The charge on each ball is approximately [tex]\( 8.08 \times 10^{-9} \) C[/tex].
Step 1
To determine the charge on each ball, let's analyze the forces acting on one of the charged balls when they are in equilibrium. The forces are:
1. Gravitational force [tex](\( F_g \))[/tex] downward.
2. Tension [tex](\( T \))[/tex] in the thread.
3. Electrostatic force [tex](\( F_e \))[/tex] between the two charges.
Step 2
Given:
Length of the thread [tex](\( L \)) = 0.26 m[/tex]
Mass of each ball [tex](\( m \)) = \( 7.75 \times 10^{-4} \) kg[/tex]
Angle between the threads = 35°, so each thread makes an angle of 17.5° with the vertical.
Step 3
First, calculate the gravitational force:
[tex]\[ F_g = mg = 7.75 \times 10^{-4} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.595 \times 10^{-3} \, \text{N} \][/tex]
The electrostatic force must balance the horizontal component of the tension:
[tex]\[ F_e = T \sin(17.5^\circ) \][/tex]
The vertical component of the tension balances the gravitational force:
[tex]\[ T \cos(17.5^\circ) = F_g \][/tex]
[tex]Solving for \( T \):[/tex]
[tex]\[ T = \frac{F_g}{\cos(17.5^\circ)} = \frac{7.595 \times 10^{-3} \, \text{N}}{\cos(17.5^\circ)} \approx 7.99 \times 10^{-3} \, \text{N} \][/tex]
Step 4
Now calculate the horizontal component of the tension:
[tex]\[ T \sin(17.5^\circ) = 7.99 \times 10^{-3} \, \text{N} \times \sin(17.5^\circ) \approx 2.41 \times 10^{-3} \, \text{N} \][/tex]
Step 5
This is the electrostatic force:
[tex]\[ F_e = \frac{k q^2}{r^2} \][/tex]
where [tex]\( r \)[/tex] is the distance between the balls.
Step 6
To find [tex]\( r \)[/tex]:
[tex]\[ r = 2L \sin(17.5^\circ) = 2 \times 0.26 \, \text{m} \times \sin(17.5^\circ) \approx 0.156 \, \text{m} \][/tex]
Using Coulomb's law:
[tex]\[ 2.41 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) q^2}{(0.156 \, \text{m})^2} \][/tex]
Step 7
Solving for [tex]\( q \)[/tex]:
[tex]\[ q^2 = \frac{2.41 \times 10^{-3} \, \text{N} \times (0.156 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \approx 6.53 \times 10^{-17} \, \text{C}^2 \][/tex]
[tex]\[ q \approx \sqrt{6.53 \times 10^{-17}} \approx 8.08 \times 10^{-9} \, \text{C} \][/tex]
Discuss whether the kinetic energy or the momentum of the ""bullet"" is more important in causing damage to the bulletproof vest.
Answer:
Answered
Explanation:
The Kinetic energy of a bullet does not remain constant. It dissipates as the bullet travel through the air. It is partly converted into heat, sound, air resistance etc.
Whereas the momentum of a travelling body remains conserved and hence constant. therefore a bullet with more momentum will cause more damage. that is momentum is more important than kinetic energy.
One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is μk = 0.400. You apply a constant force F⃗ to the block. F⃗ has magnitude 88.0 N and is directed 3 toward the wall. The spring is compressed 80.0 cm. (a) What is the speed of the block? (b) What is the magnitude of the block’s acceleration? (c) What is the direction of the block’s acceleration?
Final answer:
The speed of the block is 4.08 m/s, the magnitude of the block’s acceleration is 25.41 m/s^2, and the direction of the block’s acceleration is toward the wall.
