Final answer:
Upon removing 12000 electrons from a neutral balloon, it will acquire a positive charge of 1.92 microcoulombs (1.92 µC). This is calculated by multiplying the number of electrons by the elemental charge and converting to the appropriate unit.
Explanation:
When 12000 electrons are removed from a neutral balloon, it obtains a positive charge because electrons carry a negative charge. To determine the charge the balloon now carries, we multiply the number of electrons removed by the elementary charge of an electron.
The charge (Q) on the balloon can be calculated using the formula:
Q = n × e
Where:
n is the number of removed electronse is the elementary charge per electron (1.6 × 10⁻¹⁹ C)Let's plug in the values into the formula:
Q = 12000 × 1.6 × 10⁻¹⁹ C
Q = 1.92 × 10⁻¹⁵ C
Converting coulombs (C) to microcoulombs (µC) by multiplying by 10¶ gives:
Q = 1.92 × 10⁻¹⁵ C × 10¶ µC/C
Q = 1.92 µC
The balloon will carry a charge of 1.92 µC after 12000 electrons have been removed.
How far apart are two conducting plates that have an electric field strength of 4.5 × 103V/m between them, if their potential difference is 12.5 kV?
Answer:
Explanation:
Given
Electric Field Strength [tex]E=4.5\times 10^{3}\ V/m[/tex]
Potential Difference between Plates is given by [tex]V=12.5\ kV[/tex]
In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates
The potential difference is given by
[tex]\Delta V=Ed[/tex]
where E=Electric Field strength
d=distance between Plates
[tex]d=\frac{\Delta V}{E}[/tex]
[tex]d=\frac{12.5\times 10^3}{4.5\times 10^{3}}[/tex]
[tex]d=2.78\ m[/tex]
Which of the following statements about insulating materials is correct?a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.b.The electric field from a charged object is able to penetrate through an insulating material.c.Insulating materials exhibit a linear relationship between voltage and current.d.Silver and copper are good insulators.
Answer:
a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.
Explanation:
A conductor has low resistance, while an insulator has much higher resistance. Devices called resistors control amounts of resistance into an electrical circuits. Electric charges inside an insulator are bound to the individual atoms or molecules, not being able to move inside the material.
When you turn up the resistance, the electric current flowing through the circuit is reduced
Ohm's law:
V = I * R
R = V/I
An Ohmic conductor would have a linear relationship between the current and the voltage. With non-Ohmic conductors, the relationship is not linear. Most metals are good conductors example silver, copper etc.
The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequency ωωomega of these oscillations? Use the approximation d≪ad≪a to simplify your calculation; that is, assume that d2+a2≈a2d2+a2≈a2. Express your answer in terms of given charges, dimensions, and constants.
Answer:
[tex]\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }[/tex]
Explanation:
Additional information:
The ball has charge [tex]-q_0[/tex], and the ring has positive charge [tex]+Q[/tex] distributed uniformly along its circumference.
The electric field at distance [tex]z[/tex] along the z-axis due to the charged ring is
[tex]E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.[/tex]
Therefore, the force on the ball with charge [tex]-q_0[/tex] is
[tex]F=-q_oE_z[/tex]
[tex]F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}[/tex]
and according to Newton's second law
[tex]F=ma=m\dfrac{d^2z}{dz^2}[/tex]
substituting [tex]F[/tex] we get:
[tex]- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}[/tex]
rearranging we get:
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0[/tex]
Now we use the approximation that
[tex]z^2+a^2\approx a^2[/tex] (we use this approximation instead of the original [tex]d^2+a^2\approx a^2[/tex] since [tex]z<d[/tex], our assumption still holds )
and get
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0[/tex]
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0[/tex]
Now the last equation looks like a Simple Harmonic Equation
[tex]m\dfrac{d^2z}{dz^2}+kz=0[/tex]
where
[tex]\omega=\sqrt{ \dfrac{k}{m} }[/tex]
is the frequency of oscillation. Applying this to our equation we get:
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}[/tex]
[tex]\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}[/tex]
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 9.1 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
In order for the net effect at point A to be a downward force, the tension in cable AC (T) should be equal to the tension in cable AB (9.1 kN). The magnitude of the resulting downward force (R) would be the sum of the tensions in both cables, thus 2 * 9.1 kN = 18.2 kN.
Explanation:To understand this scenario, it is essential to apply the principles of equilibrium and vector sum in Physics. The tension in the wires can be considered as forces experienced by point A. According to the question, the net effect of these forces should be a downward force, implying that they should negate the opposite upward force.
To find the tension T in cable AC, it's logical to assume that the force due to this tension needs to be equal and opposite to the force exerted by the tension in wire AB, which is 9.1 kN. Therefore, T should also be 9.1 kN for the net effect at point A to be a downward force.
The magnitude R of the downward force can be determined by considering the combined effect of tensions in cables AB and AC. Since point A is in equilibrium, R will be the result of the total upward forces. Hence, R is equal to 2 times the tension in any one cable (as they are equal), which gives us R = 2 * 9.1 kN = 18.2 kN.
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Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that load in 1 6 the time it takes the other. How much more power does the faster crane have?
