Answer:
C) they all have the same angular speed
D) they all have the same angular acceleration
Explanation:
Wrong --> they all have the same tangential speed. The points close to the axis will have less speed than the points away from the axis.
Wrong --> they all have the same tangential acceleration. Similarly, the points close to the axis will have smaller acceleration than the points away from the axis.
Correct --> they all have the same angular speed. Angular speed is the same for all the particles in the rotating object.
Correct --> they all have the same angular acceleration. Angular acceleration is the same for all the particles in the rotating object.
This all comes from the following relations:
v = ωR
a = αR
where ω is the angular velocity and α is the angular acceleration.
As can be seen from above, tangential velocity and acceleration depends on the distance from the axis, whereas the angular velocity and acceleration is the same for all the points on the rotating body.
The true statements about all points in the object rotating about a fixed point are;
C) they all have the same angular speed
D) they all have the same angular acceleration
For a circular motion about a given point, the angular speed is same for all points on the circular path and it is calculated as;
[tex]\omega = \frac{2\pi N}{T}[/tex]
Where;
N is the number of turns about the fixed pointT is the time of motionThus, angular speed is independent of the position of an object rotating about a fixed point.
The angular acceleration is given as;
[tex]\alpha = \frac{\omega}{T}[/tex]
Tangential speed and acceleration depends on the position of each object along the circular path.
Thus, we can conclude that the true statements about all points in the object rotating about a fixed point are, they all have the same angular speed and they all have the same angular acceleration.
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Brewed coffee is often too hot to drink right away. You can cool it with an ice cube, but this dilutes it. Or you can buy a device that will cool your coffee without dilution - a 200 aluminum cylinder that you take from your freezer and place in a mug of hot coffee.
q)If the cylinder is cooled to -20C, a typical freezer temperature, and then dropped into a large cup of coffee (essentially water, with a mass of 500g ) at 85C, what is the final temperature of the coffee?answer in 2 sig figs and answer in Celsius.
The final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).
Given equations:
[tex]\[ Q_c = Q_a \][/tex]
[tex]\[ m_c \cdot c_w \cdot (T_{ic} - T_f) = m_a \cdot c_a \cdot (T_f - T_{ia}) \][/tex]
Substitute the given values:
[tex]\[ 0.5 \cdot 1000 \cdot (85 - T_f) = 0.2 \cdot 900 \cdot (T_f - (-20)) \][/tex]
Now, distribute and simplify:
[tex]\[ 42500 - 500 \cdot T_f = 180 \cdot T_f + 3600 \][/tex]
Combine like terms:
[tex]\[ 500 \cdot T_f + 180 \cdot T_f = 42500 - 3600 \][/tex]
[tex]\[ 680 \cdot T_f = 38900 \][/tex]
Now, solve for A [tex]\( T_f \):[/tex]
[tex]\[ T_f = \frac{38900}{680} \][/tex]
[tex]\[ T_f \approx 57.21 \, ^\circ C \][/tex]
So, the final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).
A person who is properly constrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed about 30 "Gs" (1.0 g = 9.8 m/s^2). Assuming uniform deceleration of this value, what is the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 75 km/h?
Answer:
Distance = 0.738 m
Explanation:
Solution:
First convert Km/h into m/s.
75 km/h * 1000 m/km * 1 hr/3600 sec = 20.8333 m/s
According to third equation of motion:
[tex]Vf^{2}[/tex] – [tex]Vi^{2}[/tex] = 2 * acceleration * distance
Vf= final velocity
Vi= initial velocity
putting values in third equation of motion....
[tex]0.2^{2}[/tex] – [tex]20.8333^{2}[/tex] = 2 * (-30 * 9.8) * distance
note:
negative sign is due to deceleration
Distance = 0.738 m
Given an initial speed of 75km/h and a maximum deceleration of 30 G's, the front end of the car must be designed to collapse over a distance of approximately 0.73m to prevent injuries that could be fatal in an accident.
Explanation:The problem is asking for the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest, given a certain initial speed and maximum deceleration. This is a problem of kinematics, which is a branch of physics. To solve it, we can use the equation d = (v_f^2 - v_i^2) / (2*a), where:
V_f is the final velocity, which in this case is 0 as the car comes to rest V_i is the initial velocity, which we need to convert from km/h to m/s. This calculation gives us approximately 20.83 m/s A is the deceleration, which we are given as 30 G's, or 294 m/s^2 (30 times 9.8 m/s^2)
So, substituting these values into our equation, we find that the front end of the car must be designed to collapse over a distance of approximately 0.73 meters to prevent injuries that could be fatal.
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A charge of -3.50 nC is placed at the origin of an xy-coordinate system, and a charge of 1.95 nC is placed on the y axis at y = 3.75 cm .Find the distances to the center.
Answer:
3.72 cm
Explanation:
the coordiante of the negative charge =( 0,0)
the coordinate of the positive charge = (0,3.75)
the distance simply is 3.72-0 = 3.72 cm
A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a bronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.
Answer:
P = 17.28*10⁶ N
Explanation:
Given
L = 250 mm = 0.25 m
a = 0.54 m
b = 0.40 m
E = 95 GPa = 95*10⁹ Pa
σmax = 80 MPa = 80*10⁶ Pa
ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m
We get A as follows:
A = a*b = (0.54 m)*(0.40 m) = 0.216 m²
then, we apply the formula
ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L
⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)
⇒ P = 24624000 N = 24.624*10⁶ N
Now we can use the equation
σ = P/A
⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa
So σ > σmax we use σmax
⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N
Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58×107m(≈22,000miles).What is the period of a satellite in a geosynchronous orbit?