Explanation:
(a) To find the speed of the block, we can use the principle of conservation of mechanical energy. The potential energy stored in the spring when it is compressed is converted into the kinetic energy of the block when it is released. The potential energy stored in the spring is given by:
PE = 0.5 * k * x^2
where k is the force constant of the spring and x is the compression of the spring. Plugging in the values, we get:
PE = 0.5 * 130.0 N/m * 0.80 m * 0.80 m = 41.60 J
The kinetic energy of the block when it is released is given by:
KE = 0.5 * m * v^2
where m is the mass of the block and v is its speed. Equating the potential and kinetic energies, we have:
PE = KE
41.60 J = 0.5 * 3.00 kg * v^2
Solving for v, we get:
v = √(41.60 J / (0.5 * 3.00 kg)) = 4.08 m/s
(b) The magnitude of the block's acceleration can be calculated using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force applied to the block minus the force of friction. The force applied to the block is given by F = 88.0 N. The force of friction can be calculated using the equation:
f = μk * m * g
where μk is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity. Plugging in the values, we get:
f = 0.400 * 3.00 kg * 9.8 m/s^2 = 11.76 N
The net force is therefore:
net force = F - f = 88.0 N - 11.76 N = 76.24 N
Using Newton's second law, we have:
76.24 N = 3.00 kg * a
Solving for a, we get:
a = 76.24 N / 3.00 kg = 25.41 m/s^2
(c) The direction of the block's acceleration can be determined by considering the net force acting on the block. In this case, the applied force and the force of friction are in opposite directions, resulting in a net force in the direction of the applied force. Therefore, the direction of the block's acceleration is toward the wall.
When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of acquiring this charge? Express your answer as a number of electrons.
The number of electrons lost by the by the honeybee in acquiring the charge of +20 pC is;
n = 1.25 × 10^(8) electrons
We are given;
Charge of honeybee; Q = 20 pC = 20 × 10^(-12) C
Now, formula for number of electrons is;
n = Q/e
Where;
e is charge on electron = 1.6 × 10^(-19) C
Thus;
n = (20 × 10^(-12))/(1.6 × 10^(-19))
n = 1.25 × 10^(8) electrons
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You toss a ball straight up at 6.8 m/s ; it leaves your hand at 2.0 m above the floor. Suppose you had tossed a second ball straight down at 6.8 m/s (from the same place 2.0 m above the floor). When would the second ball hit the floor?
Answer:0.249 s
Explanation:
Given
Ball is tossed down with a velocity of 6.8 m/s downward
height from ground=2 m
therefore time to reach ground is
[tex]s=ut+\frac{gt^2}{2}[/tex]
[tex]2=6.8\times t+\frac{9.81\times t^2}{2}[/tex]
[tex]9.81t^2+13.6t-4=0[/tex]
[tex]t=\frac{-13.6\pm \sqrt{13.6^2+4\times 4\times 9.81}}{2\times 9.81}[/tex]
[tex]t=\frac{-13.6+18.49}{19.62}=0.249 s[/tex]
Find the frequency (in hertz) of the first overtone for a 1.75 m pipe that is closed at only one end. Use 350 m/s for the speed of sound in air. Also, would increasing the air temperature raise, lower, or have no effect upon this frequency?
Answer:
The first overtone frequency is 100 Hz.
Solution:
According to the question:
Length of pipe, l = 1.75 m
Speed of sound in air, v_{sa} = 350 m/s
Frequency of first overtone, [tex]f_{1}[/tex] is given by:
[tex]f_{1} = \frac{v_{sa}}{2l}[/tex]
[tex]f_{1} = \frac{350}{2\times 1.75} = 100 Hz[/tex]
Since, the frequency, as clear from the formula depends only on the speed
and the length. It is independent of the air temperature.
Thus there will be no effect of air temperature on the frequency.
Represent 0.783 kg with Sl units having an appropriate prefix Express your answer to three significant figures and include the appropriate units. P: A Value RO ? Units 0.783 kg
Answer:
783 grams
Explanation:
Here a time of 0.783 kg is given.
Some of the prefixes of the SI units are
1 gram = 10⁻³ kilogram
1 milligram = 10⁻⁶ kilogram
1 microgram = 10⁻⁹ kilogram
The number is 0.783
Here, the only solution where the number of significant figures is three is gram
[tex]1\ kilogram = 1000\ gram[/tex]
[tex]\\\Rightarrow 0.783\ kilogram=0.783\times 1000\ gram\\ =783\ gram[/tex]
So, 0.783 kg = 783 grams
The electric field due to two point charges is found by: a) finding the stronger field. The net field will just be equal to the strongest field.
b) determining the electric field due to each charge and adding them together as vectors.
c) adding the magnitude of the two fields together.
d) adding the direction of the two fields together.
e) None of the above.
Answer:
b)determining the electric field due to each charge and adding them together as vectors.
Explanation:
The electric Field is a vector quantity, in other words it has a magnitude and a direction. On the other hand, the electric field follows the law of superposition. The electric field produced by two elements is equal to the sum of the electric fields produced by each element when the other element is not present. in other words, the total electric field is solved determining the electric field due to each charge and adding them together as vectors.