Answer:
Explanation:
Given
Load [tex]W=20000\ N[/tex]
height to which load is raised [tex]h=250\ m[/tex]
Another crane take [tex]\frac{1}{6}[/tex] th time to lift the load
Energy required required to lift the Weight
[tex]E=W\times h[/tex]
[tex]E=20000\times 250[/tex]
[tex]E=5,000,000\ J[/tex]
Suppose [tex]P_1[/tex] and [tex]P_2[/tex] is the Power required to lift the weight in t and [tex]\frac{t}{6}[/tex] time
[tex]E=P_1\times t[/tex]
[tex]E=P_2\times \frac{t}{6}[/tex]
[tex]P_1\times t=P_2\times \frac{t}{6}[/tex]
thus
[tex]P_2=6P_1[/tex]
Second Crane requires 6 times more power than the slow crane
The power of the faster crane is 6 times greater than the power of the slower crane.
Explanation:To calculate the power of the cranes, we need to use the formula:
Power = Work/Time
The work done by each crane is equal to the maximum load lifted multiplied by the height lifted, so:
Work = Load x Height
Let's assume the time taken by the slower crane is t. Therefore, the time taken by the faster crane is t/6.
Now, let's calculate the power of each crane:
Power of slower crane = Work/Time = (20000 N x 250 m) / t = 5000000 Nm/t
Power of faster crane = Work/Time = (20000 N x 250 m) / (t/6) = 30000000 Nm/t
The power of the faster crane is 6 times greater than the power of the slower crane.
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An exploration submarine should be able to descend 1200 m down in the ocean. If the ocean density is 1020 kg/m3, what is the maximum pressure on the submarine hull?
Answer:
11995200 N/m²
Explanation:
Pressure: The is the ratio of force to the surface area in contact. The S.I unit of pressure is N/m².
Generally pressure in fluid can be expressed as
P = ρgh......................... Equation 1
Where P = maximum pressure on the submarine hull, ρ = Density of ocean, h = depth of ocean, g = acceleration due to gravity.
Given: h = 1200 m, ρ = 1020 kg/m³
Constant: g = 9.8 m/s²
Substitute into equation 1
P = 1200(1020)(9.8)
P = 11995200 N/m²
Hence the maximum pressure on the submarine hull = 11995200 N/m²
Final answer:
The maximum pressure on a submarine hull that descends 1200 m in the ocean, with ocean density of 1020 kg/m³, is calculated using the formula P = [tex]P_{o}[/tex] + ρgh. With an atmospheric pressure of 101325 Pa, the final pressure would be 12116485 Pa at that depth.
Explanation:
To calculate the maximum pressure on a submarine hull that can descend 1200 m in the ocean, we use the formula for the pressure exerted by a fluid at a certain depth. The formula is P = [tex]P_{o}[/tex] + ρgh, where P is the pressure at depth, [tex]P_{o}[/tex] is the atmospheric pressure on the surface, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Atmospheric pressure [tex]P_{o}[/tex] is approximately 101325 Pa (or 1 atm). The density of sea water is given as 1020 kg/m³ and the depth is 1200 m. Taking g as 9.8 m/s², the standard acceleration due to gravity, we can substitute the values into the formula:
P = 101325 Pa + (1020 kg/m³)×(9.8 m/s²)×(1200 m)
P = 101325 Pa + 12003360 Pa
P = 12116485 Pa
Therefore, the maximum pressure on the submarine hull at a depth of 1200 m would be 12116485 Pascal (Pa).
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later, the bicyclist hops on his bike and accelerates at 2.11 m/s2 until he catches his friend. a. How much time does it take until he catches his friend?(b) How far has he traveled in this time? (c) What is his speed when he catches up?
Answer:
a. [tex]t=3.44s[/tex]
b. [tex]x=12.45m[/tex]
c. [tex]v_f=7.26\frac{m}{s}[/tex]
Explanation:
The bicyclist's friend moves with constant speed. So, we have:
[tex]x=vt[/tex]
Th bicyclist moves with constant acceleration and starts at rest ([tex]v_0=0[/tex]). So, we have:
[tex]x=v_0t+\frac{at^2}{2}\\x=\frac{at^2}{2}[/tex]
a. When he catches his friend, both travels the same distance, thus:
[tex]vt=\frac{at^2}{2}\\t=\frac{2v}{a}\\t=\frac{2(3.63\frac{m}{s})}{2.11\frac{m}{s^2}}\\t=3.44s[/tex]
b. We can use any of the distance equations, since both travels the same distance:
[tex]x=vt\\x=3.63\frac{m}{s}(3.44s)\\x=12.45m[/tex]
c. The bicyclist final speed is:
[tex]v_f=v_0+at\\v_f=at\\v_f=2.11\frac{m}{s^2}(3.44s)\\v_f=7.26\frac{m}{s}[/tex]
A power supply maintains a potential difference of 52.3 V 52.3 V across a 1570 Ω 1570 Ω resistor. What is the current in the resistor?
Answer:
0.033 A
Explanation:
Current: This can be defined as the rate of flow of electric charge in a circuit.