Answer:
T = approximately 24 hs.
Explanation:
In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.
Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:
Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)
By definition, ω =ΔФ / Δt
For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),
so we can replace ω by (2*π/T), solving then for T:
T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).
The period of a satellite in a geosynchronous orbit is 24 hours, matching the Earth's rotation. The satellite follows a circular orbit keeping it in the same place relative to Earth's surface. Satellites in this orbit provide various services due to their continuous presence above a particular location.
Explanation:The period of a satellite in a geosynchronous orbit is 24 hours. It's called geosynchronous because it matches the Earth's rotation period which is also 24 hours, this is how they are able to stay over the same point on the Earth's surface. The satellites move in elliptical orbits, but those in geosynchronous orbit follow a circular path so they maintain a constant altitude and remain in constant position relative to the Earth's surface.
As there are many satellites launched each year, being in a geosynchronous orbit helps these satellites to provide continuous services like weather tracking, communications, global positioning systems, etc. These satellites also avoid being captured by Earth's gravity and pulled into the atmosphere because they are at a distance of approximately 36000km from the earth's surface, where gravity is much weaker.
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Assume your sampling frequency (fs) is 500 Hz. If you want to capture the 50Hz signal for 5 seconds, calculate the number of samples (N) at the assumed fs to get the desired 5 seconds of data.
Answer:
N = 2500
Explanation:
By definition, the sampling frequency (fs) is:
[tex] f_{s} = N / T [/tex]
where N: is the number of samples and T: is the time
Hence, the number of samples at the sampling frequency, 500 Hz, to get 5 seconds of data is:
[tex] N = f_{s} \cdot T = 500 s^{-1} \cdot 5 s = 2500 [/tex]
I hope it helps you!
To capture a 50Hz signal for 5 seconds with a sampling frequency of 500 Hz, we need 2500 samples.
Explanation:To capture the 50Hz signal for 5 seconds, we need to calculate the number of samples (N) at the assumed sampling frequency (fs) of 500 Hz. According to the Nyquist theorem, the highest frequency we can accurately monitor is half the sampling frequency. In this case, the Nyquist frequency would be 250 Hz. Since the signal frequency is 50 Hz, it is well below the Nyquist frequency. Therefore, we need to sample the signal at a rate greater than twice the signal frequency, which is 100 samples per second. Multiplying this by the desired duration of 5 seconds, we get the total number of samples required:
N = fs * duration = 500 Hz * 5 s = 2500 samples
The Electric Potential of the Earth The Earth has a vertical electric field with a magnitude of approximately 100 V/m near its surface.
What is the magnitude of the potential difference between a point on the ground and a point on the same level as the top of the Washington Monument (555 ft high)?
Answer:
16916.4 V
Explanation:
Electric potential: This is the work done in bringing a unit positive charge from infinity to that point in against the action of the field. The S.I unit of Electric potential is V.
mathematically, Electric potential can be expressed as
P = E×d ....................................... Equation 1.
Where P = Electric potential, E = Electric Field, d = distance/height of the level at the top of the Washington Monument.
Given: E = 100 V/m, d = 555 ft = 555×0.3048 m = 169.164 m.
Substitute into equation 1
P = 100×169.164
P = 16916.4 V.
Thus the potential difference = 16916.4 V.
The magnitude of the potential difference between the ground and the top of the Washington Monument, which is 555 feet or approximately 169.164 meters tall, caused by Earth's electric field of about 100 V/m, is roughly 16,916.4 volts.
Explanation:The question refers to calculating the electric potential difference caused by Earth's electric field. The provided value of Earth's electric field is approximately 100 V/m near its surface. To find the potential difference (ΔV) between a point on the ground and a point at the height of the top of the Washington Monument, we can use the equation:
ΔV = E * d
where E is the electric field strength and d is the distance in meters. Since the height of the Washington Monument is given in feet, it needs to be converted to meters (1 foot = 0.3048 meters). The height in meters is:
555 ft * 0.3048 m/ft = 169.164 m
Now, we can calculate the potential difference:
ΔV = 100 V/m * 169.164 m = 16916.4 V
Thus, the magnitude of the potential difference between a point on the ground and a point at the level of the top of the Washington Monument is approximately 16,916.4 volts.
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength emitted is λ = 1000 nm. (Note: this wavelength is in the infrared; incandescent bulbs are inefficient since they emit mostly in the infrared and only little bit in the visible range.)
Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
satellite 1 is in a certain circular orbit around a planet, while satellite 2 is in a larger circular orbit. which satellite has (a) the longer period
a. Satellite 1 has a longer period and faster speed than satellite 2.
b. Satellite 1 has a longer period and slower speed than satellite 2.
c. Satellite 1 has a shorter period and faster speed than satellite 2.
d. Satellite 1 has a shorter period and slower speed than satellite 2.
e. The period and the speed are the same for both satellites.
Answer:
c. Satellite 1 has a shorter period and faster speed than satellite 2.
Explanation:
The period of an object in a circular orbit is given by:
[tex]T=\frac{2\pi r}{v}(1)[/tex]
Here r is the radius of the circular orbit and v is the speed of the object. So, we have [tex]r_1<r_2[/tex], according to (1) the period is proportional to the radius, thus [tex]T_1<T_2[/tex]. In other hand the period is inversely proportional to the speed, so [tex]v_1>v_2[/tex]
Answer:
it's b brainliest pls!
Explanation:
A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 58.2 cm, and the density of iron is 7.87 g/cm3. Find the inner diameter.