The electric field due to two-point charges is found by determining the electric field of each charge and adding them together as vectors, considering both their magnitudes and directions.
Explanation:The correct way to find the electric field due to two-point charges is option (b), determining the electric field due to each charge and adding them together as vectors. The electric field is a vector quantity, meaning it has both magnitude and direction. Thus, when two or more electric fields exist in the same region, you add them as vectors, taking into account both their magnitudes and directions. For example, if you have two positive charges, the fields they generate will be in different directions. Therefore, you would add these fields together as vectors, resulting in the net electric field in that region.
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Find the work done "by" the electric field on a positively charged point particle with a charge of 2.1x10^-6 C as it is moved from a potential of 15.0 V to one of 8.0 V. (Include the sign of the value in your answer.)
Answer
Work done will be [tex]14.7\times 10^{-6}J[/tex] and it will be positive
Explanation:
We have given charge [tex]2.1\times 10^{-6}C[/tex]
We have to find work done in moving the charge from 15 volt to 8 volt
Let [tex]V_1=15V\ and\ V_2=8volt[/tex]
So potential difference [tex]V=V_1-V_2=15-8=7volt[/tex]
We know that work done [tex]W=QV[/tex], here Q is charge and V is potential difference
So work done [tex]W=QV=2.1\times 10^{-6}\times 7=14.7\times 10^{-6}J[/tex]
It will be positive work done because work is done in moving charge from higher potential to lower potential
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order visible spectrum produced by a plane grating that has 900 grooves/mm with the light incident normally on the grating.
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 × [tex]10^{-9}[/tex] m
wavelength λ = 700 nm = 700 × [tex]10^{-9}[/tex] m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = [tex]\frac{1}{900*10^3 m}[/tex]
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
[tex]\frac{1}{900*10^3}[/tex] sinθ = 1 × 380 × [tex]10^{-9}[/tex]
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
[tex]\frac{1}{900*10^3}[/tex] sinθ = 1 × 700 × [tex]10^{-9}[/tex]
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
How many protons are required to generate a charge of +25.0 C?
Answer:
[tex]N=1.56*10^{20}[/tex] protons
Explanation:
The total charge Q+=+25.0C is the sum of the charge of the N protons, with charge q :
[tex]Q_{+}=N*q_{p}[/tex]
q_{p}=1.6*10^{-19}C charge of a proton
We solve to find N:
[tex]N=Q_{+}/q_{p}=25/(1.6*10^{-19})=1.56*10^{20}[/tex]
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.70 x 10^5 m/s^2 for 9.60 x 10^−4 s. What is its muzzle velocity (in m/s) (that is, its final velocity)? (Enter the magnitude.)
Final answer:
To calculate the muzzle velocity, we use the equation for constant acceleration (v = u + at) with an initial velocity (u) of 0, the given acceleration (a) of 5.70 x 10^5 m/s^2, and the time (t) of 9.60 x 10^-4 s to find the final velocity (v), which is approximately 547 m/s.
Explanation:
The question concerns calculating the muzzle velocity of a bullet as it's accelerated from the firing chamber to the end of the barrel of a gun.
Using the formula v = u + at, where v is the final velocity, u is the initial velocity (which is zero before the gun is fired), a is the acceleration, and t is the time for which the acceleration is applied, we can find the muzzle velocity. The bullet experiences an acceleration (a) of 5.70 x 105 m/s2 for a time (t) of 9.60 x 10−4 seconds.
Plugging the values into the formula, we get:
v = 0 + (5.70 x 105 m/s2)(9.60 x 10−4 s)
v = 5.47 x 102 m/s
Therefore, the final muzzle velocity of the bullet as it leaves the barrel is approximately 547 m/s.
When children draw human figures by sketching a circle with dangly lines for legs the children are demonstrating which developmental stage in art?
A. Drawing
B. Schematic
C. pre schematic
D. circular
Answer:
C
Explanation:
the correct answer is c) Pre Schematic.
It is in the Pre-schematic stage of development that the children learn to draw human figures by sketching a circle with dangly lines for legs. It remains till the age of six. After pre schematic stage comes the schematic stage, where in children can draw complex structures.
Children drawing human figures with a circle and lines for legs are in the Pre-schematic stage of art development, indicating an early attempt at representational drawing before achieving greater accuracy and detail in later stages.