The S.I unit of current is Ampere (A)
From Ohm's law.
V = IR ............................ Equation 1
Where V = Potential difference, I = current, R = resistance.
Making I the subject of the equation,
I = V/R................... Equation 2
Given: V = 52.3 V, R = 1570 Ω
Substitute into equation 2
I = 52.3/1570
I = 0.033 A.
Hence the current in the resistor = 0.033 A
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their getaway car, but aren't entirely sure if they can make the jump or need to head through the building. a) If one of them drops a pebble off the edge of the roof and it hits the ground two seconds later, how fast will they hit the ground if they jump? Give answers in terms of meters per second. b) How high up are they? Give answers in terms of meters. c) Is this a safe jump?
Answer:
a) They will hit the ground with a speed of 19.6 m/s.
b) They are at a height of 20 m.
c) It is not a safe jump.
Explanation:
Hi there!
a) The equations of height and velocity in function of time of a free falling body are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).
v = velocity of the object at time t.
Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:
v = v0 + g · t (v0 = 0)
v = g · t
v = -9.8 m/s² · 2.0 s
v = -19.6 m/s
They will hit the ground with a speed of 19.6 m/s.
b)Now, we have to use the equation of height:
h = h0 + v0 · t + 1/2 · g · t²
If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m
0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²
h0 = 1/2 · 9.8 m/s² · (2.0 s)²
h0 = 20 m
They are at a height of 20 m.
c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.
A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibrationa. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelengthe. The length of the string is equal to four times the wavelength
Answer:
d. The length of the string is equal to one-half of a wavelength
Explanation:
A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration
a. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelength
e. The length of the string is equal to four times the wavelength
A stretched string of length L fixed at both ends is vibrating in its third harmonic H
How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration
d. The length of the string is equal to one-half of a wavelength
There are two points during vibration , the node and the antinode
the node is the point where the amplitude is zero.
from the third harmonics, there are two nodes. The first node is half of the wavelength which is the closest to the fixed point.
for third harmonics=3/2lamda
Finally, consider the expression (6.67×10^−11)(5.97×10^24)/(6.38×10^6)^2. Determine the values of a and k when the value of this expression is written in scientific notation. Enter a and k, separated by commas.
Answer:
[tex]x=9.78\times 10^0[/tex]
Explanation:
In this case, we need to find the value of expression :
[tex]x=\dfrac{(6.67\times 10^{-11})\times (5.97\times 10^{24})}{(6.38\times 10^6)^2}[/tex]
On solving, we get the value of given expression as :
x = 9.7827
In scientific notation, we get the value of x as :
[tex]x=a\times 10^k[/tex]
[tex]x=9.78\times 10^0[/tex]
a = 9.78
k = 0
Hence, this is the required solution.
To express the given expression in scientific notation, we find a = 9.78 and k = -1 after evaluating the expression using the laws of exponentiation and division.
Explanation:The student's question involves evaluating an expression using scientific notation and expressing the result in proper scientific notation.
To solve (6.67×10−11)(5.97×1024) / (6.38×106)2, we need to use the laws of exponentiation and multiplication. Simplify the expression within the numerator and denominator separately before dividing them.
Numerator: (6.67×10−11)(5.97×1024) = 39.8309×1013
Denominator: (6.38×106)2 = 40.7044×1012
Divide the simplified numerator by the simplified denominator:
39.8309×1013 / 40.7044×1012 = 0.978×101
To express this in proper scientific notation, rewrite 0.978×101 as 9.78×10−0.
Therefore, the values of a and k when the value of this expression is written in scientific notation are a = 9.78 and k = −0.
A lightbulb with an intrinsic resistance of 270 \OmegaΩ is hooked up to a 12-volt battery. How much power is output by the lightbulb? Give your answer out to the thousandths place in units of watts (W).
Answer:
P = 0.533 W
Explanation:
given,
Resistance of the bulb, R = 270 Ω
Potential of the battery, V= 12 V
Power output of the bulb = ?
we know,
P = I² R
also, V = IR
[tex]P = \dfrac{V^2}{R}[/tex]
[tex]P =\dfrac{12^2}{270}[/tex]
[tex]P =\dfrac{144}{270}[/tex]
P = 0.533 W
Hence, the Power delivered by the bulb is equal to 0.533 W.
The density of liquid oxygen at its boiling point is 1.14 kg/Lkg/L , and its heat of vaporization is 213 kJ/kgkJ/kg . How much energy in joules would be absorbed by 2.0 LL of liquid oxygen as it vaporized? Express your answer to two significant figures and include the appropriate units.
Answer:
heat absorbed = 4.9 × [tex]10^{5}[/tex] J
Explanation:
given data
density of liquid oxygen = 1.14 kg/L
volume = 2 L
heat of vaporization = 213 kJ/kg
solution
first we get here mass of liquid that is
mass of liquid = density × volume ......................1
mass of liquid = 1.14 × 2
mass of liquid = 2.28 kg
so here we get now heat absorbed that is
heat absorbed = mass × heat of vaporization
heat absorbed = 2.28 kg × 213 kJ/kg
heat absorbed = 485.640 kJ
heat absorbed = 4.9 × [tex]10^{5}[/tex] J
To find the energy absorbed by 2.0 L of liquid oxygen as it vaporizes, you first convert the volume to mass using the given density and then multiply by the heat of vaporization. The resulting energy is approximately 4.86 x 10^5 J or 486 kJ.
