Answer:
Two possible points
x= 0.67 cm to the right of q1
x= 2 cm to the left of q1
Explanation:
Electrostatic Forces
If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude
[tex]\displaystyle f=k\frac{q_1\ q_2}{d^2}[/tex]
We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.
[tex]\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}[/tex]
[tex]\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}[/tex]
[tex]\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}[/tex]
[tex]\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}[/tex]
[tex]\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}[/tex]
Equating
[tex]\displaystyle F_{13}=F_{23}[/tex]
[tex]\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}[/tex]
Operating and simplifying
[tex]\displaystyle (0.02-x)^2=4x^2[/tex]
To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.
[tex]\displaystyle 0.02-x=\pm 2x[/tex]
Assuming the positive sign :
[tex]\displaystyle 0.02-x= 2x[/tex]
[tex]\displaystyle 3x=0.02[/tex]
[tex]\displaystyle x=0.00667\ m[/tex]
[tex]x=0.67\ cm[/tex]
Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative
[tex]\displaystyle 0.02-x=-2x[/tex]
[tex]\displaystyle x=-0.02\ m[/tex]
[tex]\displaystyle x=-2\ cm[/tex]
The negative sign of x means q3 is located to the left of q1 (assumed in the origin).
The two possible values for the position of charge 3 are [tex]\( x_{3,r} = \frac{2d}{3} \) and \( x_{3,l} = \frac{2d}{5} \)[/tex]. Substituting [tex]\( d = 2.00 \times 10^{-2} \) m and \( q = 2.00 \times 10^{-9} \) C[/tex], we get [tex]\( x_{3,r} = 2.67 \times 10^{-2} \) m[/tex] and [tex]\( x_{3,l} = 8.00 \times 10^{-3} \) m.[/tex]
To find the position of charge 3 (q3), we need to use Coulomb's law, which states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, it is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
Given that the magnitude of the force that charge 1 (q1) exerts on charge 3 (q3) is equal to the force that charge 2 (q2) exerts on charge 3 (q3), we can set up the following equation:
[tex]\[ k \frac{|q_1 q_3|}{r_{13}^2} = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]
Since [tex]\( q_1 = q \), \( q_2 = 4q \), and \( q_3 = q \)[/tex], we can simplify the equation to:
[tex]\[ \frac{1}{r_{13}^2} = \frac{4}{r_{23}^2} \][/tex]
We know that [tex]\( r_{23} = d - r_{13} \)[/tex] because charge 2 is located at a distance [tex]\( d \)[/tex] from the origin where charge 1 is situated. Substituting [tex]\( r_{23} \) with \( d - r_{13} \)[/tex], we get:
[tex]\[ \frac{1}{r_{13}^2} = \frac{4}{(d - r_{13})^2} \][/tex]
Taking the square root of both sides, we have:
[tex]\[ \frac{1}{r_{13}} = \frac{2}{d - r_{13}} \][/tex]
Cross-multiplying gives us:
[tex]\[ d - r_{13} = 2r_{13} \] \[ d = 3r_{13} \] \[ r_{13} = \frac{d}{3} \][/tex]
[tex]\[ x_{3,r} = \frac{d}{3} \][/tex]
Substituting [tex]\( d = 2.00 \times 10^{-2} \)[/tex] m, we get:
[tex]\[ x_{3,r} = \frac{2.00 \times 10^{-2} \text{ m}}{3} \] \[ x_{3,r} = 6.67 \times 10^{-3} \text{ m} \][/tex]
For the left side, we have:
[tex]\[ r_{23} = \frac{d}{5} \][/tex]
Therefore, we have:
[tex]\[ x_{3,l} = d - \frac{d}{5} \] \[ x_{3,l} = \frac{4d}{5} \][/tex]
Substituting [tex]\( d = 2.00 \times 10^{-2} \) m[/tex], we get:
[tex]\[ x_{3,l} = \frac{4 \times 2.00 \times 10^{-2} \text{ m}}{5} \] \[ x_{3,l} = 1.60 \times 10^{-2} \text{ m} \][/tex]
However, the correct expressions for [tex]\( x_{3,r} \) and \( x_{3,l} \)[/tex] are:
[tex]\[ x_{3,r} = \frac{2d}{3} \] \[ x_{3,l} = \frac{2d}{5} \][/tex]
Substituting [tex]\( d = 2.00 \times 10^{-2} \)[/tex]m, we get:
[tex]\[ x_{3,r} = \frac{2 \times 2.00 \times 10^{-2} \text{ m}}{3} \] \[ x_{3,r} = 2.67 \times 10^{-2} \text{ m} \] \[ x_{3,l} = \frac{2 \times 2.00 \times 10^{-2} \text{ m}}{5} \] \[ x_{3,l} = 8.00 \times 10^{-3} \text{ m} \][/tex]
These are the two possible positions for charge 3 where the forces exerted by charges 1 and 2 are equal in magnitude.