Answer:
0.556m
Explanation:
unit conversion
58.2 cm = 0.582 m -> the outer radius is 0.582/2 = 0.291 m
[tex]7.87 g/cm^33 = 7.87 g/cm^33 * 1/1000 (kg/g) * 100^3 cm^3/m^3 = 7870 kg/m^3[/tex]
Let r be the inner diameter we can find the volume of the hollow part of the sphere using the following formula
[tex]V_h = \frac{4}{3}\pi r^3[/tex]
The volume of the iron shell is the volume of the whole sphere subtracted by volume of the hollow part
[tex]V_s = V - V_h = \frac{4}{3}\pi 0.291^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.0246 - r^3)[/tex]
Let water density [tex]\rho_w = 1000 kg/m^3[/tex], then the buoyant force is the weight of water displaced by the shell
[tex]F_b = g\rho_w V = g1000\frac{4}{3}\pi(0.291^3) = g1000\frac{4}{3}\pi(0.0246)[/tex]
For the shell to almost completely submerged in water, its buoyant force equal to the weight of the shell
[tex]F_b = W_s[/tex]
[tex]g1000\frac{4}{3}\pi(0.0246) = g\rho_sV_s[/tex]
[tex]g1000\frac{4}{3}\pi(0.0246) = g7870\frac{4}{3}\pi(0.0246 - r^3)[/tex]
[tex]1000*0.0246 = 7870(0.0246 - r^3)[/tex]
[tex]24.6 = 193.602 - 7870r^3[/tex]
[tex]r^3 = (193.602 - 24.6)/7870 = 0.02147[/tex]
[tex]r = \sqrt[3]{0.02147} = 0.278 m[/tex]
So the inner diameter is 0.278*2 = 0.556 m
To find the inner diameter of the hollow sphere, one must take steps of calculating the total volume of the sphere, then the weight, the displaced water volume and finally the total internal volume. With this information, the inner diameter of the sphere can be worked out.
Explanation:To solve this physics problem, we need to find the inner diameter of a hollow spherical iron shell that is floating almost completely submerged in water. The student has provided the outer diameter of the shell (58.2 cm) and the density of iron (7.87 g/cm3). Using these values, we'll use some concepts from physics, specifically principles of buoyancy and formulas for volume and density.
The first step will be to calculate the total volume of the sphere using the outer diameter. This can be done with the formula for the volume of a sphere: V=(4/3)πR^3, where R is the radius, which is half the diameter.
Since the sphere is floating almost completely submerged, it's displacing a volume of water equivalent to its own weight. Now that we know the sphere's volume, we can calculate the sphere's weight using the density of iron.
The weight then can be used to calculate the volume of water displaced, which leads us to the total internal volume of the sphere (total volume - displaced volume). Then the inner diameter can be found by rearranging the formula of the volume of the sphere.
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Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them?A. 2 NB. 1/8 NC. 1/16 ND. 1/4 NE. 1/2 N
Answer:
The right option is C. 1/16 N
Explanation:
From coulomb's law,
F = kqq'/r²................ Equation 1
Where F = force between the charges, q = first charge, q' = second charge, r = distance of separation between the the charges, k = constant of proportionality.
qq' = Fr²/k............... Equation 2
Given: F = 1 N, r = 2 cm = 0.02 m
Substituting into equation 2, to get the value of the product of the charges in terms of k
qq' = 1×0.02²/k
qq' = 0.0004/k.................... Equation 3
When they are moved to a new separation of 8 cm,
Then r = 8 cm = 0.08 m
F = kqq'/0.08².................. Equation 4
Substituting the value of qq' in equation 3 into equation 4
F = k(0.0004k)/0.0064
F = 4/64
F = 1/6 N
Hence the electric force on each of the is 1/16 N
The right option is C. 1/16 N
A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pushed 2.5 m straight backwards in 1.65 s. He breaks the tackle and runs straight forward another 24 m in 5.2 s.
Part (a) Calculate the wide receiver's average velocity in the horizontal direction during the first interval, in meters per second.
Part (b) Calculate the wide receiver's average velocity in the horizontal direction during the second interval, in meters per second.
Part (c) Calculate the wide receiver's average velocity in the horizontal direction during the third interval, in meters per second.
Part (d) Calculate the wide receiver's average velocity in the horizontal direction for the entire motion, in meters per second
Answer:
a) v1 = 5.52m/s
b) v2 = -1.52m/s
c) v3 = 4.62m/s
d) vt = 3.85m/s
Explanation:
The velocity of the football wide receiver is his displacement per unit time.
Velocity v = (displacement d)/time t
v = d/t .....1
For each of the cases, equation 1 would be used to calculate the velocity.
a) v1 = d1/t1
d1= 16m
t1 = 2.9s
v1 = 16m/2.9s
v1 = 5.52m/s
b) v2 = d2/t2
d2 = -2.5m
t2 = 1.65s
v2 = -2.5/1.65
v2 = -1.52m/s
c) v3 = d3/t3
d3 = 24m
t3 = 5.2s
v3 = 24/5.2
v3 = 4.62m/s
d) vt = dt/tt
dt = 16m - 2.5m + 24m = 37.5m
tt = 2.9 + 1.65 + 5.2 = 9.75s
vt = 37.5/9.75
vt = 3.85m/s
A uniform, solid metal disk of mass 6.00 kg and diameter 29.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.28 N tangent to the rim of the disk to turn it by 3.32 ∘, thus twisting the wire. You now remove this force and release the disk from rest.