Explanation:When children draw human figures by sketching a circle with dangly lines for legs, they are demonstrating the Pre-schematic developmental stage in art. This stage is distinct from the earlier scribbling phase, signaling that children have begun to recognize and attempt to represent the human form in a more organized albeit simplified manner. However, this representation lacks the complexity and accuracy found in later stages, such as the Schematic stage, where more detailed and proportionate figures are drawn.
The Pre-schematic stage typically occurs in children aged between 4 and 7 years old. During this time, their cognitive and motor skills are developing rapidly, but they are still learning how to translate what they see and imagine onto paper with greater fidelity. Thus, the circle and lines approach is a developmentally appropriate method for them to begin exploring human figures in their art.
With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as high as 4 m/s. If we image it with 7 MHz sound and the speed of sound is approximately 1500 m/s in human tissue, what is the frequency shift between the opening and closing of the valve?
Answer:
|Δf| = 37.3 kHz
Explanation:
given,
peak velocity = 4 m/s
speed of the sound = 1500 m/s
frequency = 7 MHz
[tex]v = C\dfrac{\pm \dlta f}{2 f_0}[/tex]
[tex]\delta f = \pm 2 f_0 (\dfrac{V}{C})[/tex]
[tex]\delta f = \pm 2\times 7 (\dfrac{4}{1500})[/tex]
[tex]=\pm 0.0373 MHz[/tex]
= 37.3 kHz
|Δf| = 37.3 kHz
hence, frequency shift between the opening and closing valve is 37.3 kHz
A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the road is .521. What is the maximum speed in m/s at which the car can take the curve?
Answer:
The maximum speed of car will be 20.5m/sec
Explanation:
We have given mass of car = 1100 kg
Radius of curve = 82.3 m
Static friction [tex]\mu _s=0.521[/tex]
We have to find the maximum speed of car
We know that at maximum speed centripetal force will be equal to frictional force [tex]m\frac{v^2}{r}=\mu _srg[/tex]
[tex]v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec[/tex]
So the maximum speed of car will be 20.5m/sec
Answer:20.51 m/s
Explanation:
Given
Mass of car(m)=1100 kg
radius of curve =82.3 m
coefficient of static friction([tex]\mu [/tex])=0.521
here centripetal force is provided by Friction Force
[tex]F_c(centripetal\ force)=\frac{mv^2}{r}[/tex]
Friction Force[tex]=\mu N[/tex]
where N=Normal reaction
[tex]\frac{mv^2}{r}=\mu N[/tex]
[tex]\frac{1100\times v^2}{82.3}=0.521\times 1100\times 9.81[/tex]
[tex]v^2=0.521\times 9.81\times 82.3[/tex]
[tex]v=\sqrt{420.63}=20.51 m/s [/tex]
What is an electric field? (choose all that are true) a) A vector field created by a charged particle.
b) The field created when one charged particle is brought close to another charged particle.
c) A scalar field created by a charged particle.
d) A vector field created only by positive charges.
e) None of the above.
Answer:
option A
Explanation:
the correct answer is option A
An electric field is a vector field created by charged particle.
Electric field is mathematically defined as the ratio of the vector component of the force to the charge in coulomb.
Si unit of Electric field is N/C
Electric field is created by the electric charge or by the varying magnetic field.
Electric field is responsible for the attractive force between the electron and atomic nucleus.
Xylene (a common solvent in the petroleum industry) boils at 281.3°F at one atmosphere pressure. At what temperature does Xylene boil in °C, R and K?
Answer:
Xylene boils at 138.5 °C, 740.97 R and 411.65 K
Explanation:
To convert the temperature in Fahrenheit to Celsius you need to use this formula[tex]T_{\°C}=(T_{\°F}-32)\cdot \frac{5}{9}[/tex]
We know that temperature is 281.3 °F so in °C is:
[tex]\°C=(281.3-32)\cdot \frac{5}{9}= 138.5 \°C[/tex]
To convert Fahrenheit to Rankine you need to use this formula[tex]T_{R}=T_{\°F}+459.67\\T_{R}=281.3\°F+459.67=740.97 R[/tex]
To convert Fahrenheit to Kelvin you need to use this formula[tex]T_{K}=(T_{\°F}+459.67)\cdot \frac{5}{9} \\T_{K}=(281.3 \°F+459.67)\cdot \frac{5}{9} \\T_{K}=411.65K[/tex]
The output of an ac generator connected to an RLC series combination has a frequency of 12 kHz and an amplitude of 28 V. If R = 4.0 Ohms, L = 30 μH, and C = 800 nF, find a. The impedance b. The amplitude for current c. The phase difference between the current and the emf of the generator
Answer:
(a) 14.88 ohm
(b) 1.88 A
(c) -74.4°
Explanation:
Vo = 28 V
f = 12 kHz = 12000 Hz
R = 4 ohm
L = 30 micro henry = 30 x 10^-6 H
C = 800 nF = 800 x 10^-9 F
(a)
The inductive reactance,
XL = 2 π f L = 2 x 3.14 x 12000 x 30 x 10^-6 = 2.26 ohm
The capacitive reactance
[tex]X_{c}=\frac{1}{2\pi fC}=\frac{1}{2 \times 3.14 \times 12000 \times 800 \times 10^{-9}}[/tex]
Xc = 16.59 ohm
Let the impedance is Z.