Explanation:The energy absorbed by a volume of liquid oxygen as it vaporizes can be calculated using the formula Q = mLv. In this equation, 'm' represents mass, 'Lv' represents the heat of vaporization, and 'Q' represents the total energy absorbed.
Firstly, convert the volume of liquid oxygen to mass. The density of liquid oxygen at its boiling point is 1.14 kg/L, so the mass of 2.0 L of liquid oxygen would be (1.14 kg/L) * (2.0 L) = 2.28 kg.
Then, use the heat of vaporization and the calculated mass to find the total energy. The heat of vaporization of oxygen is 213 kJ/kg, so Q = (2.28 kg) * (213 kJ/kg) = 485.64 kJ. This needs to be expressed in joules by multiplying by 10^3, resulting in 485640 J. Therefore, the energy absorbed by 2.0 L of liquid oxygen as it vaporizes is approximately 4.86 x 10^5 J.
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Two point charges, +2.20 μC and -8.00 μC, are separated by 2.60 m. What is the electric potential midway between them? Number Units
Answer:
Electric potential, [tex]V=-4.01\times 10^4\ volts[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=2.2\ \mu C[/tex]
Point charge 2, [tex]q_2=-8\ \mu C[/tex]
Distance between charges, d = 2.6 m
We need to find the electric potential midway between them. The electric potential is given by :
[tex]V=\dfrac{kq}{r}[/tex]
In this case, r = 1.3 m (midway between charges)
[tex]V=\dfrac{kq_1}{r}-\dfrac{kq_2}{r}[/tex]
[tex]V=\dfrac{k}{r}(q_1-q_2)[/tex]
[tex]V=\dfrac{9\times 10^9}{1.3}(2.2\times 10^{-6}-8\times 10^{-6})[/tex]
[tex]V=-40153.84\ volts[/tex]
or
[tex]V=-4.01\times 10^4\ volts[/tex]
So, the electric potential midway between the charges is [tex]V=-4.01\times 10^4\ volts[/tex]. Hence, this is the required solution.
Final answer:
The electric potential midway between two point charges of +2.20 μC and -8.00 μC, separated by 2.60 m, is calculated separately for each charge using the formula V = kq/r and summed up. The total electric potential at the midpoint is -4.00 × 10⁴ V.
Explanation:
To find the electric potential midway between two point charges, we need to consider the contribution from each charge separately and then sum them up.
The electric potential V due to a single point charge q at a distance r is given by the formula:
V = kq/r
where k is the Coulomb's constant (k ≈ 8.99 × 109 Nm²/C²).
In this case, we have two charges, +2.20 μC and -8.00 μC, and they are separated by 2.60 m. So the distance from the midpoint to each charge is 1.30 m (half of 2.60 m).
Calculating the potential due to the +2.20 μC charge:
V₁ = (8.99 × 109)(+2.20 × 10⁻⁶) / 1.30 = 1.53 × 10⁴ V
And for the -8.00 μC charge:
V₂ = (8.99 × 10⁹)(-8.00 × 10⁻⁶) / 1.30 = -5.53 × 10⁴ V
The total electric potential at the midpoint is the sum of V₁ and V₂:
Vtotal = V₁ + V₂ = 1.53 × 10⁴ V - 5.53 × 10⁴ V = -4.00 × 10⁴V
The electric potential midway between the two charges is -4.00 × 10⁴V.
How much stronger is the gravitational pull of the Sun on Earth, at 1 AU, than it is on Saturn at 10 AU?
How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr?
The reaction produces -4.95 kJ of heat when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr.
The equation of the reaction is;
CO(g) + H2(g) -------> CH2O(g)
The heat of reaction is obtained from;
Enthalpy of products - Enthalpy of reactants = (-116kJ/mol) - (-110.5 kJ/mol)
= -5.5 kJ/mol
Number of moles of CO is obtained from;
PV = nRT
P = 112 kPa or 1.1 atm
T = 85°C + 273 = 358 K
n = ?
R = 0.082 atmLK-1mol-1
V = 15.0 L
n = PV/RT
= 1.1 atm × 15.0 L/ 0.082 atmLK-1mol-1 × 358 K
= 0.56 moles
Number of moles of H2
n = PV/RT
P= 744 torr or 0.98 atm
V = 14.4 L
T = 75°C + 273 = 348 K
n = 0.98 atm × 14.4 L/0.082 atmLK-1mol-1 × 348 K
n = 0.49 moles
We can see that H2 is the limiting reactant here hence 0.49 moles of formaldehyde is produced.
