Answer:maximum
Explanation:
A 0.410 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 35.0 pC charge on its surface. What is the potential (in V) near its surface
Answer:
The potential is [tex]V = 153.659 \ V[/tex]
Explanation:
From the question we are told that
The diameter of the plastic sphere is [tex]d = 0.410 \ cm = 0.0041 \ m[/tex]
The magnitude of the charge is [tex]q = 35.0 pC = 35.0 *10^{-12} \ C[/tex]
The radius of the plastic sphere is mathematically evaluated as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.0041}{2}[/tex]
[tex]r = 0.00205 \ m[/tex]
The potential near the surface is mathematically represented as
[tex]V = \frac{k * q}{r }[/tex]
Where k is the Coulombs constant with value [tex]9 *10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]V = \frac{9*10^9 * 35 *10^{-12}}{0.00205}[/tex]
[tex]V = 153.659 \ V[/tex]
A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upward speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s.
Required:
What is the average force exerted by the ground on the ball during this time? Also explain whether it's upwards or downwards.
Answer:
22 N upward
Explanation:
From the question,
Applying newton's second law of motion
F = m(v-u)/t....................... Equation 1
Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact
Note: Let upward be negative and downward be positive
Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s
Substitute into equation 1
F = 0.14(-1-1.2)/0.014
F = 0.14(-2.2)/0.014
F = 10(-2.2)
F = -22 N
Note the negative sign shows that the force act upward
what is quantic fisic
Answer:
it is the physics that explains how everything works. The best description we have of the. nature of the particles that make up matters and the forces with which they interact. It underlines how atoms work, and so why chemistry and biology work as they do
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.652 kJkJ of heat. It shrinks on cooling, and the atmosphere does 389 JJ of work on the balloon. Express your an
Question:
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.652 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer in Joules (J)
Answer:
-263J
Explanation:
Though its difficult and infact impossible to measure the internal energy of a system, the change in internal energy ΔE, can however be determined. This change when it is accompanied by work(W) and transfer of heat(Q) in or out of the system, can be calculated as follows;
ΔE = Q + W ----------------(i)
Q is negative if heat is lost. It is positive otherwise
W is negative if work is done by the system. It is positive otherwise.
From the question;
Q = -0.652kJ = -652J {the negative sign shows heat loss}
W = +389J {the positive sign shows work done on the system(balloon)}
Substitute these values into equation (i) as follows;
ΔE = -652 + 389
ΔE = -263J
Therefore the change in internal energy is -263J
PS: The negative sign shows that the process is exothermic. This means that the system (balloon) lost some energy to the environment.
A 300-foot cable weighing 5 pounds per foot is hanging from a winch 300 feet above ground level. Find the work (in ft-lb) done in winding up the cable when there is a 300-pound load attached to the end of the cable.
Answer:
315,000 ft·lb
Explanation:
At 300 ft and 5 lb/ft, the weight of the cable is (300 f)(5 lb/ft) = 1500 lb. The work done to raise it is equivalent to the work done to raise the cable's center of mass. Since the cable is of uniform density, its center of mass is half the cable length below the winch.
total work done = work to raise cable + work to raise load
= (1500 lb)(150 ft) +(300 lb)(300 ft) = 315,000 ft·lb
A quartz crystal vibrates with a frequency of 35,621 Hz. What is the period of the crystal's motion?
Period = 1 / frequency
Period = 1 / (35,621 /s)
Period = 2.8073... x 10⁻⁵ sec
Period = 28.07 microseconds
or
Period = 0.0281 millisecond
The period (T) of the crystal's motion is equal to [tex]2.81 \times 10^{-5}\;seconds[/tex].
Given the following data:
Frequency of quartz crystal = 35,621 Hertz.To calculate the period of the crystal's motion:
The formula for the period of oscillation.Mathematically, the period of oscillation of an object is given by this formula:
[tex]T=\frac{1}{F}[/tex]
Where:
T is the period of oscillation.F is the frequency.Substituting the given parameters into the formula, we have;
[tex]T=\frac{1}{35621} \\\\T=2.81 \times 10^{-5}\;seconds[/tex]
Therefore, the period (T) of the crystal's motion is equal to [tex]2.81 \times 10^{-5}\;seconds[/tex].
Read more on period here: https://brainly.com/question/14024265
Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus
Answer:
D). Uranus.
Explanation:
Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.
As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.
Two charged particles of equal magnitude (+Q and +Q) are fixed at opposite corners of a square that lies in a plane. A test charge +q is placed at the third corner of the square. What is the direction of force on the test charge due to other two charges?