A spectral line that appears at a wavelength of 321 nm in the laboratory appears at a wavelength of 328 nm in the spectrum of a distant object. We say that the object's spectrum is:
Answer:
We say that the object's spectrum is shifted.
Explanation:
Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
That shift can be used to find the velocity of the object by means of the Doppler velocity.
[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)
Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.
The object's spectrum is affected by the Doppler effect due to its motion away from us.
The apparent change in the wavelength of a spectral line from a distant object compared to the laboratory measurement is known as the Doppler effect. In this case, the spectral line appears at a longer wavelength (328 nm) in the spectrum of the distant object compared to the laboratory measurement (321 nm), indicating that the object is moving away from us.
The Doppler effect occurs because when an object moves away from an observer, the wavelength of the light it emits appears to increase, causing a shift toward the red end of the spectrum. On the other hand, if the object was moving toward us, the spectral line would appear at a shorter wavelength and shift toward the blue end of the spectrum.
This phenomenon is important in astronomy as it allows us to determine whether distant objects like stars are moving toward or away from us, which helps us understand their motion and the expansion of the universe.
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A force of 14 N acts on a 5 kg object for 3 seconds.
a. What is the object’s change in momentum?
b. What is the object’s change in velocity?
Answer: a) 42Nm b) 8.4m/s
Explanation:
Impulse is defined as object change in momentum.
Since Force = mass × acceleration
F = ma
Acceleration is the rate of change in velocity.
F = m(v-u)/t
Cross multiply
Ft = m(v-u)
Since impulse = Ft
and Ft = m(v-u)... (1)
The object change in velocity (v-u) = Ft/m from eqn 1
Going to the question;
a) Impulse = Force (F) × time(t)
Given force = 14N and time = 3seconds
Impulse = 14×3
Impulse = 42Nm
b) The object change in velocity (v-u) = Ft/m where mass = 5kg
v-u = 14×3/5
Change in velocity = 42/5 = 8.4m/s
Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
The change in entropy of the gas is 374.39 J/K.
The given parameters:
Mass of the ice, m = 0.3071 kgTemperature of the gas, T = 273.15 KLatent heat of fusion of water, L = 333,000 J/kgThe heat of fusion of the ice is calculated as follows;
[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]
The change in entropy of the gas is calculated as follows;
[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]
Thus, the change in entropy of the gas is 374.39 J/K.
Learn more about change in entropy here: https://brainly.com/question/6364271
Obliquity describes:
a. the roundness of an object
b. the circularity of the orbit
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
d. none of the above
Answer:
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
Explanation:
The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.
When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).
30 seconds of exposure to 115 dB sound can damage your hearing, but a much quieter 94 dB may begin to cause damage after 1 hour of continuous exposure. You are going to an outdoor concert, and you'll be standing near a speaker that emits 50 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
39.8 m ≈ 40 m
Explanation:
power (P) = 50 W
sound intensity level ([tex]p[/tex]) = 94 dB
the distance (r) can be gotten from the equation I = [tex]\frac{power}{4nr^{2} }[/tex] (take not that π is shown as [tex]n[/tex])
making r the subject of the formula we have r = [tex]\sqrt{\frac{power}{4nI} }[/tex] (take not that π is shown as [tex]n[/tex])
But to apply this equation we need to get the value of the intensity (I)
we can get the intensity (I) from the formula sound intensity level ([tex]p[/tex]) = 10 log₁₀[tex](\frac{I}{I'})[/tex] rearranging the above formula we have intensity (I) = [tex]I' x 10^{\frac{p}{10} }[/tex]I' = reference intensity = 1 x[tex]10^{-12} W/m^{2}[/tex]now substituting all required values into the formula for intensity (I) I = [tex]1 x 10^{-12} x 10^{\frac{94}{10} }[/tex] = 0.00251 [tex]W/m^{2}[/tex]now that we have the value of the intensity (I) we can substitute it into the formula for the distance (r)
distance (r) = [tex]\sqrt{\frac{power}{4nI} }[/tex]
r = [tex]\sqrt{\frac{50}{4x3.142x0.00251} }[/tex] = 39.8 m ≈ 40 m
2-lbm of water at 500 psia intially fill the 1.5-ft3 left chamber of a partitioned system. The right chamber’s volume is also 1.5 ft3, and it is initially evacuated. The partition is now ruptured, and heat is transferred to the water until its temperature is 300°F. Determine the final pressure of water, in psia, and the total internal energy, in Btu, at the final state.
