the flea as it leaves the ground is approximately 1.95 m/s and the flea is in the air for approximately 0.40 seconds.
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A particle m is thrown vertically upward with an initial velocity v. Assuming a resisting medium proportional to the velocity, where the proportionality factor is c, calculate the velocity with which the particle will strike the ground upon its return if there is a uniform gravitational field.
Answer:[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]
Explanation:
Given
v=initial velocity
resisting acceleration =cv
also gravity is opposing the upward motion
Therefore distance traveled during upward motion
[tex]v^2_f-v^2=2as[/tex]
Where a=cv+g
[tex]0-v^2=2(cv+g)s[/tex]
[tex]s=\frac{v^2}{2(cv+g)}[/tex]
Now let v_0 be the velocity at the ground
[tex]v^2_0-0=2(g-vc)s[/tex]
substituting s value
[tex]v^2_0=v^2\frac{g-vc}{(cv+g)}[/tex]
[tex]v_0=v\sqrt{\frac{g-vc}{g+vc}}[/tex]
The mass of a high speed train is 4.5×105 kg, and it is traveling forward at a velocity of 8.3×101 m/s. Given that momentum equals mass times velocity, determine the values of m and n when the momentum of the train (in kg⋅m/s) is written in scientific notation. Enter m and n, separated by commas.
Answer : The value of momentum is [tex]3.735\times 10^7kg.m/s[/tex] and m = 3.735 and n = 7
Explanation : Given,
Mass of high speed train = [tex]4.5\times 10^5kg[/tex]
Velocity of train = [tex]8.3\times 10^1m/s[/tex]
Formula used :
[tex]p=mv[/tex]
where,
p = momentum
m = mass
v = velocity
Now put all the given values in the above formula, we get:
[tex]p=(4.5\times 10^5kg)\times (8.3\times 10^1m/s)[/tex]
[tex]p=3.735\times 10^7kg.m/s[/tex]
Therefore, the value of momentum is [tex]3.735\times 10^7kg.m/s[/tex] and m = 3.735 and n = 7
Vector C has a magnitude of 21.0 m and points in the −y‑ direction. Vectors A and B both have positive y‑ components, and make angles of α=43.4° and β=27.7° with the positive and negative x- axis, respectively. If the vector sum A+B+C=0 , what are the magnitudes of A and B?
The magnitudes of vectors A and B can be determined by resolving each vector into its x and y components, setting up equations for their sum, and solving the equations simultaneously.
Explanation:To find the magnitudes of vectors A and B when their sum with vector C results in a zero vector (A + B + C = 0), we can break down each vector into its x and y components and then solve the component equations separately.
Since vector C points in the negative y-direction and has a magnitude of 21.0 m, its components are Cx = 0 m and Cy = -21.0 m.
Vector A has a positive y-component and makes an angle of α = 43.4° with the positive x-axis. Thus, the components of A are Ax = A cos(α) and Ay = A sin(α).
Vector B also has a positive y-component and makes an angle of β = 27.7° with the negative x-axis, which is the same as 180° - β with the positive x-axis. So, B's components are Bx = -B cos(180° - β) = -B cos(β) and By = B sin(180° - β) = B sin(β).
The equations for the x and y components of the vector sum being zero are:
Substituting the component expressions in, we get:
These are two equations with two unknowns (the magnitudes A and B), which can be solved simultaneously to find the values of A and B. The solution involves elementary algebra and the use of trigonometric identities.
A ball is thrown straight up and reaches a maximum height of 36 m above the point from which it was thrown. With what speed was the ball thrown?
Final answer:
To find the initial speed of a ball thrown to a maximum height of 36 m, we use kinematic equations that factor in the acceleration due to gravity. The ball's initial speed can be calculated using the formula for objects under constant acceleration, considering that the final velocity at the max height is 0 m/s.