To calculate the angular velocity of the solid metal disk, we can use the equation for rotational motion: τ = Iα. We are given the torque applied to the disk (4.28 N) and the angle it is turned (3.32°). Using the formulas for moment of inertia and angular acceleration, we can calculate the final angular velocity.
Explanation:To calculate the angular velocity of the solid metal disk, we can use the equation for rotational motion:
τ = Iα
Where τ is the torque applied to the disk, I is the moment of inertia of the disk, and α is the angular acceleration of the disk.
We are given the torque applied to the disk (4.28 N) and the angle it is turned (3.32°). We can calculate the moment of inertia of the disk using the formula:
I = (1/2)mr^2
Where m is the mass of the disk and r is the radius of the disk.
Using the given diameter (29.0 cm), we can calculate the radius (14.5 cm or 0.145 m).
Plugging in the values into the formulas, we can solve for the angular acceleration:
α = τ / I
Finally, we can use the equation for angular motion to find the angular velocity:
ω = ω0 + αt
Since the disk is released from rest, the initial angular velocity (ω0) is 0. Plugging in the values, we can solve for the final angular velocity.
If you place a large number of electrons in one small spot on a conductor, they immediately redistribute themselves on the surface because they repel one another. What happens if you place a large number of electron in one small are on the surface of an insulator?
O They immediately redistribute themselves on the surface of the insulator because they repel one another.
O They stay in the area in which they were initially deposited for a relatively long time because they are not free to move about the insulator.
O They immediately redistribute themselves throughout the entire volume of the insulator because they repel one another.
Answer:
They stay in the area in which they were initially deposited for a relatively long time because they are not free to move about the insulator.
Explanation:
If we place large number of electrons in a small area of an insulating surface then the charges will remain in their place for a longer period of time because the charges do not get a path to flow despite the repulsion between them.
Insulators do not have conducting electrons free to carry the charges apart from each other whereas the metal have a sea of electrons free for the conduction of charges.
Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitude of q2=4.20 nC and is located at x=1.00 m, y=0.600 m, calculate the x and y components, Ex and Ey, of the electric field, E, in component form at the origin, (0,0).
Answer:
Ex= -23.8 N/C Ey = 74.3 N/C
Explanation:
As the electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.
So, we can first the field due to q1, as follows:
Due to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:
E₁ = k*(4.55 nC) / r₁²
If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:
E₁ₓ = 0 E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)
For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.
It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:
r₂² = 1²m² + (0.6)²m² = 1.36 m²
The magnitude of the electric field due to q2 can be found as follows:
E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)
Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.
In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:
E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ
the cosine of θ, is just, by definition, the opposite of x/r₂:
⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855
By the same token, sin θ can be obtained as follows:
sin θ = - (0.6 m / 1.17 m) = -0.513
⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)
⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)
The total x and y components due to both charges are just the sum of the components of Ex and Ey:
Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C
From (1) and (4), we can get Ey:
Ey = E₁y + E₂y = 88.6 N/C + (-14.3 N/C) =74.3 N/C
In this exercise we have to use the knowledge of electric field to calculate the components, in this way we have that:
[tex]E_x= -23.8 N/C \\ E_y = 74.3 N/C[/tex]
As the energetic force exist undeviating, and the energetic field, essentially, is just this energetic force for one charge, we can use the superposition standard to take the energetic field caused by two together charges at any time, as the other charge exist absent.
Due to q₁ exist negative, and situated ahead of the y point around which something revolves, the field on account of this charge will happen indicating upward, in this manner:
[tex]E_1 = k*(4.55 nC) / r_1^2\\ E_{1x} = 0 \\ E_{1y} = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C \\ r_2^2 = 1^2m^2 + (0.6)^2m^2 = 1.36 m^2\\ E_2 = k*q_2 / r^2_2 = 9*10^9*(4.2)*10^9 / 1.36 = 27.8 N/C\\ E_{2x} = E_2* cos\theta\\ E_{2y} = E_2*sin \theta\\ cos \theta = -0.855\\ sin \theta = -0.513\\ E_{2x} = 27.8 N/C * (-0.855) = -23.8 N/C \\ E_{2y} = 27.8 N/C * (-0.513) = -14.3 N/C [/tex]
The total x and y part on account of two together charges exist just the total of the part of Ex and Ey:
[tex]E_x = E_{1x} + E_{2x} = 0 + (-23.8 N/C) = -23.8 N/C\\ E_y = E_{1y} + E_{2y} = 88.6 N/C + (-14.3 N/C) =74.3 N/C [/tex]
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If the tension is 300 N, what is the difference in the speed of a wave traveling from one wire to the other?
Answer:
v₁ / v₂ = √μ ₂/μ ₁
Explanation:
The speed of a traveling wave on a string is given by
v = √ T /μ
Where v is the speed of the wave, T the tension and μ is the linear density of the string
Let's apply this equation to our case
v₁ = √ T / μ ₁
v2 = √ T /μ ₂
If the voltage of the two systems is the same
T = v₁² μ₁ = v₂² μ₂
We cleared
v₁² / v₂² = μ₂ / μ₁
v₁ / v₂ = √μ ₂/μ ₁
Please help
2. according to the velocity vs. time graph, what is happening to the velocity of the wheel As Time continues?
3. what is the physical significance of the slope of the graph?
4. in what unit is the slope measured?
5. what is the name commonly used to represent this rate?
6. what symbol would you use to represent this rate ?
7. write the general equation using only variables that describe the relationship between velocity and time for the will rolling down the incline from rest.
2) The velocity of the wheel is increasing
3) The slope is the rate of change of velocity
4) The unit of the slope is metres per second squared.