If 1 mole of formaldehyde produces -5.5 kJ of heat
0.49 moles of formaldehyde produces -5.5 kJ × 0.49 moles / 1 mole
= -4.95 kJ of heat
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A 12,000-lb bus collides with a 2800-lb car. The velocity of the bus before the collision is vB = 18i (ft/s) and the velocity of the car is vC =33j (ft/s). The two vehicles become entangled and remain together after the collision. The coefficient of kinetic friction between the vehicles’ tires and the road is μk = 0.6. Find (a) velocity of combined center of mass immediately after the collision, (b) final position (assume brakes are on and you have skidding, not rolling). (c) If the collision lasted 0.5 seconds, what was the impulsive force of the bus on the car?
Answer:
(a) 20.84 ft/s
(b) 11.24 ft
(c) -68160 N
Explanation:
Parameters given:
Mass of Bus, Mb = 12000 lb
Mass of car, Mc = 2800 lb
Initial speed of bus before collision, u(b) = 18 ft/s
Initial speed of car before collision, u(c) = 33 ft/s
Coefficient of friction, μk = 0.6
(a)Combined velocity after collision.
Since the bus and car are entangled and move together after the collision, they have the same velocity after collision.
Using the law of conservation of momentum, we have:
Mb*u(b) + Mc*u(c) = Mb*v(b) + Mc*v(c)
Where v(b) = velocity of the bus,L after collision,
v(c) = velocity of the car after collision.
Since v(b) = v(c) = v,
Mb*u(b) + Mc*u(c) = (Mb + Mc)*v
=> (12000 * 18) + (2800 * 33) = (12000 + 2800) * v
=> 308400 = 14800 * v
=> v = 20.84 ft/s
Combined velocity after collision is 20.84 ft/s
(b) The force acting on the bus and car after collision is given as:
F = m*a
Where F = force,
m = combined mass of bus and car,
a = acceleration of the bus and car after collision.
We know that the only force acting on the combined mass of the bus and car is the Frictional force, Fr and it is given as:
Fr = μk*m*g
Where g = acceleration due to gravity
Hence,
Fr = ma
=> - μk*m*g = m*a
The negative sign signifies that the fictional force is acting in the opposite direction to the lotion.
=> a = - μk*g
a = - 0.6 * 32.2
a = - 19.32 ft/s^2
Using one of the equations of linear motion, we can find the distance moved by the car and bus after collision:
v*v = u*u + 2*a*s
Where s = distance moved
The final velocity, v, of the car and bus is 0 because they come to rest and the initial velocity, u, is 20.84 ft/s, the velocity of the car and bus after collision.
Hence,
0 = (20.84*20.84) + (-2*19.32*s)
=> s = -434.3056/-38.64
s = 11.24 ft
(c) Impulsive force is the force that two bodies which are colliding exert on one another. It is given mathematically as
I. F. = (momentum change)/time
Momentum change of the bus is:
Momentum change = final momentum - initial momentum
Momentum change = Mb*v(b) - Mb*u(b)
Momentum change = (12000*20.84) - (12000*18)
Momentum change = 250080 - 216000
Momentum change = - 34080 kgft/s
=> I. F. = - 34080/0.5
I. F. = - 68160 N
The Impulsive force of the bus on the car is - 68160N
The negative sign means the force is acting opposite to the motion of the car.
There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2.0 × 107 m/s in a direction opposite to the field. Its speed increases to 4.0 × 107 m/s over a distance of 1.2 cm. What is the magnitude of the electric field?
Answer:
Explanation:
Given
speed of Electron [tex]u=2\times 10^7\ m/s[/tex]
final speed of Electron [tex]v=4\times 10^7\ m/s[/tex]
distance traveled [tex]d=1.2\ cm[/tex]
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=Final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}[/tex]
[tex]a=5\times 10^{16}\ m/s^2[/tex]
acceleration is given by [tex]a=\frac{qE}{m}[/tex]
where q=charge of electron
m=mass of electron
E=electric Field strength
[tex]5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}[/tex]
[tex]E=248.3\ kN/C[/tex]
A small, charged, spherical object at the origin of a Cartesian coordinate system contains 2.60 × 10 4 more electrons than protons. What is the magnitude of the electric field it produces at the position (2.00 mm, 1.00 mm)?
Answer:
E = 7.77 N/C
Explanation:
The charge of a single electron is 1.6 x 10^{-19} C. The net charge of the object is therefore the multiplication of the number of excess electrons and the charge of a single electron:
[tex]Q = (2.6\times 10^4) \times 1.6\times 10^{-19} = 4.16 \times 10^{-15}~C[/tex]
The electric field can be found by the following formula
[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]
where 'r' can be calculated as
[tex]r = \sqrt{(2\times 10^{-3})^2 + (1\times 10^{-3})^2} = 0.0022~m\\r^2 = 4.84\times 10^{-6}[/tex]
Finally, the electric field at the position (2.00 mm, 1.00 mm) is
[tex]E = \frac{1}{4\pi(8.8\times 10^{-12})}\frac{4.16\times 10^{-15}}{4.84\times 10^{-6}} = 7.77~N/C[/tex]
The magnitude of the electric field it produces at the position is 7.5 N/C.