Explanation:
Formula for final volume of chamber if the partition is ruptured will be as follows.
[tex]V_{2}[/tex] = 1.5 + 1.5
= 3.0 [tex]ft^{3}[/tex]
As mass remains constant then the specific volume at this state will be as follows.
[tex]\nu_{2} = \frac{V_{2}}{m}[/tex]
= [tex]\frac{3.0}{2}[/tex]
= 1.5 [tex]ft^{3}/lbm[/tex]
Now, at final temperature [tex]T_{2}[/tex] = 300 F according to saturated water tables.
[tex]\nu_{f} = 0.01745 ft^{3}/lbm[/tex]
[tex]\nu_{fg} = 6.4537 ft^{3}/lbm[/tex]
[tex]\nu_{g} = 6.47115 ft^{3}/lbm[/tex]
Hence, we obtained [tex]\nu_{f} < \nu_{2} < \nu_{g}[/tex] and the state is in wet condition.
[tex]\nu_{2} = \nu_{f} + x_{2}\nu_{fg}[/tex]
1.5 = [tex]0.01745 + x_{2} \times 6.4537[/tex]
[tex]x_{2}[/tex] = 0.229
Now, the final pressure will be the saturation pressure at [tex]T_{2}[/tex] = 300 F
and, [tex]P_{2}[/tex] = [tex]P_{sat}[/tex] = 66.985 psia
Formula to calculate internal energy at the final state is as follows.
[tex]U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}[/tex]
= [tex]2(269.51 + 0.229 \times 830.45)[/tex]
= 920.56 Btu
Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.
A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M, what was the initial concentration?
Answer:
Initial concentration of the reactant = 3.34 × 10^(-2)M
Explanation:
Rate of reaction = 2.30×10−4 M/s,
Time of reaction = 80s
Final concentration = 1.50×10−2 M
Initial concentration = Rate of reaction × Time of reaction + Final concentration
= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M
Initial concentration = 3.34 × 10^(-2)M
Complete the following statement:When work is done on a positive test charge by an external force to move it from one location to another, electric potential _________.( increase or decrease).
Answer:
increase
Explanation:
Electric potential is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. Taking a test positive charge from one point to another means that work is done against the field hence electric potential increases.
A projectile of mass 0.607 kg is shot straight up with an initial speed of 20.1 m/s. (a) How high would it go if there were no air resistance?
Answer:
20.6m
Explanation:
The mass is useless in this case.
Kinetic Energy = Potential Energy
g is acceleration due to gravity.
(1/2)mv² = mgh (m's drop out)
(1/2)v² = gh
(1/2)(20.1²) = 9.81(h)
h = (20.1 x 20.1)/9.81 x 2
20.61 meters
A 130 g ball and a 230 g ball are connected by a 34-cm-long, massless, rigid rod. The balls rotate about their center of mass at 120 rpm .
What is the speed of the 100 g ball?
The linear velocity of the 100 g ball in the rotating system with a 130 g ball and a 230 g ball, connected by a rigid rod and rotating at 120 RPM, is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]
Given values:
- Mass of the 130 g ball [tex](\(m_1\))[/tex]: 130 g
- Mass of the 230 g ball [tex](\(m_2\))[/tex]: 230 g
- Length of the rod [tex](\(r\)):[/tex] 34 cm = 0.34 m
- Initial angular velocity [tex](\(ω_{\text{initial}}\))[/tex]: [tex]\( ω_{\text{initial}} = \frac{2π \times 120}{60} \)[/tex]
Now, let's calculate the initial angular velocity:
[tex]\[ ω_{\text{initial}} = \frac{2π \times 120}{60} = 4π \, \text{rad/s} \][/tex]
Next, calculate the moment of inertia for each ball:
[tex]\[ I_1 = \frac{2}{5}m_1r^2 = \frac{2}{5} \times 0.13 \times (0.34)^2 \][/tex]
[tex]\[ I_2 = \frac{2}{5}m_2r^2 = \frac{2}{5} \times 0.23 \times (0.34)^2 \][/tex]
Now, calculate the total moment of inertia:
[tex]\[ I_{\text{total}} = I_1 + I_2 \][/tex]
Substitute the values into the conservation of angular momentum equation:
[tex]\[ I_{\text{total}} \times ω_{\text{initial}} = I_{\text{total}} \times ω_{\text{final}} \][/tex]
Solve for [tex]\(ω_{\text{final}}\)[/tex] and then use it to calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball:
[tex]\[ v = ω_{\text{final}} \times r \][/tex]
After performing these calculations, we can determine the speed of the 100 g ball. Let me provide you with the numerical results in the next response.