[tex]Z=\sqrt{4^{2}+\left ( 2.26-16.59 \right )^{2}}[/tex]
Z = 14.88 ohm
(b)
The formula for the amplitude of current
[tex]I_{o}=\frac{V_{o}}{Z}=\frac{28}{14.88}[/tex]
Io = 1.88 A
(c)
Let the phase difference is Ф
[tex]tan\phi =\frac{X_{L}-X_{C}}{R}=\frac{2.26-16.59}{4}=-3.5825[/tex]
Ф = -74.4°
A large asteroid of mass 33900 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 610 kg, is in a circular orbit about the first at a distance of 146 meters as a result of their mutual gravitational attraction. What is the speed of the second asteroid? Now suppose that the first and second asteroids carry charges of 1.18 C and -1.18 C, respectively. How fast would the second asteroid have to be moving in order to occupy the same circular orbit as before?
Answer:
a) 1.2*10^-4 m/s
b) 375 m/s
Explanation:
I assume the large asteroid doesn't move.
The smaller asteroid is affected by an acceleration determined by the universal gravitation law:
a = G * M / d^2
Where
G: universal gravitation constant (6.67*10^-11 m^3/(kg*s^2))
M: mass of the large asteroid (33900 kg)
d: distance between them (146 m)
Then:
a = 6.67*10^-11 * 33900 / 146^2 = 10^-10 m/s^2
I assume the asteroid in a circular orbit, in this case the centripetal acceleration is:
a = v^2/r
Rearranging:
v^2 = a * r
[tex]v = \sqrt{a * r}[/tex]
v = \sqrt{10^-10 * 146} = 1.2*10^-4 m/s
If the asteroids have electric charges of 1.18 C and -1.18 C there will be an electric force of:
F = 1/(4π*e0)*(q1*q2)/d^2
Where e0 is the electrical constant (8.85*10^-12 F/m)
F = 1/(4π*8.85*10^-12) (-1.18*1.18)/ 146^2 = -587 kN
On an asteroid witha mass of 610 kg this force causes an acceleration of:
F = m * a
a = F / m
a = 587000 / 610 = 962 m/s^2
With the electric acceleration, the gravitational one is negligible.
The speed is then:
v = \sqrt{962 * 146} = 375 m/s
One common application of conservation of energy in mechanics is to determine the speed of an object. Although the simulation doesn’t give the skater's speed, you can calculate it because the skater's kinetic energy is known at any location on the track. Consider again the case where the skater starts 7 m above the ground and skates down the track. What is the skater's speed when the skater is at the bottom of the track? Express your answer numerically in meters per second to two significant figures.
The skater's speed at the bottom of the track can be determined using the law of conservation of energy.
Explanation:The skater's speed at the bottom of the track can be determined by applying the law of conservation of energy. At the top of the track, the skater has only potential energy, which is converted to kinetic energy as the skater moves down the track. The potential energy at the top of the track can be calculated using the equation:
mgh = (1/2)mv2
where m is the mass of the skater, h is the height, and v is the speed of the skater. At the bottom of the track, all the potential energy is converted to kinetic energy, so we can set the potential energy equal to the kinetic energy:
mgh = (1/2)mv2
Rearranging the equation gives:
v = √(2gh)
Plugging in the values for mass (which is not given in the question) and height, you can calculate the skater's speed at the bottom of the track.
Hot air enters a rectangular duct (20cm wide, 25cm high, and 5m long) at 100 kPa and 60 degrees C at an average velocity of 5 m/s. While air flows the duct, it gets cool down (loses energy) so that air leave the duct at 54 degrees C. Determine the rate of heat loss from the air under steady condition
Answer:
1.57 kW
Explanation:
The rate of heat loss is given by:
q = Gm * Cp * (tfin - ti)
Where
q: rate of heat loss
Gm: mass flow
Cp: specific heat at constant pressure
The Cp of air is:
Cp = 1 kJ/(kg*K)
The mass flow is the volumetric flow divided by the specific volume
Gm = Gv / v
The volumetric flow is the air speed multiplied by the cruss section of the duct.