The given parameters:
Number of excess electron, n = 2.6 x 10⁴Position of the excess electron, x = (2.00 mm, 1.00 mm)The position of the charged object is calculated as follows;
[tex]r^2 = (2.0 \times 10^{-3})^2 + (1.0 \times 10^{-3})^2\\\\r^2 = 5\times 10^{-6} \ m^2[/tex]
The charge of the electron is calculated as follows;
[tex]Q = nq\\\\Q = 2.6 \times 10^4 \times 1.6\times 10^{-19}\\\\Q =4.16 \times 10^{-15} \ C[/tex]
The magnitude of the electric field it produces at the position is calculated as follows;
[tex]E = \frac{F}{Q}= \frac{kQ}{r^2} = \frac{9\times 10^9 \times 4.16 \times 10^{-15}}{5\times 10^{-6}} \\\\E = 7.5 \ N/C[/tex]
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Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?
To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.
The volume of a sphere can be expressed as
[tex]V = \frac{4}{3} \pi r^3[/tex]
Here r is the radius of the sphere and V is the volume of Sphere
Using the expression of the density we know that
[tex]\rho = \frac{m}{V} \rightarrow V = \frac{m}{\rho}[/tex]
The density is given as
[tex]\rho = (19.5g/cm^3)(\frac{10^3kg/m^3}{1g/cm^3})[/tex]
[tex]\rho = 19.5*10^3kg/m^3[/tex]
Now replacing the mass given and the actual density we have that the volume is
[tex]V = \frac{60kg}{19.5*10^3kg/m^3 }[/tex]
[tex]V = 3.0769*10^{-3} m ^3[/tex]
The radius then is,
[tex]V = \frac{4}{3} \pi r^3[/tex]
[tex]r = \sqrt[3]{\frac{3V}{4\pi}}[/tex]
Replacing,
[tex]r = \sqrt[3]{\frac{3(3.0769*10^{-3})}{4\pi}}[/tex]
The radius of a sphere made of this material that has a critical mass is 9.02 cm.
At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph ?
Answer:
T= 89.25 K
Explanation:
Given that
V= 590 mph
We know that
1 mph = 0.44 m/s
That is why ,V= 263.75 m/s
We know that speed of the gas molecule is given as
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
R= 8.314 J/mol.k
M= 32 g/mol = 0.032 kg/mol
T=Temperature in Kelvin unit
[tex]V^2=\dfrac{3RT}{M}[/tex]
[tex]T=\dfrac{V^2\times M}{3R}[/tex]
[tex]T=\dfrac{263.76^2\times 0.032}{3\times 8.314}\ K[/tex]
T= 89.25 K
Therefore the average temperature ,T = 89.25 K
Answer:
temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph = 89.24 K
Explanation:
Average speed of oxygen molecule is given by
[tex]v= \sqrt{\frac{3RT}{M} }[/tex]
[tex]590\times0.44704 = 263.75[/tex] m/s
R= 8.314 J/mol K = universal gas constant
M= molecular weight of oxygen = 32 g/mol =0.032 Kg/mol
now plugging these values to find T we get
[tex]263.75=\sqrt{\frac{3(8.314)(T)}{0.032}}[/tex]
solving the above equation we get
T= 89.24 K
The distance from the earth to the sun is about 1.50×1011 m . Find the total power radiated by the sun.
Answer:
Power, [tex]P=3.93\times 10^{26}\ W[/tex]
Explanation:
Given that,
The distance from the earth to the sun is about, [tex]d=1.5\times 10^{11}\ m[/tex]
let us assume that the average intensity of solar radiation at the upper atmosphere of earth,. [tex]I=1390\ W/m^2[/tex]
We need to find the total power radiated by the sun. The intensity is defined as the total power divided by the area of a sphere of radius equal to the average distance between the earth and the sun. It is given by :
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
P is total power
[tex]P=4\pi r^2\times I[/tex]
[tex]P=3.93\times 10^{26}\ W[/tex]
So, the total power radiated by the sun is [tex]P=3.93\times 10^{26}\ W[/tex]. Hence, this is the required solution.
Final answer:
The total power radiated by the Sun is calculated using the area of the Sun and the power radiated per square meter at its surface. The Sun's total power output is found to be 3.82×1026 W. This value is important for understanding the energy Earth receives from the Sun, known as the solar constant (1360 W/m²).
Explanation:
The distance from the Earth to the Sun is about 1.50×1011 meters. The total power radiated by the Sun can be determined by using the following physics principle: The Sun, behaving as a perfect black body with an emissivity of exactly 1, radiates power uniformly across its surface area. The power radiated per square meter on the Sun's surface is found to be 6.3×107 W/m². The sun's radius is approximately 7.00×108 meters. Using the formula Power = Area × Intensity, we calculate the Sun's total power output.
The area of a sphere is given by 4πR2, where R is the radius of the sphere. Therefore, the total power output of the Sun is 4πR2σT4, which calculates to 3.82×1026 W. This immense amount of power is what we refer to as the solar luminosity.
At the distance of the Earth, this power is spread over a spherical area with a radius equal to the Earth-Sun distance. We can find the power per square meter at this distance, known as the solar constant, which is approximately 1360 W/m².