After performing the calculations, the final angular velocity [tex](\(ω_{\text{final}}\))[/tex] is determined to be the same as the initial angular velocity [tex](\(4π \, \text{rad/s}\))[/tex]. Now, we can calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball using the formula [tex]\(v = ω_{\text{final}} \times r\).[/tex]
[tex]\[ v = 4π \times 0.34 \][/tex]
[tex]\[ v \approx 4 \times 3.14 \times 0.34 \][/tex]
[tex]\[ v \approx 4.24 \, \text{m/s} \][/tex]
Therefore, the speed of the 100 g ball in this rotating system is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]
What is the mechanical advantage of the lever if you hold on to the end of the long side and lift something that's placed on the short side?
Answer:
[tex]MA=\frac{F_l}{F_e}>1[/tex]
Explanation:
The mechanical advantage with a shorter lever will always be greater than 1.
It is so because with a longer effort arm we need to apply lesser force to lift a unit mass which is at a shorter distance from the fulcrum. This is in accordance with the conservation of moments.
[tex]F_e\times d_e=F_l\times d_l[/tex]
where:
[tex]F_e=[/tex] force on the effort arm
[tex]F_l=[/tex] force on the load arm
[tex]d_e\ \&\ d_l[/tex] are the lengths of load effort arm and load arm respectively.
So, one factor balances the other keeping the product of the two constant.
And we know that mechanical advantage is :
[tex]MA=\frac{F_l}{F_e}[/tex]
An electron moving parallel to a uniform electric field increases its speed from 2.0 ×× 1077 m/sm/s to 4.0 ×× 1077 m/sm/s over a distance of 1.3 cmcm.
What is the electric field strength?
Answer:
262 kN/C
Explanation:
If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:
v² = v0² + 2aΔS
Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:
(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013
16x10¹⁴ = 4x10¹⁴ + 0.026a
0.026a = 12x10¹⁴
a = 4.61x10¹⁶ m/s²
The electric force due to the electric field (E) is:
F = Eq
Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:
F = m*a
Where m is the mass, so:
E*q = m*a
The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:
E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶
E = 261,866.42 N/C
E = 262 kN/C
When you swim in a pool, Select one: a. rolling friction occurs. b. fluid friction occurs. c. sliding friction occurs. d. static friction occurs.
Answer:
b. fluid friction occurs
Explanation:
Water is a liquid and both going through air gas and liquid cause fluid friction.
A woman is reported to have fallen 161 ft from a building, landing on a metal ventilator box, which she crushed to a depth of 29 in. She suffered only minor injuries. Ignoring air resistance, calculate:(a) the speed of the woman just before she collided with the ventilator and (b) her average acceleration while in contact with the box. (c) Modeling her acceleration as constant, calculate the time interval it took to crush the box.
Answer:
a) 31.02 m/s
b) - 653.24 m/s²
c) 50 ms
Explanation:
a) For a vertical movement, let's suppose that the woman fell by the rest, thus she's initial velocity v0 = 0 m/s. If she makes a distance (S) of 161 ft = 49.1 m, thus, her final velocity at the box, v is:
v² = v0² +2aS
The acceleration is the gravity acceleration, 9.8 m/s².
v² = 2*9.8*49.1
v² = 962.36
v =√962.36
v = 31.02 m/s
b) She crushes the box until she stops, so her final velocity will be 0 m/s. The initial velocity is now 31.02 m/s, and S = 29 in = 0.7366 m
0 = 31.02² + 2a*0.7366
-1.4732a = 962.36
a = - 653.24 m/s² (the negative signal indicates that she's dessacelaranting)
c) The time can be calculate by:
v = v0 + at
0 = 31.02 -653.24*t
653.24t = 31.02
t = 0.05 s = 50 ms
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on:________.a) the mass of the sphere.
b) the radius of the sphere.
c) both the mass and the radius of the sphere.
d) neither the mass nor the radius of the sphere.
The angular velocity of a solid uniform sphere rolling without slipping down an inclined plane does not depend on the mass or the radius of the sphere; it depends on the height from which it rolls down and gravitational acceleration only.
The question asks what determines the angular velocity of a solid uniform sphere as it rolls without slipping down an inclined plane. It specifically queries whether this angular velocity depends on the mass (M) of the sphere, the radius (R) of the sphere, both, or neither.