Gv = s * h * w (I name speed s because I have already used v)
The specific volume is obtained from the gas state equation:
p * v = R * T
60 C is 333 K
The gas constant for air is 287 J/(kg*K)
Then:
v = (R * T)/p
v = (287 * 333) / 100000 = 0.955 m^3/kg
Then, the mass flow is
Gm = s * h * w / v
And rthe heat loss is of:
q = s * h * w * Cp * (tfin - ti) / v
q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)
Leady oxide is a material that?s usually composed of A. 25% free lead and 75% lead oxide. B. 10% free lead and 90% lead oxide. C. 60% free lead and 40% lead oxide. D. 50% free lead and 50% lead oxide.
Answer:
option A is correct
25% free lead and 75% lead oxide
Explanation:
we have given free lead and lead oxide %
so here we know lead oxide material usually composed in the range of for free lead and lead oxide as
lead oxide material contain free lead range is 22 % to 38 %
so here we have only option 1 which contain in free lead in the range of 22 % to 38 % i.e 25 % free lead
so
option A is correct
25% free lead and 75% lead oxide
A bicyclist pedals at a speed of 5.0 km/h. How far does he travel in 80 minutes? A. 0.08 km В. 300 m C. 400 m D. 6.7 km
Answer:d-6.7 km
Explanation:
Given
Bicyclist pedals at a speed of 5 km/h
so his speed in meter per second
[tex]5\times \frac{5}{18}=\frac{25}{18} m/s[/tex]
In 80 minutes he would travel
Distance traveled[tex]=\frac{25}{18}\times 80\times 60=6666.667 m\approx 6.67 km[/tex]
For which of the following charge distributions would Gauss’s law not be useful for calculating the electric field? A. a uniformly charged sphere of radius R
B. a spherical shell of radius R with charge uniformly distributed over its surface
C. a right circular cylinder of radius R and height h with charge uniformly distributed over its surface
D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface
E. Gauss’s law would be useful for finding the electric field in all of these cases.
Answer:
The correct answer is option 'c'.
Explanation:
Gauss's law theoretically can be used to calculate the electric field by any shape of conductor
According to Gauss's law we have
[tex]\oint _s\overrightarrow{E}\cdot \widehat{ds}=\frac{q_{in}}{\epsilon _o}[/tex]
Now the integral on the right hand side of the above relation is solved easily if there is a high degree of symmetry in the electric field which is possible in cases when the object is highly symmetric.
The cases of high symmetry include electric fields due to charged spheres, infinite line charge, point charge, infinite plane charged sheet, infinite cylindrical conductor.
But the for the case of limited height of cylinder the symmetry cannot be utilized thus the integral becomes complex to solve, thus cannot be used.
Assume that a projectile is fired horizontally from a "gun" located 1 meter above the ground. If the ball strikes the ground a distance of 2 meters from the end of the "gun" determine the muzzle velocity of the gun.
Answer:
[tex]v_{o}=8.85m/s[/tex]
Explanation:
To determine the muzzle velocity of the gun, we must know how long does it take the ball to strikes the ground
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
Since the ground is at y=0 and [tex]v_{oy}=0[/tex]
[tex]0=1-\frac{1}{2}(9.8)t^{2}[/tex]
Solving for t
[tex]t=0.4517s[/tex]
Now, to determine the muzzle velocity we need to find its acceleration first
[tex]x=x_{o}+v_{ox}t+\frac{1}{2}at^{2}[/tex] (1)
[tex]v=v_{ox}+at[/tex] (2)
If we analyze the final velocity is 0. From (2) we have that
[tex]v_{ox}=-at[/tex] (3)
Replacing (3) in (1)
[tex]2=-at^{2}+\frac{1}{2}at^{2}[/tex]
[tex]2=a(0.4517)^{2} (\frac{1}{2}-1)[/tex]
[tex]a=-19.60m/s^{2}[/tex]
Solving (3)
[tex]v_{ox}=-at=-(19.60m/s^{2} )(0.4517s)=8.85m/s[/tex]
A monument has a height of 348 ft, 8 in. Express this height in meters. Answer in units of m.