How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 30.0°C greater than when they were laid? Their original length is 30.0 m.
Answer:
1.2 cm
Explanation:
Thermal Expansion
It's the tendency that materials have to change its size and/or shape under changes of temperature. It can be in one (linear), two (surface) or three (volume) dimensions.
The formula to compute the expansion of a material under a change of temperature from [tex]T_o[/tex] to [tex]T_f[/tex] is given by.
[tex]\Delta L=L_o.\alpha .(T_f-T_o)[/tex]
Where Lo is the initial length and [tex]\alpha[/tex] is the linear temperature expansion coefficient, which value is specific for each material. The data provided in the problem is as follows:
[tex]L_o=30\ m,\ T_f-T_o=30^oC,\ \alpha=13\times 10^{-6}\ ^oC^{-1}[/tex]
Computing the expansion we have
[tex]\Delta L=30\times 13\times 10^{-6}(30)=0.0117=1.17\ cm[/tex]
The expansion gap should be approximately 1.2 cm
A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , where x is in cm and t is in seconds. What is the frequency of the oscillation?
Answer:
1/2 Hz
Explanation:
A simple harmonic motion has an equation in the form of
[tex]x(t) = Acos(\omega t - \phi)[/tex]
where A is the amplitude, [tex]\omega = 2\pi f[/tex] is the angular frequency and [tex]\phi[/tex] is the initial phase.
Since our body has an equation of x = 5cos(π t + π/3) we can equate [tex]\omega = \pi[/tex] and solve for frequency f
[tex]2\pi f = \pi[/tex]
f = 1/2 Hz
0.5Hz
Explanation:The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;
x = A cos (ωt ± ∅) --------------------------(i)
where;
A = amplitude of the wave
ω = angular velocity of the wave
∅ = phase constant of the wave
From the question;
x = 5cos(π t + π/3) -----------------------------(ii)
Comparing equations (i) and (ii), the following deductions among others can be made;
A = 5cm
ω = π
But the angular velocity (ω) of the wave is related to its frequency (f) as follows;
ω = 2 π f --------------------(iii)
Substitute the value of ω = π into equation (iii) as follows;
π = 2 π f
Divide through by π;
1 = 2f
Solve for f;
f = 1/2
f = 0.5
Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz
How strong is the attractive force between a glass rod with a 0.700 μC charge and a silk cloth with a –0.600 μC charge, which are 12.0 cm apart, using the approximation that they act like point charges?
Answer:
[tex]F=0.26N[/tex]
Explanation:
Assuming that thet act like point charges, the attractive force is given by Coulomb's law:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
Where k is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the magnitudes of the point charges and d is the distance of separation between them. Thus, we replace the given values and get how strong is the attractive force between them:
[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(0.7*10^{-6}C)(-0.6*10^{-6}C)}{(12*10^{-2}m)^2}\\F=0.26N[/tex]
A ball is thrown vertically upward with a speed of 19.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started?
Answer:
Explanation:
initial speed, u = 19 m/s
(a) Let it rises upto height h.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity and it is zero.
0 = 19 x 19 - 2 x 9.8 x h
h = 18.4 m
(c) Let the ball takes time t to reach to the maximum height.
use first equation of motion
v = u - gt
0 = 19 - 9.8 x t
t = 1.94 s
(c) The time taken by the ball to reach to the ground = 2 x time to reach to maximum height
T = 2 x t = 2 x 1.94 = 3.88 s
(d) When the ball reaches the ground, let the velocity is v.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity
v² = 0 + 2 x 9.8 x 18.4
v = 19 m/s
(a) The ball reaches a height of approximately 18.68 meters. (b) It takes approximately 1.94 seconds to reach its highest point (c) The time taken to hit the ground is 1.94 seconds. When the ball returns to the level from which it started, its velocity is -19.0 m/s.
Explanation:(a) To find the height that the ball reaches, we can use the kinematic equation:
Δy = v2 / (2g)
where Δy is the change in height, v is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have:
Δy = (19.0 m/s)2 / (2 * 9.8 m/s2)
Calculating, we find that the ball rises to a height of approximately 18.68 meters.
(b) The time it takes for the ball to reach its highest point can be calculated using the equation:
t = v / g
where t is the time, v is the initial velocity, and g is the acceleration due to gravity. Substituting the given values, we get:
t = (19.0 m/s) / (9.8 m/s2)
Simplifying, we find that it takes approximately 1.94 seconds for the ball to reach its highest point.
(c) Since the time it takes for the ball to reach its highest point is the same as the time it takes for it to fall back down, the time it takes for the ball to hit the ground after reaching its highest point is also 1.94 seconds.
(d) When the ball returns to the level from which it started, its velocity is equal in magnitude but opposite in direction to its initial velocity. Therefore, the velocity when it returns is -19.0 m/s.
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Given two vectors A⃗ =4.00i^+7.00j^ and B⃗ =5.00i^−2.00j^ , find the vector product A⃗ ×B⃗ (expressed in unit vectors).