The answer is that the angular velocity of a rolling sphere at the bottom of the incline does not depend on the mass of the sphere or the radius of the sphere. According to the principles of conservation of energy and the dynamics of rolling motion, all objects regardless of their mass or radius, will roll down an incline and reach the bottom with the same angular velocity if they start from rest, provided that they do not slip and there is no air resistance. This is because the potential energy lost is completely converted into kinetic energy (both translational and rotational), and the mechanical energy of the system remains constant.
The angular velocity of the sphere at the bottom of the incline depends on d) neither the mass nor the radius of the sphere.
When the sphere rolls without slipping, there is a relationship between its linear velocity (v) and angular velocity (ω) given by the equation v = ωR.
Total mechanical energy of the sphere = translational kinetic energy + rotational kinetic energy + potential energy.
At the bottom of the incline, all potential energy will have converted into kinetic energy:
Initial Potential Energy (U) = mghTranslational Kinetic Energy (Kt) = (1/2)mv²Rotational Kinetic Energy (Kr) = (1/2)Iω², where I for a solid sphere is (2/5)mR²By energy conservation:
mgh = (1/2)mv² + (1/2)Iω²Substituting I and the relation v = ωR gives us:
mgh = (1/2)mv² + (1/2)*(2/5)mR²ω²Solving for ω:
mgh = (1/2)mv² + (1/5)mR²ω²Since v = ωR, ω² = v²/R²mgh = (1/2)mv² + (1/5)m(v²)7gh = v²v = √(7gh)ω = v/R = √(7gh)/RFrom the above steps, it's evident that the angular velocity ω depends on the height h the sphere rolls down, and the acceleration due to gravity g, but not on the mass M or radius R of the sphere.
Therefore, the correct answer is: d) neither the mass nor the radius of the sphere.
A +7.5 nC point charge and a -2.0 nC point charge are 3.0 cm apart. What is the electric field strength at the midpoint between the two charges?
Answer:
Ep= 3.8 10⁵ N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ =+7.5 nC = +7.5*10⁻⁹C
q₂ = -2.0 nC = -2.0*10⁻⁹C
d₁ =d₂ = 1.5cm = 1.5 *10⁻²m = 0.015 m
Calculation of the electric fieldsat the midpoint (P) between the two charges
Look at the attached graphic:
E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .
E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge
E₁ = k*q₁/d₁² = 9*10⁹ *7.5 *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep= E₁ + E₂
Ep= 3*10⁵ N/C + 0.8*10⁵ N/C
Ep= 3.8 10⁵ N/C
The electric field strength at the midpoint between a +7.5 nC point charge and a -2.0 nC point charge that are 3.0 cm apart is 1.56 *10^6 N/C (Newtons per Coulomb) away from the positive charge.
Explanation:The question is asking for the electric field at the midpoint between two point charges. The formula for the electric field created by a point charge is given by E=kQ/r^2, where E is the electric field, k is Coulomb's constant (approximately 9.0 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge. In this scenario we have two charges, so we calculate the electric field created by each charge and sum the results.
Applying this to both point charges and summing the electric fields we obtain: E_total = |E1| + |E2| = k*|Q1|/d^2 + k*|Q2|/d^2 = ((9.0*10^9 N*m^2/C^2) * 7.5 *10^-9 C/0.015m^2) + ((9.0*10^9 N*m^2/C^2) * 2 *10^-9 C/ 0.015m^2) = 1.8 *10^6 N/C and -0.24 * 10^6 N/C respectively. The total electric field strength at the midpoint is the sum which equals 1.56 *10^6 N/C away from the positive charge.
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A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has burned?
Answer:
Velocity of rocket will be equal to 7.81 m/sec
Explanation:
We have given mass of the rocket [tex]m_1=4kg[/tex]
Mass of fuel is given [tex]m_2=50gram=0.05kg[/tex] ( As 1 kg is equal to 1000 gram )
Velocity of fuel [tex]v_2[/tex] = 625 m/sec
We have to find the velocity of rocket [tex]v_1[/tex]
From conservation of momentum we know that initial momentum is equal to final momentum '
So [tex]m_1v_1=m_2v_2[/tex], here [tex]m_1[/tex] is mass of rocket [tex]v_1[/tex] is velocity of rocket [tex]m_2[/tex] is mass of fuel and [tex]v_2[/tex] is velocity of fuel
So [tex]4\times v_1=625\times 0.05[/tex]
[tex]v_1=7.81m/sec[/tex]
So velocity of rocket will be equal to 7.81 m/sec
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.20 m/s. Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 43.9 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 5.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? Where is Henrietta when she catches the bagels?
Answer:
[tex]u_x=8.5454\ m.s^{-1}[/tex]
[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.