Answer:
The test charge will take the south-west direction indicated in option 6.
Explanation:
The image is shown below.
Since all the charges are positively charged, they will all repel each other. If we consider the force on +q due to +Q and +Q, then we can proceed as follows
The +Q particle at the top left corner of the cube will exert a vertical downward force on +q in the -ve y-axis.
The +Q particle at the bottom right corner of the cube will exert a force on +q towards the horizontal left on the -ve x-axis.
Both of these forces will act at angle of 90°, and therefore, the resultant force will act at an angle of 45° to horizontal and vertical forces.
The result is that the +q charge will move in a south-west direction of the cube.
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices
Answer:
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
Explanation:
The power dissipation by an electrical circuit is given by the following formula:
Power Dissipation = (Voltage)(Current)
P = VI
but, from Ohm's Law, we know that:
Voltage = (Current)(Resistance)
V = IR
Substituting this in formula of power:
P = (IR)(I)
P = I²R ---------------- equation 1
Now, if we double the current , then the power dissipated by that circuit will be:
P' = I'²R
where,
I' = 2 I
Therefore,
P' = (2 I)²R
P' = 4 I²R
using equation 1
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
An elevator filled with passengers has a mass of 1,700 kilograms and accelerates upward from rest at a rate of 1 meters/seconds 2 for 1.8 seconds. Calculate the tension in the cable (in Newtons) supporting the elevator during this time.
Answer:
The tension in the cable is 18371.9 newtons.
Explanation:
Physically speaking, the tension can be calculated with the help of the Second Newton's Law. The upward acceleration means that magnitude of tension must be greater than weight of elevator, whose equation of equilibrium is described below:
[tex]\Sigma F = T - m\cdot g = m \cdot a[/tex]
Where:
[tex]T[/tex] - Tension in the cable, measured in newtons.
[tex]m[/tex] - Mass of the elevator, measured in kilograms.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
[tex]a[/tex] - Net acceleration of the elevator, measured in meters in square second.
Now, tension is cleared and resultant expression is also simplified:
[tex]T = m \cdot (a + g)[/tex]
If [tex]m = 1700\,kg[/tex], [tex]a = 1\,\frac{m}{s^{2}}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the tension in the cable is:
[tex]T = (1700\,kg)\cdot \left(1\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T = 18371.9\,N[/tex]
The tension in the cable is 18371.9 newtons.
Red light from three separate sources passes through a diffraction grating with 5.60×105 slits/m. The wavelengths of the three lines are 6.56 ×10−7m (hydrogen), 6.50 ×10−7m (neon), and 6.97 ×10−7m (argon).
Part A
Calculate the angle for the first-order diffraction line of first source (hydrogen).
Express your answer using three significant figures.
θH = ∘
Part B
Calculate the angle for the first-order diffraction line of second source (neon).
Express your answer using three significant figures.
θNe = ∘
Part C
Calculate the angle for the first-order diffraction line of third source (argon).
Express your answer using three significant figures.
θAr = ∘
Answer:
I can help
Explanation:
An electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. What is the direction of the magnetic force on the electron?
Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.
Explanation:
The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:
F = BqVSinØ
If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field
According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.
What would be the Roche limit (in units of Earth radii) if the Earth had the same mass, but its radius was increased to 1.5 Earth radii?
First calculate the density of this new, larger, Earth. Now use this new density and the new radius in the calculator above to determine the Roche limit for this new larger 'Earth.
Answer:
Roche limit = 1.89 of earth radius
Explanation:
We know that,
Mass of earth = 5.972 × 10²⁷ g
New radius = 1.5(old radius) = 1.5(6.371 × 10⁸) = 9.5565 × 10⁸
Density of earth = 5.5132 g/cm³
New density of earth = Mass of earth / (4/3)πr³
New density of earth = 5.972 × 10²⁷ kg / (4/3)(22/7)( 9.5565 × 10⁸)³
New density of earth = 1.634 g/cm³
Roche limit = [2(Density of earth)/(New density of earth)]¹/³r
Roche limit = 1.89 of earth radius
Two 2.0-cm-diameter insulating spheres have a 6.70 cm space between them. One sphere is charged to +70.0 nC, the other to -40.0 nC. What is the electric field strength at the midpoint between the two spheres?