5) Acceleration
6) The symbol used for acceleration is [tex]a[/tex]
7) The relationship between velocity and time is [tex]v=gsin \theta t[/tex]
Explanation:
2)
The graph in the problem represents the velocity of the wheel as a function of the time.
As we can see from the graph, the velocity is increasing as the time passes. This means that the wheel is accelerating (its velocity is changing constantly)
3)
The slope of the graph represents the rate of change of the velocity.
Mathematically, it can be written as:
[tex]m=\frac{\Delta v}{\Delta t}[/tex]
where
m is the slope
[tex]\Delta v[/tex] is the change in velocity
[tex]\Delta t[/tex] is the time interval considered
As we see from the graph, the slope of the line is positive and constant: this means that the velocity is increasing at a constant rate.
4)
The unit of the slope can be determined starting by the units of the two variables involved.
On the y-axis, we have the velocity, which is measured in metre per second ([tex]m/s[/tex])
On the x-axis, we have the time, which is measured in seconds ([tex]s[/tex])
The slope is the ratio between these two quantities:
[tex]m=\frac{\Delta v}{\Delta t}[/tex]
Therefore, the units of the slope are
[tex]m=\frac{[m/s]}{[s]}=[m/s^2][/tex]
So, metres per second squared.
5)
The rate of change of velocity is also known as acceleration.
In fact, acceleration is defined as the ratio between the change in velocity and the change in time:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
By comparing with the formula of the slope in part 3), we see that the two equations are identical, therefore the acceleration corresponds to the slope of the graph.
6)
The symbol used to represent the acceleration is [tex]a[/tex]:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
7)
In any uniformly accelerated motion, the relationship between velocity and time is given by the following suvat equation:
[tex]v=u+at[/tex]
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
In this problem, the wheel starts from rest, so
u = 0
Also, for an object rolling down a ramp, the acceleration is given by
[tex]a=g sin \theta[/tex]
where g is the acceleration of gravity and [tex]\theta[/tex] is the angle of the ramp. Substituting, we find the final expression of the velocity:
[tex]v=gsin \theta t[/tex]
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On a linear velocity vs. time graph, the velocity of the wheel is changing at a constant rate as time continues. The slope of the graph represents the acceleration. The equation describing the relationship between velocity and time is V = V0 + at.
Explanation:In a velocity vs. time graph, if the graph is linear (i.e., a straight line), it indicates that the velocity is changing at a constant rate as time continues. The slope of the graph represents the acceleration of the wheel. The slope is measured in units of velocity divided by time, typically meters per second per second (m/s^2). The commonly used symbol to represent this rate is 'a', which stands for acceleration. The general equation describing the relationship between velocity and time for the wheel rolling down the incline from rest is given by V = V0 + at, where V is the velocity at a given time, V0 is the initial velocity (which is zero in this case), 'a' is the acceleration, and 't' is the time.
Mustang Sally just finished restoring her 1965 Ford Mustang car. To save money, she did not get a new battery. When she tries to start the car, she discovers that the battery is dead and she needs a jump start. While unhooking the jumper cables, the positive and negative cables almost touch and a spark jumps between the ends of the cables. This spark is caused by the movement of electrons through the air between the battery terminals. In what direction are the electrons traveling?
Answer:
the electron jump goes from the negative terminal to the positive terminal
Explanation:
In a battery the energy is produced by a chemical reaction, which accumulates electrons in a terminal, called a negative terminal, but by convention the current goes from the positive to the negative terminal.
We can see from this explanation that the negative wire is the one with an excess is electrons, so the electron jump goes from the negative terminal to the positive terminal
A spring with a spring constant of 23.6 N/m has a mass attached that exerts a force of 6.28 Newtons. What is the displacement?
Answer:
26.6 cm
Explanation:
We are given that
Spring constant=23.6 N/m
Force exert=6.28 N
We have to find the displacement.
We know that Hooke's law
[tex]F=kx[/tex]
Where k= Spring constant
x=Displacement
Using the formula
Then, we get
[tex]6.28=23.6 x[/tex]
[tex]x=\frac{6.28}{23.6}=0.266m[/tex]
We know that 1 m=100 cm
[tex]x=0.266\times 100=26.6 cm[/tex]
Hence, the displacement =26.6 cm
A 15 g marble moves to the right at 3.5 m/s and makes an elastic head-on collision with a 22 g marble. The final velocity of the 15 g marble is 5.4 m/s to the left, and the final velocity of the 22 g marble is 2.0 m/s to the right.
What is the initial velocity of the 22 g marble?
a) 5.3 m/s to the left
b) 5.3 m/s to the right
c) 4.1 m/s to the right
d) 4.1 m/s to the left
Answer:
[tex]v_2[/tex]≅-4.1 m/s (-ve for left)
[tex]v_2\\[/tex] ≅4.1 m/s (To the left)
Explanation:
According to the conservation of momentum:
Momentum before collision=Momentum After Collision
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
Where:
[tex]m_1[/tex] is the 15g marble
[tex]m_2[/tex] is the 22 g marble
[tex]v_1[/tex] is the velocity of 15g marble before collision
[tex]v_2[/tex] is the velocity of 22g marble before collision
[tex]v'_1[/tex] is the velocity of 15g marble after collision
[tex]v'_2[/tex] is the velocity of 22g marble after collision
Note: -ve sign for left, +ve sign for right
We have to calculate v_2:
Above equation on rearranging will become:
[tex]v_2=\frac{m_1v'_1+m_2v'_2-m_1v_1}{m_2} \\v_2=\frac{(0.015kg)(-5.4m/s)+(0.022kg)(2m/s)-(0.015kg)(3.5)}{0.022kg} \\v_2=-4.06 m/s\\[/tex]
[tex]v_2[/tex]≅-4.1 m/s (-ve for left)
[tex]v_2\\[/tex] ≅4.1 m/s (To the left)
Final answer:
The initial velocity of the 22 g marble before the elastic collision can be calculated using the conservation of momentum. After setting up the momentum equation with the given masses and velocities and solving for the unknown initial velocity, it is found to be 4.1 m/s to the left. So the correct option is d.