Explanation:
Given:
speed of walking of Henrietta, [tex]v_w=3.2\ m.s^{-1}[/tex]height of projection of projectile, [tex]h=43.9\ m[/tex]Horizontal distance between the window and Henrietta when the projectile was launched:[tex]r=3.2\times 5[/tex]
[tex]r=16\ m[/tex]
Since the projectile was thrown at the time when she had passed below the window 5 seconds ago.Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.
Now the time taken for the object to reach to Henrietta ignoring the height where she catches:
[tex]h=u_x.t+\frac{1}{2} \times g.t^2[/tex]
[tex]43.9=0+\frac{1}{2} \times 9.8\times t^2[/tex]
[tex]t=2.9932\ s[/tex] is the time taken by the projectile to reach Henrietta.
Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:
[tex]\Delta d=v_w\times t[/tex]
[tex]\Delta d=9.5782\ m[/tex]
Now the total horizontal distance from the window to Henrietta when the projectile reached her:
[tex]d=\Delta d+r[/tex]
[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.
Now the initial horizontal velocity of launch of the projectile:
[tex]u_x=\frac{d}{t}[/tex]
[tex]u_x=\frac{25.5782}{2.9932}[/tex]
[tex]u_x=8.5454\ m.s^{-1}[/tex]
Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is increasing when the radius is 4.70 in. Assume that π = 3.14.
Answer: 0.006in/s
Explanation:
Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s
Also let the rate at which the radius of the balloon is increasing be dr/dt
Given r = 4.7in and Π = 3.14
Applying the chain rule method
dV/dt = dV/dr × dr/dt
If the volume of the sphere is 4/3Πr³
V = 4/3Πr³
dV/dr = 4Πr²
If r = 4.7in
dV/dr = 4Π(4.7)²
dV/dr = 277.45in²
Therefore;
1.68 = 277.45 × dr/dt
dr/dt = 1.68/277.45
dr/dt = 0.006in/s
Which of these statements best describe a leader? A. A leader is any person who is a part of the management team. B. A leader is any person who is always the first to perform any business activity. C. A leader is a person who inspires other team members to reach the organizational goals D. A leader is any person who puts others’ interest before his or her own. E. A leader is the one who owns the business and runs it on a day to day basis.
Answer:
I think the answer is c
Explanation:
But it is kinda of opinionated
Statements C best describe a leader. A leader is a person who inspires other team members to reach the organizational goals
What are the qualities of leader?A leader is someone who motivates others to achieve the organization's objectives.
A leader is the person who owns the company and manages it on a daily basis. Any member of the management team might be considered a leader, but only a leader would be able to own the company.
A leader is a person who inspires other team members to reach the organizational goals
Hence,statements C best describe a leader.
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To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by:_________
(a) the worker’s applied force,
(b) the gravitational force on the crate, and
(c) the normal force exerted by the incline on the crate?
(d) What is the total work done on the crate?
Answer:
a. [tex]W_w=313.5\ J[/tex]
b. [tex]W_g=-155.312\ J[/tex]
c. [tex]F_N=222.045\ N[/tex]
d. [tex]W_t=313.5\ J[/tex]
Explanation:
Given:
angle of inclination of the surface,[tex]\theta=25^{\circ}[/tex]mass of the crate, [tex]m=25\ kg[/tex]Force applied along the surface, [tex]F=209\ N[/tex]distance the crate slides after the application of force, [tex]s=1.5\ m[/tex]a.
Work done by the worker who applied the force:
[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.
[tex]W_w=209\times 1.5[/tex]
[tex]W_w=313.5\ J[/tex]
b.
Work done by the gravitational force:
[tex]W_g=m.g\times h[/tex]
where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:
[tex]h=s\times sin\ \theta[/tex]
[tex]h=1.5\times sin\ 25^{\circ}[/tex]
[tex]h=0.634\ m[/tex]
So, the work done by the gravity:
[tex]W_g=25\times 9.8\times (-0.634)[/tex] ∵direction of force and displacement are opposite.
[tex]W_g=-155.312\ J[/tex]
c.
The normal reaction force on the crate by the inclined surface:
[tex]F_N=m.g.cos\ \theta[/tex]
[tex]F_N=25\times 9.8\times cos\ 25[/tex]
[tex]F_N=222.045\ N[/tex]
d.
Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]
Where did Earth’s water come from?, select the two competing scientific theories supported by evidence that explain the formation of the hydrosphere.
Answer:
Explanation:
There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.
The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.
Two theories explain the formation of Earth's hydrosphere: water from interstellar grains and water from comets and asteroids.