Answer:
Explanation:
The distance of middle point from centres of spheres will be as follows
From each of 2 cm diameter sphere
R = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m
Expression for electric field = Q / 4πε R²
Electric field due to positive charge
E₁ = 70 x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴
= 33.3 x 10⁴ N/C
Electric field due to negative charge
E₂ = 40 x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴
= 19.02 x 10⁴ N/C
E₁ and E₂ act in the same direction so
Total field = (33.3 + 19.02 ) x 10⁴
= 52.32 x 10⁴ N/C .
Find the pressure difference (in kPa) on an airplane wing if air flows over the upper surface with a speed of 125 m/s, and along the bottom surface with a speed of 109 m/s. [Express answer in TWO decimal places]
Answer:
P= 2414.9 Pa
Explanation:
given
density of air , p = 1.29 kg/m³
speed of air over the upper surface , v₁ = 125 m/s
speed of air over the lower surface , v₂ = 109 m/s
the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)
P = 0.5 × 1.29 × ( 125² - 109²)
P= 0.645(3744)
P = 2414.9 Pa
the pressure difference on an airplane wing is 2414.9 Pa
A crate resting on a horizontal floor (\muμs = 0.75, \muμk = 0.24 ) has a horizontal force F = 93 Newtons applied to the right. This applied force is the maximum possible force for which the crate does not begin to slide. If you applied this same force after the crate is already sliding, what would be the resulting acceleration (in meters/second2) ?
Answer:
The acceleration is [tex]a = 5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The coefficient of kinetic friction is [tex]\mu_k = 0.24[/tex]
The coefficient of static friction is [tex]\mu_s = 0.75[/tex]
The horizontal force is [tex]F_h = 93 \ N[/tex]
Generally the static frictional force is mathematically represented as
[tex]F_F = \mu_s * (m * g )[/tex]
The static frictional force is the equivalent to the maximum possible force for which the crate does not begin to slide So
[tex]F_h = F_F = \mu_s * (m * g )[/tex]
=> [tex]93 = \mu_s * (m * g )[/tex]
=> [tex]m = \frac{93}{\mu_s * g }[/tex]
substituting values
[tex]m = \frac{93}{0.75 * 9.8 }[/tex]
[tex]m = 12.65 \ kg[/tex]
When the crate is already sliding the frictional force is
[tex]F_s = \mu_k *(m * g )[/tex]
substituting values
[tex]F_s = 0.24 * 12.65 * 9.8[/tex]
[tex]F_s = 29.82 \ N[/tex]
Now the net force when the horizontal force is applied during sliding is
[tex]F_{net} = F_h - F_s[/tex]
substituting values
[tex]F_{net} = 93 - 29.8[/tex]
[tex]F_{net} = 63.2 \ N[/tex]
This net force is mathematically represented as
[tex]F_{net } = m * a[/tex]
Where a is the acceleration of the crate
So
[tex]a = \frac{F_{net}}{m }[/tex]
[tex]a = \frac{ 63.2}{12.65 }[/tex]
[tex]a = 5 \ m/s^2[/tex]
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasin
Complete question:
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasing four hours later.
Answer:
The rate at which the distance between the cars is increasing four hours later is 35 mi/h.
Explanation:
Given;
speed of one car, dx/dt = 28 mi/h South
speed of the second car, dy/dt = 21 mi/h West
The distance between the cars is the line joining west to south, which forms a right angled triangle with the two positions.
Apply Pythagoras theorem to evaluate this distance;
let the distance between the cars = z
x² + y² = z² -------- equation (1)
Differentiate with respect to time (t)
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex] ----- equation (2)
Since the speed of the cars is constant, after 4 hours their different distance will be;
x: 28(4) = 112 mi
y: 21(4) = 84 mi
[tex]z = \sqrt{x^2 + y^2} \\\\z = \sqrt{112^2 + 84^2} \\\\z = 140 \ mi[/tex]
Substitute in the value of x, y, z, dx/dt, dy /dt into equation (2) and solve for dz/dt
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\\\2(112)(28) + 2(84)(21) = 2(140)\frac{dz}{dt} \\\\9800 = 280\frac{dz}{dt} \\\\\frac{dz}{dt} = \frac{9800}{280} \\\\\frac{dz}{dt} = 35 \ mi/h[/tex]
Therefore, the rate at which the distance between the cars is increasing four hours later is 35 mi/h
a wave with a high amplitude______?
. . . is carrying more energy than a wave in the same medium with a lower amplitude.