Answer:
[tex]-43\hat{k}[/tex]
Explanation:
given,
[tex]\vec{A} = 4 \hat{i} + 7 \hat{j}[/tex]
[tex]\vec{B} = 5 \hat{i} - 2 \hat{j}[/tex]
vector product [tex] \vec{A} \times \vec{B} = ?[/tex]
[tex]\vec{A} \times \vec{B}[/tex] = [tex]\begin{bmatrix}i & j & k\\ 4 & 7 &0 \\ 5 & -2 & 0\end{bmatrix}[/tex]
now, expanding the vector
[tex]\vec{A} \times \vec{B}= \hat{k}(-2\times 4 - 7\times 5)[/tex]
[tex]\vec{A} \times \vec{B}= -43\hat{k}[/tex]
the vector product is equal to [tex]-43\hat{k}[/tex]
George determines the mass of his evaporating dish to be 3.375 g. He adds a solid sample to the evaporating dish, and the mass of them combined is 26.719 g. What must be the mass of his solid sample
Explanation:
The given data is as follows.
Mass of evaporating dish = 3.375 g
Total mass = Mass of solid sample + evaporating dish
That is, Mass of solid sample + evaporating dish = 26.719 g
Therefore, we will calculate the mass of solid sample as follows.
Mass of solid sample = (Mass of solid sample + evaporating dish) - mass of evaporating dish
= 26.719 g – 3.375 g
= 23.344 g
Thus, we can conclude that mass of his solid sample must be 23.344 g.
The mass of the solid sample is 23.344 g.
Explanation:In order to find the mass of the solid sample, we need to subtract the mass of the evaporating dish from the combined mass of the dish and the sample. The mass of the solid sample can be calculated by subtracting 3.375 g (mass of the evaporating dish) from 26.719 g (combined mass of dish and sample).
Mass of solid sample = 26.719 g - 3.375 g = 23.344 g.
Therefore, the mass of the solid sample is 23.344 g.
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By what factor does the energy of a 1-nm X-ray photon exceed that of a 10-MHz radio photon? How many times more energy has a 1-nm gamma ray than a 10-MHz radio photon?
To solve this problem we will apply the concepts related to the relationship between energy and frequency, from the latter we will obtain similar expressions that relate to the wavelength to find the two energy states between the given values. Finally we will make the comparative radius between the two. The relation between energy and frequency is given as,
[tex]E = hf[/tex]
Here,
E = Energy
h = Planck's constant
The relation between the speed of the electromagnetic waves (c), frequency (f) and wavelength ([tex]\lambda[/tex] ) is,
[tex]c = f\lambda[/tex]
Rearrange the above equation for frequency f as follows
[tex]f = \frac{c}{\lambda}[/tex]
Substitute,
[tex]E = h\frac{c}{\lambda}[/tex]
The wavelength x-ray or gamma ray photon is
[tex]\lambda = 1.0nm (\frac{1nm}{10^{9}nm})[/tex]
[tex]\lambda = 10^{-9} m[/tex]
Therefore the energy would be,
[tex]E_1 = \frac{hc}{\lambda}[/tex]
[tex]E_1 = \frac{(6.63*!0^{-34}J\cdo s)(3*10^{8}m/s)}{10^{-9}m}[/tex]
[tex]E_1 = 19.89*10^{-17} J[/tex]
The frequency is given as,
[tex]f = 10MHz (\frac{10^6z}{1.0MHz})[/tex]
[tex]f = 10^7Hz[/tex]
Now the second energy would be
[tex]E_2 = hf[/tex]
[tex]E_2 = (6.63*10^{-27}J\cdot s)(10^7Hz)[/tex]
[tex]E_2 = 6.63*10^{-27}J[/tex]
Therefore the ratio between them is
[tex]\frac{E_1}{E_2} = \frac{19.89*10^{-17}J}{6.63*10^{-27}J}[/tex]
[tex]\frac{E_1}{E_2} = 3*10^{20}[/tex]
Therefore the energy of 1nm x ray or gamma ray photon is [tex]3*10^{20}[/tex] times more than energy of 10MHz radio photon
The atomic radii of a divalent cation and a monovalent anion are 0.35 nm and 0.129 nm, respectively.(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).Enter your answer for part (a) in accordance to the question statement N(b) What is the force of repulsion at this same separation distance?
Answer:
a) The force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another) is - 2.01 × 10⁻⁹ N
b) The force of repulsion at this same separation distance is 2.01 × 10⁻⁹ N
Explanation:
F = kq₁q₂/r²
r = 0.35 + 0.129 (since the ions are just touching each other)
r = 0.479 nm = 4.79 × 10⁻¹⁰ m
Since the first ion is a divalent cation, Z₁ = +2 and the monovalent anion, Z₂ = -1
q = Ze; e = 1.602 × 10⁻¹⁹ C
K = 8.99 × 10⁹ Nm²/C²
F = (8.99 × 10⁹)(1.602 × 10⁻¹⁹)²(2)(-1)/(4.79 × 10⁻¹⁰)² = - 2.01 × 10⁻⁹ N
b) At equilibrium,
Force of attraction + Force of repulsion = 0
Force of repulsion = -(Force of attraction) = 2.01 × 10⁻⁹ N