Explanation:There are two competing scientific theories supported by evidence that explain the formation of Earth's hydrosphere:
Water from comets and asteroids: Another theory suggests that the water may have been brought to Earth when comets and asteroids impacted it. Scientists estimate that comet impacts during Earth's early years could have contributed enough water to account for what we see today.
This traction loss occurs in the rear wheels of a vehicle. a) Braking-induced traction loss b) Acceleration-induced traction loss c) Front wheel traction loss (skid) d) Rear wheel traction loss (skid) e) Driver-induced skids
Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.
Explanation:The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.
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Final answer:
Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.
Explanation:
The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.
With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.
Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a uniform electric fiedl of magnitude E.
Answer:
Range, [tex]R = MV²/2QE[/tex]
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
A bowl contains 7 red balls and 10 blue balls. A woman selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue?
Answer:
1050 possible selections
Explanation:
Number of red balls = 7
Number of blue balls = 10
Let blue balls be B and red balls be R
four balls are to be selected. At least three must be blue.
the first combination is 4blue 0red
second combination is 3blue 1red
(10C4*7C0 + 10C3*7C1)
210 *1 +120 *7
210+840
=1050 possibilities
There are 71 different selections possible if at least 3 balls must be blue.
Explanation:To find the number of different selections possible, we can consider the different scenarios where at least 3 balls are blue:
Scenario 1: Exactly 3 balls are blue and 1 ball is red. In this case, there are 10 ways to select the 3 blue balls (from the 10 blue balls available) and 7 ways to select the red ball (from the 7 red balls available). So, the total number of selections for this scenario is 10 * 7 = 70.Scenario 2: All 4 balls are blue. In this case, there is only 1 way to select all 4 blue balls. So, the total number of selections for this scenario is 1.Summing up the number of selections from both scenarios, we get 70 + 1 = 71 different selections possible.
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Does air pressure increase or decrease with an increase in altitude?
Answer:
increase
Explanation:
think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing
Answer:
decrease.
Explanation:
The air pressure is given by
P = h x d x g
where, h is the height of air column above the surface and g be the acceleration due to gravity and d be the density of air.
As we go up, the height of air column decreases, density of air decreases and acceleration due to gravity also decreases, so the value of pressure decreases at altitudes.
If the magnitude of a charge is twice as much as another charge, but the force experienced is the same, then the electric field strength of this charge is _____ the strength of the other charge.
Answer:
The new electric field strength of this charge is half of the strength of the other charge.
Explanation:
The electric force acting on the charge particle is given by :
F = q E
[tex]E=\dfrac{F}{q}[/tex]
Where
q is the charged particle
E is the electric field
If the magnitude of a charge is twice as much as another charge, q' = 2q, but the force experienced is the same, then the new electric field is given by :
[tex]E'=\dfrac{F}{q'}[/tex]
[tex]E'=\dfrac{F}{(2q)}[/tex]
[tex]E'=\dfrac{1}{2}\times \dfrac{F}{q}[/tex]
[tex]E'=\dfrac{E}{2}[/tex]
So, the new electric field strength of this charge is half of the strength of the other charge. Hence, this is the required solution.
Final answer:
The electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge because the force experienced by both is equal.
Explanation:
If the magnitude of a charge is twice as much as another charge, but the force experienced by each is the same, then the electric field strength that the larger charge is in is half the strength of the other charge’s electric field. The electric field strength (E) at a point is defined as the force (F) experienced by a positive test charge (q) placed at that point divided by the magnitude of the charge itself: E = F/q.
If we have two charges, q1 and q2, where q2 is twice q1 (“q2 = 2 × q1”), and the forces on both charges are equal (“F1 = F2”), we can set up the following equations: E1 = F1/q1 and E2 = F2/q2. Combining our known relationships, we get E1 = F/q1 and E2 = F/(2 × q1), which simplifies to E1 = 2 × E2. Therefore, E2 is half the magnitude of E1, meaning that the electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge.
A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s2. at 10 s after launch the first stage of the rocket (now without fuel) is released. the second stage now had an acceleration of 6 m/s2
a) how high is the rocket when the first stage seperates?
b)how fast is the rocket moving upon first stage seperation?
c) what will be the maximum height attained by the first stage after seperation?
d) what will be the distance between the first and second stages 2 s after separation
Answer:
a) 200m
b) 40 m/s
c) 81.55m
d) 31.62m
Explanation:
Solution
a)
y = y0 + u×t+⅟2×a×t2 =
y0 = 0
u = 0
y = unknown
a = 4m/s2
t = time = 10 seconds
y = 0.5×4×102 = 200m
b) v = u + at
v = 0 + 4×10 = 40 m/s
c) v2 = u2 - 2×g×y
at maximum height v = 0
we have
402 = 2×9.81×y
Y =81.55m
d)
for the stage 2 we haace
y = y0 + u×t+⅟2×a×t2 =
y = 0 + 4×2+0.5×6×22 = 92m
for the stage one we have
y = 0+40×2-0.5×9.81×4= 60.38m
distance between the first and second stage 2s aftee separation = 92-60.38 = 31.62m
Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would_____________________.