Answer:
The height of mountain in meter will be 106.2732 m
Explanation:
We have given height of mountain = 348 ft,8 in
We know that 1 feet = 0.3048 meter
So 348 feet [tex]=348\times 0.3048=106.07meter[/tex]
And we know that 1 inch = 0.0254 meter
So 8 inch [tex]8\times 0.0254=0.2032m[/tex]
So the total height of mountain in meter = 106.07+0.2032 = 106.2732 m
The height of mountain in meter will be 106.2732 m
Two 3.5-cm-diameter disks face each other, 2.0 mm apart. They are charged to ± 11 nC . a) What is the electric field strength between the disks?
Express your answer in newtons per coulomb.
b) A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Express your answer in meters per second.
Answer:
a) 1.29*10^6 N/C
b) 0.703 *10^6 m/s
Explanation:
This is a parallel plates capacitor. In a parallel plates capacitor the electric field depends on the charge of the disks, its area and the vacuum permisivity (Assuming there is no dielectric) and can be found using the expression:
[tex]E = \frac{Q}{A*e_0} =\frac{11*10^{-9}C}{(\frac{1}{4}\pi*(0.035m)^2)*8.85*10^{-12}C^2/Nm^2} = 1.29 *10^6 N/C[/tex]
For the second part, we use conservation of energy. The change in kinetic energy must be equal to the change in potential energy. The potential energy is given by:
[tex]PE = V*q[/tex]
V is the electric potential or voltage, q is the charge of the proton. The electric potential is equal to:
[tex]V = -E*d[/tex]
Where d is the distance to the positive disk. Then:
[tex]\frac{1}{2}mv_1^2 +V_1q = \frac{1}{2}mv_2^2 +V_2q\\\frac{1}{2}m(v_1^2 - v_2^2)=(V_2-V_1)q = (r_1-r_2)Eq|r_2 = 0m, v_2=0m/s\\v_1 = \sqrt{2\frac{(0.002m)*1.29*10^6 N/C*1.6*10^{-19}C}{1.67*10^{-27}kg}}= 0.703 *10^6 m/s[/tex]
A person is standing on a level floor. His head,upper torso,
arms and hands together weigh 438N and have a centerof gravity that
is 1.17m above the floor. His upper legsweigh 144N and have a
center of gravity that is 0.835m above thefloor. Finally, his lower
legs and feet together weigh 87Nand have a center of gravity that
is 0.270m above the floor. Relative to the floor, find the location
of the center of gravityfor the entire body.
Answer:
g = 0.98 m
Explanation:
given data:
upper portion weight 438 N and its center of gravity is 1.17 m
upper legs weight 144 N and its center of gravity is 0.835 m
lower leg weight 88 N and its center of gravity is 0.270 m
position of center of gravity of whole body can be determine by using following relation
[tex]g = \frac{\sum ({m_i g_i)}}{\sum m_i}[/tex]
where [tex]m_i[/tex] is mass of respective part and [tex]g_i[/tex] is center of gravity
[tex]g = \frac{\sum{ 438\times 1.17 + 144\times 0.835 + 88\times 0.270}}{438+144+88}[/tex]
g = 0.98 m
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 13.513.5 m/s2 for a time period of 3.503.50 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 5.155.15 m/s2. After the rocket turns off, how much time does it take for the sled to come to a stop?
By the time the sled finally comes to a rest, how far has it traveled from its starting point?
1) 9.18 s
In the first part of the motion, the rocket accelerates at a rate of
[tex]a_1=13.5 m/s^2[/tex]
For a time period of
[tex]t_1=3.50 s[/tex]
So we can calculate the velocity of the rocket after this time period by using the SUVAT equation:
[tex]v_1=u+a_1t_1[/tex]
where u = 0 is the initial velocity of the rocket. Substituting a1 and t1,
[tex]v_1=(13.5)(3.50)=47.3 m/s[/tex]
In the second part of the motion, the rocket decelerates with a constant acceleration of
[tex]a_2 = -5.15 m/s^2[/tex]
Until it comes to a stop, to reach a final velocity of
[tex]v_2 = 0[/tex]
So we can use again the same equation
[tex]v_2 = v_1 + a_2 t_2[/tex]
where [tex]v_1 = 47.3 m/s[/tex]. Solving for t2, we find after how much time the rocket comes to a stop:
[tex]t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s[/tex]
2) 299.9 m
We have to calculate the distance travelled by the rocket in each part of the motion.