Let us suppose the magnitude of the original Coulomb force between the two charged spheres is FF. In this scenario, a third sphere touches the grey sphere and the red sphere multiple times, being grounded each touch. If the grey sphere is touched twice, and the red sphere is touched three times, what is the magnitude of the Coulomb force between the spheres now
Answer:
F ’= 1/32 F
We see that the value of the force is the initial force over 32
Explanation:
In this problem the sphere that is touching the others is connected to ground, after each touch,
Let's analyze the charge of the gray sphere, when you touch it for the first time, the charge is divided between the two spheres each having Q / 2, when the sphere separates and touches ground, its charge passes zero. When I touch the gray dial again, its charge is reduced by half
½ (Q / 2) = ¼ Q
For the red dial repeat the same scheme
with the first touch the charge is reduced to Q / 2
with the second touch e reduce to ½ (Q / 2) = ¼ Q
with the third toce it is reduced to ½ (¼ Q) = ⅛ Q
Now let's analyze what happens to the electric force
if the force is F for when the charge of each sphere is Q
F = k Q Q / r²
with the remaining charge strength is
F ’= k (¼ Q) (⅛ Q) / r²
F ’= 1/32 k Q Q / r²
F ’= 1/32 F
We see that the value of the force is the initial force over 32
Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m = 60 g) attached to the end of the string traveling in uniform circular motion above her head at a constant speed of 9.0 m/s. What is the length of the string between her hand and the tennis ball? You may ignore gravity in this problem (assume the motion of the tennis ball and string happen in a purely horizontal plane). A. 41 m B. 0.24 m C. 3.2 cm D. 0.41 m
Answer:
r = 0.405m = 40.5cm
Explanation:
In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:
[tex]F_c=ma_c=m\frac{v^2}{r}[/tex] (1)
Fc: centripetal acceleration (tension force on the string) = 12N
m: mass of the ball = 60g = 0.06kg
r: length of the string = ?
v: linear speed of the ball = 9.0m/s
You solve for r in the equation (1) and replace the values of the other parameters:
[tex]r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m[/tex]
The length of the string between Wanda and the ball is 0.405m = 40.5cm
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.
Answer:28.29m/s
Explanation:
In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;
([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v -----------------(i)
Where;
[tex]m_{C}[/tex] = mass of car
[tex]u_{C}[/tex] = initial velocity of car before collision
[tex]m_{D}[/tex] = mass of deer
[tex]u_{D}[/tex] = initial velocity of the deer before collision
v = common velocity with which the car and the deer move after collision
From the question;
[tex]m_{C}[/tex] = 900kg
[tex]u_{C}[/tex] = +30.0m/s (direction of the motion of the car taken positive)
[tex]m_{D}[/tex] = 150kg
[tex]u_{D}[/tex] = +18.0m/s (relative to the direction of the car, the velocity of the deer is also positive )
Substitute these values into equation (i) as follows;
(900 x 30.0) + (150 x 18.0) = (900 + 150)v
27000 + 2700 = 1050v
29700 = 1050v
v = [tex]\frac{29700}{1050}[/tex]
v = 28.29m/s
Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate
For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.
When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.
For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.
Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.
Learn more: https://brainly.com/question/5019199
A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 42.0 cm from its center.
Answer:
-1.4x10^6N/C
Explanation:
Pls see attached file
The magnitude of the electric field.
Magnitude is the size of the object in properties that is determines the size of the object. It also displays the result of the order of class of the object. The direction of the electric field tells us about the position of the field in four different directions. As per the question, the answer is 1.4x10^6N/C.
The rod of 16cm of total length is given. Has a charge of a total of -25.0uc. The rod's axis is pointed at 42.0 cm from its center and is given in the question. The rod Length will be then 0.16m and the total change will be 25x10 cm and point where the electricity will be calculated is shown by the axis of the rod at the distance of 42 cms.The magnitude and direction will be calculated based on the measure of the formula of E. This answer to the question will be 1.4x10^6N/C.Learn more about the uniformly charged.
brainly.com/question/12088419.
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:
Yes the monkey will get hit and it will not avoid the dart.
Explanation:
Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.
No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.
The position of a particle is r(t)= (4.0t'i+ 2.4j- 5.6tk) m. (Express your answers in vector form.) (a) Determine its velocity (in m/s) and acceleration (in m/s2) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.) v(t)= ________m/s a(t)= ________m/s2 (b) What are its velocity (in m/s) and acceleration (in m/s2) at time t 0? v(0) =_______ m/s a(0)=_______ m/s2
The position of a particle is r(t)= (4.0t²i+ 2.4j- 5.6tk) m. (Express your answers in vector form.)