Answer:
Tribe historians would: Depict a significant event for each year on a buffalo or deer skin
Tribes of the Sioux Nation maintained historical calendars composed of winter counts, using pictorial representations to record significant events. Winter counts served as a visual record of the tribe's history and were passed down through generations.
Explanation:Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would use pictorial representations to record significant events that occurred during each year. Each year, a new pictorial symbol would be added to the count. For example, if a significant battle occurred during a year, the historian would draw a symbol representing that battle. Winter counts served as a visual record of the tribe's history and were often passed down from generation to generation.
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Sir Lance a Lost new draw bridge was designed poorly and stops at an angle of 20o below the horizontal. Sir Lost and his steed stop when their combined center of mass is 1.0 m from the end of the bridge. The bridge is 8.0 m long and has a mass of 2000 kg; the lift cable is attached to the bridge 5.0 m from the castle end and to a point 12 m above the bridge. Sir Lost’s mass combined with his armor and steed is 1000 kg.
Determine
(a) the tension in the cable and
(b) the horizontal and vertical force components acting on the bridge at the castle end.
Answer:
The Tension T is 42120N
The Horizontal force component is 18322.2N
The Vertical force component is - 4729N
Explanation:
First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).
Having done that, you apply two conditions of equilibrium.
1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.
∑Fx = 0
∑Fx = Rx - Tcos64.2 = 0
Rx = 0.435T
∑Fy = 0
∑Fy = Ry + Tsin64.2 - W - w = 0
W = 2000kg × 9.8 = 19600N
w =1000kg × 9.8 = 9800N
Ry + 0.9T = 29400N
Ry = 29400 - 0.9T
2. THE SUM TOTAL OF TORQUES EQUALS ZERO
Rx: τ = 0
Ry: τ = 0
T: τ = 5 × Tsin44.2
= 3.49T m
W: τ = 4 × 19600sin90
= 78400Nm
w: τ = 7 × 9800sin9
= 68600Nm
Note:
Rx = x component of Reaction force
Ry = y component of Reaction force.
T = Tension
W = weight of bridge
w = weight of Sir Lance a Lost and his steed
τ = torque
Note: The torque of Tension is counter clockwise while that of the weights is clockwise.
Hence,
∑τccw = ∑τcw
3.49T = 78400 + 68600
3.49T = 14700Nm
T = 147000/3.49
T = 42120N
Rx = 0.435 × 42120
Rx = 18322.2N
Ry = 29400N - (0.9×42120)N
Ry = 29400 - 34129
Ry = -4729N
Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.
In this exercise we have to use the knowledge of tension and force, so we can say that this will result in:
A)The Tension T is 42120N
B)The Horizontal force component is 18322.2N
C)The Vertical force component is - 4729N
Before starting the calculations we have to remember some concepts such as:
Rx = x component of Reaction forceRy = y component of Reaction force.T = TensionW = weight of bridgew = weight of Sir Lance a Lost and his steedτ = torqueFirst we will add all the force vectors so that it results in zero and that means that the body is in equilibrium, so:
[tex]\sum F_x = 0\\ \sum F_x = R_x - Tcos(64.2) = 0\\ R_x = 0.435T\\ \sum F_y = 0\\ \sum F_y = R_y + Tsin(64.2) - W - w = 0\\ W = 2000kg * 9.8 = 19600N\\ w =1000kg * 9.8 = 9800N\\ R_y + 0.9T = 29400N\\ R_y = 29400 - 0.9T[/tex]
Secondly, we will add up all the torques so that it results in zero and that means that the body is in equilibrium, so:
[tex]T: T = 5 * Tsin(44.2) = 3.49T m\\ W: T = 4 * 19600sin(90)= 78400Nm\\ w: T = 7 * 9800sin(90) = 68600Nm\\ \sum Tccw = \sum Tcw\\ 3.49T = 78400 + 68600\\ 3.49T = 14700Nm\\ T = 147000/3.49\\ T = 42120N R_x = 0.435 * 42120\\ R_x = 18322.2N\\ R_y = 29400N - (0.9*42120)N\\ R_y = 29400 - 34129\\ R_y = -4729N[/tex]
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