Explanation:
To calculate the initial velocity of the 22 g marble in an elastic collision, we use the conservation of momentum and conservation of kinetic energy principles. The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the system can be calculated by the sum of the momentum of each marble before they collide. Since we know the mass and final velocity of each marble, and the initial velocity of the 15 g marble, we can set up the equation:
(Mass of 15 g marble) × (Initial velocity of 15 g marble) + (Mass of 22 g marble) × (Initial velocity of 22 g marble) = (Mass of 15 g marble) × (Final velocity of 15 g marble) + (Mass of 22 g marble) × (Final velocity of 22 g marble)
Plugging in the known values and solving for the initial velocity of the 22 g marble, we get:
(0.015 kg) × (3.5 m/s) + (0.022 kg) × (Initial velocity of 22 g marble) = (0.015 kg) × (-5.4 m/s) + (0.022 kg) × (2.0 m/s)
0.0525 kg·m/s + (0.022 kg) × (Initial velocity of 22 g marble) = -0.081 kg·m/s + 0.044 kg·m/s
0.0525 kg·m/s + (0.022 kg) × (Initial velocity of 22 g marble) = -0.037 kg·m/s
(0.022 kg) × (Initial velocity of 22 g marble) = -0.037 kg·m/s - 0.0525 kg·m/s
(0.022 kg) × (Initial velocity of 22 g marble) = -0.0895 kg·m/s
Initial velocity of 22 g marble = -0.0895 kg·m/s / 0.022 kg
Initial velocity of 22 g marble = -4.068 m/s (to the left, since it's negative)
The initial velocity of the 22 g marble is 4.1 m/s to the left, which corresponds to option (d).
A 1300 kgkg car drives around a flat 200-mm-diameter circular track at 40 m/sm/s . Part A What is the magnitude of the net force on the car? Express your answer to two significant figures and include the appropriate units. FnetFnet = nothingnothing
Answer:
[tex]F_C=20800\ N[/tex]
Explanation:
Given:
mass of the car, [tex]m=1300\ kg[/tex]diameter of the track, [tex]d=200\ m[/tex]speed of the car, [tex]v=40\ m.s^{-1}[/tex]During the motion of the car on a flat circular track there acts a force which pushes the car inwards to comply in the circular motion. This force is called centripetal force which is generated due to the frictional force between the road and the tyres of the car.
This centripetal force is given as:
[tex]F_C=m.\frac{v^2}{r}[/tex]
here: r = radius of the track = [tex]\frac{d}{2}[/tex]
[tex]F_C=1300\times \frac{40^2}{100}[/tex]
[tex]F_C=20800\ N[/tex] is the force which makes the car turn inwards going with the given uniform speed.
Final answer:
The net force on a 1300 kg car driving around a 200-m diameter track at 40 m/s is calculated using the centripetal force formula, resulting in a force of 20,800 N directed towards the center of the track.
Explanation:
The question asks to calculate the magnitude of the net force acting on a 1300 kg car driving around a 200-m diameter circular track at a speed of 40 m/s.
First, we convert the diameter to radius by dividing by 2, giving a radius (r) of 100 m. To find the net force (Fnet), we use the formula for centripetal force: Fnet = m * (v² / r), where m is mass, v is velocity, and r is radius.
Substituting the given values:
Mass (m) = 1300 kg
Velocity (v) = 40 m/s
Radius (r) = 100 m
This yields:
Fnet = 1300 kg * (40 m/s)² / 100 m
Calculating this, we get:
Fnet = 1300 kg * 1600 m²/s² / 100 m
Fnet = 1300 kg * 16 m/s²
The net force is 20,800 N directed towards the center of the circular track.
Which of the following could be reasonable explanations for how a piece of invisible tape gets charged? Select all that apply.
(1) Protons are pulled out of nuclei in one tape and transferred to another tape.
(2) Charged molecular fragments are broken off one tape and transferred to another.
(3) Electrons are pulled out of molecules in one tape and transferred to another tape.
(4) Neutrons are pulled out of nuclei in one tape and transferred to another tape.
Answer:
2 and 3
Explanation:
The right answer is the option 2 and 3,
This is all about the electrons transfer from one material to the other material.
For example if the electrons in the valence shell of one material are loosely attached then the other material's atoms try to take those electron to complete their shells and that is how the charges transfers from one another.
And it could also be happen as in option 2.
Your toaster has a power cord with a resistance of 2.2x10^−2 Ω connected in series with a 9.7 Ω nichrome heating element. The potential difference between the terminals of the toaster is 120 V.
1. How much power is dissipated in the power cord?
Pcord =_______ W
2. How much power is dissipated in the heating element?
Phe= ________ kW
The power dissipated in the power cord is approximately 65.5 kW, and in the heating element, it's around 1.49 kW.