(a) Determine its velocity (in m/s) and acceleration (in m/s²) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.)
v(t)= ________m/s
a(t)= ________m/s²
(b) What are its velocity (in m/s) and acceleration (in m/s²) at time t 0?
v(0) =_______ m/s
a(0)=_______ m/s²
Answer:
(a)
v(t)= [tex]8ti - 5.6k[/tex] m/s
a(t)= 8i m/s²
(b)
v(0) = -5.6k m/s
a(0)= 8i m/s²
Explanation:From the question, the position of the particle is given by;
r(t)= (4.0t²i+ 2.4j- 5.6tk) -----------------(i)
(a)
(i)To get the velocity, v(t), of the particle, we'll take the first derivative of the position of the particle (given by equation (i)) with respect to time, t, as follows;
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]\frac{d(4.0t^2i + 2.4j - 5.6tk)}{dt}[/tex]
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]8ti +0j - 5.6k[/tex]
v(t) = [tex]8ti - 5.6k[/tex] --------------------(ii)
(ii) To get the acceleration, a(t), of the particle, we'll take the first derivative of the velocity of the particle (given by equation (ii)) with respect to time, t, as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex] = [tex]\frac{d(8ti - 5.6k)}{dt}[/tex]
a(t) = 8i --------------------(iii)
(b)
(i) To get the velocity of the particle at time t = 0, substitute the value of t = 0 into equation (ii) as follows;
v(t) = [tex]8ti - 5.6k[/tex]
v(0) = 8(0)i - 5.6k
v(0) = 0 - 5.6k
v(0) = -5.6k
(ii) To get the acceleration of the particle at time t = 0, substitute the value of t = 0 into equation (iii) as follows;
a(t) = 8i
a(0) = 8i
A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field of 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y
Answer:
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
Explanation:
- To find the direction of the conventional current in the wire you use the following formula:
[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex] (1)
i: current in the wire = ?
F: magnitude of the magnetic force on the wire = 15.1N
B: magnitude of the magnetic field = 6.1T
l: length of the wire that is affected by the magnetic field = 0.45m
The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).
The direction of the current must be in the +y direction (+^j). In fact, you have:
^j X ^k = ^i
The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):
[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
At the lowest point in a vertical dive (radius = 0.58 km), an airplane has a speed of 300 km/h which is not changing. Determine the magnitude of the acceleration of the pilot at this lowest point. Group of answer choices
Answer:
The centripetal acceleration is [tex]a = 11.97 \ m/s^2[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 0.58 \ km = 0.58 * 1000 = 580 \ m[/tex]
The speed is [tex]v = 300\ km /hr = \frac{300 *1000}{1 * 3600 } = 83.33 \ m/s[/tex]
The centripetal acceleration of the pilot is mathematically represented as
[tex]a = \frac{v^2 }{r}[/tex]
substituting values
[tex]a = \frac{(83.33)^2 }{580}[/tex]
[tex]a = 11.97 \ m/s^2[/tex]
Two 10-cm-diameter charged rings face each other, 18.0 cmcm apart. Both rings are charged to 30.0 nCnC . What is the electric field strength
Answer:
E=7453.99 V/m
Explanation:
The electric field on the charged is given by
E= Kqx/(r^2 +x^2)^3/2
Where;
K= constant of Coulomb's law
q= magnitude of charge= 30.0×10^-9 C
r= radius of the rings= 5 cm or 0.05m
x= distance between the rings = 18cm = 0.18 m
Substituting values;
E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2
E= 48.6/(2.5×10^-3 + 0.0324)^3/2
E= 48.6/(0.0025 + 0.0324)^3/2
E= 48.6/6.52×10^-3
E=7453.99 V/m
When a spinning bike wheel is placed horizontally, hung from a pivot at one end, the axis of rotation of the wheel will swing in a horizontal circle. In which direction does it turn?a) upwardb) downwardc) horizontally, CWd) horizontally, CCW
Answer:
answer is D
Explanation:
horizontally, CCW
If a key is pressed on a piano, the frequency of the resulting sound will determine the ________, and the amplitude will determine the ________ of the perceived musical note.
Answer:
If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.
Explanation:
The frequency of a vibrating string is primarily based on three factors:
The sounding length (longer is lower, shorter is higher)
The tension on the string (more tension is higher, less is lower)
The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)
To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.
Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.
And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.
Many things can influence the amplitude.
What is producing the sound?
How far are you from the source of the sound? The farther away the smaller the amplitude.
Intervening material. Sound does not travel through walls as well as air.
Depends on what is detecting the wave sound. Ear vs. microphone.
Answer:
The frequency will determine the pitch
the amplitude will determine the loudness
Explanation:
The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.
The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.