Explanation:
Calculating the Launch Speed of the Ball
To determine the initial speed at which the ball was thrown to reach a maximum height of 36 m, we can use the principles of kinematics under the influence of gravity. In the absence of air resistance, a ball thrown upwards will decelerate at a rate equal to the acceleration due to gravity until it comes to a stop at its maximum height. We use the following kinematic equation for an object under constant acceleration:
s = ut + 1/2at^2
Where:
s is the displacement (maximum height in this case, which is 36 m)
u is the initial velocity (what we want to find)
a is the acceleration due to gravity (-9.81 m/s^2, the negative sign indicates acceleration is in the direction opposite to the initial velocity)
t is the time taken to reach the maximum height (not needed in this calculation)
At the maximum height, the final velocity (v) is 0 m/s, so we use the following equation:
v^2 = u^2 + 2as
Plugging in the known values:
0 = u^2 + 2(-9.81 m/s^2)(36 m)
u^2 = 2(9.81 m/s^2)(36 m)
u = √(2(9.81 m/s^2)(36 m))
The initial speed u can be calculated from this equation to find out with what speed the ball was thrown to achieve a 36 m height.
The equation for bouyancy force on a fully submerged object of volume V and mass M is given by: OF) = PwVg OF) = Pwg/V OF, = Mg
Answer:
[tex]OF=\rho _wVg[/tex]
Explanation:
The bouyancy force an object feels (OF) when submerged ina fluid is always the weight of the liquid the object displaces. This weight will be the mass of fluid displaced ([tex]m_d[/tex]) multiplied by the acceleration of gravity g. The mass displaced will be the density of the fluid [tex]\rho_w[/tex] multiplied by the volume of fluid displaced [tex]V_d[/tex]. If the object is fully submerged, then this volume will be the same as the volume of the object V. We write all these steps in equations:
[tex]OF=m_dg=\rho _wV_dg=\rho _wVg[/tex]
If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3000 m to an altitude of 420 m , where she will pull her ripcord?
Answer:
22.93 seconds
Explanation:
s = Displacement = 3000 - 420 = 2580 m = The distance she will free fall
a = Acceleration due to gravity = 9.81 m/s²
u = Initial velocity = 0
v = Final velocity
t = Time taken
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2580=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2580\times 2}{9.81}}\\\Rightarrow t=22.93\ s[/tex]
Time taken by the skydiver to cover the distance between 3000 m to an altitude of 420 m neglecting air resistance is 22.93 seconds
A series RLC circuit has a resistance of 44.0 Ω and an impedance of 71.0 Ω. What average power is delivered to this circuit when ΔVrms = 210 V?
Answer:
power = 384.92 W
Explanation:
given data
resistance = 44.0 Ω
impedance = 71 Ω
ΔVrms = 210 V
to find out
average power
solution
we know here power formula that is
power = [tex]\frac{V^2rms}{impedance}[/tex]cos∅ .........1
we know here cos∅ = [tex]\frac{resistance}{impedance}[/tex]
cos ∅ = [tex]\frac{44}{71}[/tex] = 0.6197
so from equation 1
power = [tex]\frac{210^2}{71}[/tex] × 0.6197
power = 384.92 W
The route followed by a hiker consists of three displacement vectors A with arrow, B with arrow, and C with arrow. Vector A with arrow is along a measured trail and is 1550 m in a direction 26.0° north of east. Vector B with arrow is not along a measured trail, but the hiker uses a compass and knows that the direction is 41.0° east of south. Similarly, the direction of vector vector C is 37.0° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or A with arrow + B with arrow + C with arrow = 0. Find the magnitudes of vector B with arrow and vector C with arrow.
Answer:
Magnitude of vector B = 6643 m
Magnitude of vector C = 7201 m
Explanation:
Knowing that the sum of the internal angles of a triangle is 180°, we can obtain the internal angles of the triangle formed by the three displacement vectors (see the attached figure, the calculated angles are in red and the given angles in black).