The distance travelled in the first part is given by:
[tex]d_1 = ut_1 + \frac{1}{2}a_1 t_1^2[/tex]
Using the numbers found in part a),
[tex]d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m[/tex]
The distance travelled in the second part of the motion is
[tex]d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2[/tex]
Using the numbers found in part a),
[tex]d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m[/tex]
So, the total distance travelled by the rocket is
d = 82.7 m + 217.2 m = 299.9 m
An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees the car, the locomotive is 300 m from the crossing and its speed is 18 m/s. If the engineer’s reaction time is 0.45 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2 .
The magnitude of the minimum deceleration needed to avoid the accident is –0.55 m/s²
To solve the question given above, we'll begin by calculating the distance travelled during the reaction time. This can be obtained as follow:
Speed = 18 m/s
Time = 0.45 s
Distance =?Speed = distance / time
18 = distance / 0.45
Cross multiply
Distance = 18 × 0.45
Distance = 8.1 mThus, the engineer travelled a distance of 8.1 m during the reaction time.
Next, we shall the distance between the engineer and the car. This can be obtained as follow:
Total distance = 300 m
Distance during the reaction time = 8.1 m
Distance between the engineer and the car =?Distance between the engineer and the car = (Total distance) – (Distance during the reaction time)= 300 – 8.1
Distance between the engineer and the car = 291.9 mFinally, we shall determine the magnitude of the deceleration needed to avoid the accident. This can be obtained as follow:
Initial velocity (u) = 18 m/s
Final velocity (v) = 0 m/s
Distance (s) = 291.9 m
Deceleration (a) =?v² = u² + 2as0² = 18² + (2 × a × 291.9)
0 = 324 + 583.8a
Collect like terms
0 – 324 = 583.8a
–324 = 583.8a
Divide both side by 583.8
a = –324 / 583.8
a = –0.55 m/s²Therefore, the magnitude of the deceleration needed to avoid the accident is –0.55 m/s²
Learn more: https://brainly.com/question/2797154
To avoid an accident, the engineer must decelerate the train at a minimum rate of approximately -0.55 m/s^2, taking into account the reaction time which contributed an additional 8.1 m to the stopping distance.
Explanation:An engineer in a locomotive sees a car stuck on the track and needs to calculate the necessary deceleration to avoid an accident. The engineer has a reaction time of 0.45 seconds, and the initial speed of the train is 18 m/s. The total stopping distance must include the distance traveled during the reaction time plus the actual braking distance.
The distance covered during the engineer's reaction time is found using the formula d = vt, where d is distance, v is velocity, and t is time. We get d = 18 m/s * 0.45 s = 8.1 m.
So, the stopping distance minus the reaction distance is 300 m - 8.1 m = 291.9 m. To find the deceleration, we can use the formula v^2 = u^2 + 2as, where v is the final velocity (which is 0), u is the initial velocity, a is the acceleration (deceleration in this case), and s is the distance. Solving for a, we get a = -u^2 / (2s). Substituting the values, a = -(18 m/s)^2 / (2 * 291.9 m), resulting in a deceleration of approximately -0.55 m/s^2.
What is the the electric field at a point midway between a -7.67 μC and a +5.03 μC charge 3.38 cm apart? Take the direction towards the positive charge to be positive.
Answer:
-4.01 × 10⁸ N/C.
Explanation:
Given : Charge Q = -7.67 × 10⁻⁶ C
Charge q = + 5.03×10⁻⁶ C
Distance between the charges = 0.0338 m
Let d be the distance from the charge Q and charge q to the mid point.
Midway distance = d = 0.0169 m
Electric field due to the negative charge = E₁ = k Q / d²
Electric field due to the positive charge = E₂ = k q / d²
Here k = 9 × 10⁹ N m²/C² is the Coulomb's constant.
E₁ = (9 × 10⁹)(7.67 × 10⁻⁶) / (0.0169)² = 2.42 × 10⁸ N/C
E₂ = (9 × 10⁹)( 5.03×10⁻⁶) /(0.0169)² = 1.585 × 10⁸N/C
Assume that the negative charge is towards the left of the mid point and the positive charge is towards the right of the mid point,
The electric field due to the negative charge is inwards towards the charge(left) and the electric field due to positive charge is outwards and so towards the negative X direction.
Total electric field = E = E₁ + E₂ = -2.42 × 10⁸ - 1.585 × 10⁸
= - 4.01 × 10⁸ N/C
Direction towards the left (towards the negative charge).