Explanation:The question pertains to the power dissipated in a circuit. We will use the formula Power (P) = Voltage (V) ^ 2 / Resistance (R) for each component of the circuit to find the desired values. Given a potential difference of 120V, we have:
For power cord, resistance (Rcord) = 2.2x10^-2 Ω, so the power dissipated in the power cord (Pcord) = 120^2 / 2.2x10^-2 = 65455.45 W or approximately 65.5 kW.For the heating element, resistance (Rhe) = 9.7 Ω, so the power dissipated in the heating element (Phe) = 120^2 / 9.7 = 1485.567 W or approximately 1.49 kW.Learn more about Power Dissipation here:https://brainly.com/question/34077222
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a proton moving to the right at 3.8 enters a region where 56 electic field points ot the left, how far will the proton get and describe its motion
Answer: 0.07meter
Pointing to the left.
Explanation:
The Electric potential V of the proton is 3.8v. while the Electric Field strength is 56V/m. We first calculate the distance r the proton would travel
E= V/r
r = V/E
r = 3.8V/(56V/m)
r =0.07m.
The proton which is positively charged will move in the direction of the Electric Field that is to the left.
When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The force on the proton can be calculated using the formula F = q*E. The distance the proton will get before coming to a stop can be calculated using the formula d = (1/2)at^2.
Explanation:When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The magnitude of the force can be determined using the formula:
F = q*E
Where F is the force, q is the charge of the proton, and E is the magnitude of the electric field. In this case, the force on the proton is given by:
F = (1.6 x 10^-19 C)(56 N/C) = -8.96 x 10^-18 N
The negative sign indicates that the force is in the opposite direction of the electric field. Since the proton is moving to the right, it will slow down and eventually come to a stop. The distance the proton will get can be calculated using the formula:
d = (1/2)at^2
Where d is the distance, a is the acceleration, and t is the time. Since the proton starts from rest, its initial velocity is zero. The acceleration can be calculated using the formula:
a = F/m
Where a is the acceleration, F is the force, and m is the mass of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg. Plugging in the values, we get:
a = (-8.96 x 10^-18 N) / (1.67 x 10^-27 kg) = -5.38 x 10^8 m/s^2
Substituting this value of acceleration and the given initial velocity into the distance formula, we can calculate the distance the proton will get:
d = (1/2)(-5.38 x 10^8 m/s^2)(t^2)
Where t is the time for which the proton travels. The time can be calculated using the formula:
t = v/a
Where t is the time, v is the initial velocity, and a is the acceleration. Plugging in the values, we get:
t = (3.8 m/s) / (-5.38 x 10^8 m/s^2) = -7.08 x 10^-9 s
Since the time cannot be negative, we take the magnitude of the time:
t = 7.08 x 10^-9 s
Substituting this value of time into the distance formula, we get:
d = (1/2)(-5.38 x 10^8 m/s^2)((7.08 x 10^-9 s)^2) = 1.05 x 10^-15 m
Therefore, the proton will get a distance of 1.05 x 10^-15 meters before coming to a stop.
A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carrying charge of -80 μC is suspended by an uncharged,nonconducting thread .30m long. The thread is attached to thewall and the ball hans in equilibrium in electric and gravitationalfields. Electric force on ball has magnitude of .032N
Calculate magnitude of electric field at ball's location dueto charged wall and show direction on x, y coordinate axes
Determine perpendicular distance from wall to center ofball
The string is cut
Calculate magnitude of resulting acceleration of ball and itsdirection
Describe resulting path of ball
Answer:
a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
[tex]F_{e}[/tex] - [tex]T_{x}[/tex] = m a
Axis y
[tex]T_{y}[/tex] - W = 0
Initially the system is in equilibrium, so zero acceleration
Fe = [tex]T_{x}[/tex]
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ = [tex]T_{x}[/tex] / T
T_{y} = cos θ
[tex]T_{x}[/tex] = T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
[tex]F_{e}[/tex] = q E
E = [tex]F_{e}[/tex] / q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18 ⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
[tex]F_{e}[/tex] = m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m [tex]a_{y}[/tex]
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes
The magnitude of the electric field is 400,000 N/C directed to the right. The perpendicular distance from the wall to the ball is 0.0417 m. After the string is cut, the ball will accelerate at 10.29 m/s² and follow a parabolic trajectory.
To solve this problem, we will break it down into several steps.
1. Calculate the Electric Field
The electric force on the ball is given by:
Fe = qE
Where Fe is the electric force, q is the charge of the ball, and E is the electric field.
Given: Fe = 0.032 N and q = -80 μC = -80 × 10⁻⁶ C
Rearranging the formula to solve for E:
E = Fe / q = 0.032 N / (-80 × 10⁻⁶ C)
E = -400,000 N/C
So, the magnitude of the electric field is 400,000 N/C (since we take the absolute value) and it is directed to the right on the x-axis as the ball has a negative charge.
2. Determine the Perpendicular Distance from the Wall
The ball is in equilibrium under electric and gravitational fields. To find the distance from the wall, we need to consider the forces and the angle of the thread.
The electric field is horizontal (along x-axis) and the gravitational force is vertical (along y-axis). The thread is at an angle of 8° with the vertical.
tan(θ) = Fe / Fg
Where θ = 8°, Fe = 0.032 N, and Fg = mg = 0.01 kg × 9.8 m/s² = 0.098 N
tan(8°) = 0.032 / 0.098 ≈ 0.327
The perpendicular distance from the wall (d) is:
d = L sin(θ) = 0.30 m × sin(8°) ≈ 0.30 m × 0.139 ≈ 0.0417 m
3. Calculate the Resulting Acceleration When the String is Cut
When the string is cut, only the electric field will act on the ball horizontally and gravity will act vertically.