The angles were calculated as follows (see figure):
angle BC = 180°- 90° - 41° - 37 ° = 12°
angle AB = 180° - 90° - 26° +41° = 105°
angle AC = 180° - 105° - 12 = 63°
Once we obtain the internal angles, we can use the sine rule:
sin a/ A = sin b/ B = sin c/ C where "A" is the side opposite to the angle "a", "B" is the side opposite to the angle "b" and "C" is the side the opposite to the angle "c".
Then:
sin 12° / A = sin 63°/ B = sin (105°) / C
sin 12° / 1550 m = sin 63° / B
B = sin 63° * (1550 m / sin 12°) = 6643 m
sin 12° /1550 m = sin 105° /C
C = sin 105° * (1550 m / sin 12°) = 7201 m
In attempting to
pass thepuck to a teammate, a hockeyplayer
gives it an initialspeedof 1.7
m/s. However, this speed is inadequate tocompensate for
the kinetic frictionbetween the puck and theice. As
a result, the puck travels only half the
distancebetweenthe players before slidingto a
halt. What minimum initial speed should the puck
havebeengiven so that it reached
theteammate, assuming that the same force of
kineticfrictionacted on the puck
everywherebetween the two players?
Answer:
2.04 m/s
Explanation:
Given:
[tex]u[/tex] = initial inadequate speed of the puck = 1.7 m/s[tex]v[/tex] = final velocity of the puck while reaching half the distance of the targeted teammate = 0 m/sAssumptions:
[tex]m[/tex] = mass of the puck[tex]U[/tex] = minimum initial speed of the puck so that it reaches the target[tex]x[/tex] = distance of the targeted teammate [tex]f_k[/tex] = kinetic friction between the puck and the iceWork-energy theorem: For the various forces acting on an object, the work done by all the forces brings a change in kinetic energy of an object which is equal to the total work done.
For the initial case, the puck travels half the distance of the target teammate. In this case, the change in kinetic energy of the puck will be equal to the work done by the friction.
[tex]\therefore \dfrac{1}{2}m(v^2-u^2)=f_k\dfrac{x}{2}\\\Rightarrow \dfrac{1}{2}m(0^2-1.7^2)=f_k\dfrac{x}{2}[/tex]...........eqn(1)
Now, again using the work energy theorem for the puck to reach the targeted teammate, the change in kinetic energy of the puck will be equal to the work done by the kinetic friction.
[tex]\therefore \dfrac{1}{2}m(v^2-U^2)=f_k\dfrac{x}{2}\\\Rightarrow \dfrac{1}{2}m(0^2-U^2)=f_kx[/tex]...........eqn(2)
On dividing equation (1) by (2), we have
[tex]\dfrac{-1.7^2}{-U^2}=\dfrac{1}{2}\\\Rightarrow \dfrac{1.7^2}{U^2}=\dfrac{1}{2}\\\Rightarrow U^2= 2\times 1.7^2\\\Rightarrow U^2 = 5.78\\\Rightarrow U=\pm \sqrt{5.78}\\\Rightarrow U=\pm 2.04\\\textrm{Since the speed is always positive.}\\\therefore U = 2.04\ m/s[/tex]
Hence, the puck must be kicked with a minimum initial speed of 2.04 m/s so that it reaches the teammate.
A long wave travels in 20m depth of water. If the amplitude of this wave in 20m depth water is 1.1 m , what is the amplitude of the same wave in 6.2 m depth? Assume the wave is traveling perpendicular to the coast.
Answer:
A = 3.55 m
Explanation:
given,
depth of water = 20 m
Amplitude of wave at 20 m depth = 1.1 m
Amplitude at the depth of 6.2 m
amplitude is inversely proportional to depth of water
so,
[tex]\dfrac{A_1}{A_2} = \dfrac{d_2}{d_1}[/tex]
[tex]A_2= A_1\dfrac{d_1}{d_2}[/tex]
[tex]A_2= 1.1\dfrac{20}{6.2}[/tex]
A = 3.55 m
hence, the amplitude of the wave at the depth of 6.2 m is 3.55 m.