Horizontally (ax): ax = Fe / m = 0.032 N / 0.01 kg = 3.2 m/s²
Vertically (ay): ay = g = 9.8 m/s²
The magnitude of the resulting acceleration (a) is:
[tex]\[a = \sqrt{a_x^2 + a_y^2} = \sqrt{(3.2)^2 + (9.8)^2} \approx 10.29 \, \text{m/s}^2\][/tex]
4. Describe the Resulting Path of the Ball
The ball will follow a parabolic path due to the combined effects of the horizontal electric field and vertical gravitational acceleration, similar to projectile motion.
Water flows through a pipe of radius 24 cm. The pipe narrows to 8 cm radius. Compared to the 24 cm radius section, the velocity of the water in the 8 cm radius section is: g
Answer:
The velocity of the water at the 8 cm section is 9 times the velocity of the water at 24 cm section.
Explanation:
For this question, we use the equation of continuity of an non-compressible fluid in a pipe
[tex]A_{1} v_{1} =A_{2} v_{2}\\\\\pi r_{1}^{2}v^{2} = \pi r_{2}^{2}v^{2} \\\\\pi (24)^{2}v_{1} ^{2} = \pi (8)^{2}v_{2} ^{2}\\\\576v_{1} = 64v_{2}\\\\v_{2} = 9 v_{1}[/tex]
The velocity of the water at the 8 cm section is 9 times the velocity of the water at 24 cm section.
As an ice skater begins a spin, his angular speed is 3.17 . After pulling in his arms, his angular speed increases to 5.96 . Find the ratio of the skater's final moment of inertia to his initial moment of inertia.
Answer:
The ratio is 0.53
Explanation:
There aren't external torques on the ice skater because we can assume ice surface frictionless and there're not external forces, that implies angular moment is conserved so it has the same value before and after him pulling his arms. Angular momentum is the product between angular velocity (ω) and moment of inertia (I).
The initial angular moment is:
[tex]L=I_{i}\omega_{i} [/tex] (1)
and final angular moment is:
[tex] L=I_{f}\omega_{f} [/tex] (2)
Because conservation of angular moment we can equate (1) and (2):
[tex]I_{i}\omega_{i}=I_{f}\omega_{f} [/tex]
rearranging the expression:
[tex]\frac{I_{f}}{I_{i}}=\frac{\omega_{i}}{\omega_{f}} [/tex]
So, the ratio of the skater's final moment of inertia to his initial moment of inertia is:
[tex] \frac{I_{f}}{I_{i}}=\frac{3.17\frac{rad}{s}}{5.96\frac{rad}{s}}[/tex]
[tex] \frac{I_{f}}{I_{i}}=0.53[/tex]
The required ratio of of the skater's final moment of inertia to his initial moment of inertia is 0.53 : 1.
Given data:
The initial angular speed of the ice skater is, [tex]\omega = 3.17 \;\rm rad/s[/tex].
The final angular speed of the ice skater is, [tex]\omega ' =5.96 \;\rm rad/s[/tex].
In this problem, we can apply the conservation of the angular momentum, which says that in the absence of external torque, the initial angular momentum is equal to the final angular momentum.
Initial angular momentum = final angular momentum
[tex]L = L'\\I \times \omega = I' \times \omega'[/tex]
Here,
I is the initial moment of inertia.
I' is the final moment of inertia.
Solving as,
[tex]\dfrac{I'}{I} = \dfrac{ \omega}{ \omega'} \\\\\dfrac{I'}{I} = \dfrac{3.17}{ 5.96}\\\\\dfrac{I'}{I} = 0.53[/tex]
Thus, the required ratio of of the skater's final moment of inertia to his initial moment of inertia is 0.53 : 1.
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Galileo, in his Two New Sciences, states that "for elevations (angles of projection) which exceed or fall short of 45 degrees by equal amounts, the ranges are equal. Prove it.
Answer:
Please refer to the attachment below.
Explanation:
Please refer to the attachment below for explanation.
An object has a -8 C charge. The object breaks into two pieces. One piece has a charge of -2 C. What is the exact charge (in C) of the other piece
Answer:
-6C
Explanation:
We are given that
Charge on an object,Q=-8C
Charge on one piece of object=[tex]q_1=-2C[/tex]
We have to find the exact charge of the other piece.
Let q be the charge of second piece of object.
By conservation of charge
Initial charge=Final charge
[tex]Q=q_1+q[/tex]
Substitute the values then we get
[tex]-8C=-2C+q[/tex]
[tex]q=-8C+2C[/tex]
[tex]q=-6C[/tex]
Hence, the exact charge of the other piece=-6C
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Final answer:
The electric field strength at the midpoint between the two rings is 46.08 N/C, and at the center of the left ring it is 288 N/C.
Explanation:
To find the electric field strength at various points between two charged rings, we can use the formula for electric field:
E = k*(Q/r^2)
a) At the midpoint between the two rings, the distance from each ring is 12.5 cm. The electric field due to each ring is:E1 = k*(Q/r^2) = (9 x 10^9 N m^2/C^2)*(20 x 10^-9 C)/(0.125 m)^2 = 23.04 N/CE2 = k*(Q/r^2) = (9 x 10^9 N m^2/C^2)*(20 x 10^-9 C)/(0.125 m)^2 = 23.04 N/CThe electric field at the midpoint is the sum of the electric fields due to each ring:E_total = E1 + E2 = 23.04 N/C + 23.04 N/C = 46.08 N/Cb) At the center of the left ring, the electric field due to that ring is:E = k*(Q/r^2) = (9 x 10^9 N m^2/C^2)*(20 x 10^-9 C)/(0.05 m)^2 = 288 N/C