Explanation:
Work done by gravity is given by the formula,
W = [tex]\rho A (h_{1} - h)g (h - h_{2})[/tex] ......... (1)
It is known that when levels are same then height of the liquid is as follows.
h = [tex]\frac{h_{1} + h_{2}}{2}[/tex] ......... (2)
Putting value of equation (2) in equation (1) the overall formula will be as follows.
W = [tex]\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})[/tex]
= [tex]\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})[/tex]
= 0.689 J
Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.
PART ONE
A square plate is produced by welding together four smaller square plates, each of side
a. The weight of each of the four plates is
shown in the figure.
Find the x-coordinate of the center of gravity (as a multiple of a).
Answer in units of a.
(PICTURED)
PART TWO
Find the y-coordinate of the center of gravity
(as a multiple of a).
Answer in units of a
Explanation:
Make a table, listing the x and y coordinates of each square's center of gravity and its mass. Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.
[tex]\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right[/tex]
The x-coordinate of the center of gravity is 15/14 a.
The y-coordinate of the center of gravity is 47/42 a.
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and particle 1 experiences an attractive force of 3.44 N.
What is the magnitude and sign of q2?
Answer:
Charge of particle 2, [tex]q_2=-7.13\ \mu C[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=3.11\ \mu C=3.11\times 10^{-6}\ C[/tex]
The distance between charges, r = 0.241 m
Force experienced by particle 1, F = 3.44 N
We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]q_2=\dfrac{Fr^2}{kq_1}[/tex]
[tex]q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}[/tex]
[tex]q_2=7.13\times 10^{-6}\ C[/tex]
or
[tex]q_2=7.13\ \mu C[/tex]
So, the magnitude of electric charge 2 is [tex]q_2=7.13\ \mu C[/tex]. Since, the force is attractive then the magnitude of charge 2 must be negative.
Final answer:
To determine the magnitude and sign of q2, Coulomb's Law is applied, revealing that q2 has a negative charge of approximately -4.48 μC, due to the attractive force observed.
Explanation:
To find the magnitude and sign of q2, we will use Coulomb's Law, which quantifies the electric force between two charged particles. Coulomb's Law is given by the formula F = k |q1*q2| / r^2, where F is the magnitude of the force between the charges, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
Given that q1 = +3.11 μC (or 3.11 x 10^-6 C), r = 0.241 m, and F = 3.44 N, and knowing that the force is attractive (indicating that q2 must have an opposite sign to q1), we can solve for q2 as follows:
F = k |q1*q2| / r^2
3.44 = (8.99 x 10^9) |(3.11 x 10^-6) * q2| / (0.241)^2
Solving for q2, we get q2 ≈ -4.48 x 10^-6 C. The negative sign indicates that q2 is negatively charged, and the magnitude of this charge is 4.48 μC.
What is the magnitude of the net gravitational force on the m1=20kgm1=20kg mass? Assume m2=10kgm2=10kg and m3=10kg.
Answer:
The net force on the 20kg mass m1 is equal to 6.09 x 10^-7 N. This force is due to the sun of the vertical components of the forces F1 and F2 of the masses m2 and m3 respectively on the mass m1.
The law of gravitational force has been applied and the use of the Pythagorean theorem also used.
Explanation:
The full solution can be found in the attachment below.
Thank you for reading this post and I hope it is helpful to you.
The magnitude of the net gravitational force on the masses can be determined using Newton's law when the distance between the two masses is known.
Magnitude of gravitational force
The magnitude of gravitaional force on the two given masses can be determined by applying Newton's law of universal gravitation as shown below;
F = Gm₁m₂/r²
where;
G is universal gravitation constantr is the distance between the two massesm₁ is the first mass m₂ is the second massThus, the magnitude of the net gravitational force on the masses can be determined using Newton's law when the distance between the two masses is known.
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Air is contained in a rigid well-insulated tank with a volume of 0.6 m3. The tank is fitted with a paddle wheel that transfers energy to the air at a constant rate of 4 W for 1 h. The initial density of the air is 1.2 kg/m3.
If no changes in kinetic or potential energy occur, determine:
(a) the specific volume at the final state, in m^3/kg
(b) the energy transfer by work, in kJ.
(c) the change in specific internal energy of the air, in kJ/kg.
Answer:
0.833 m³/kg
-14.4 kJ
20 kJ/kg
Explanation:
[tex]\rho[/tex] = Density = 1.2 kg/m³
[tex]V[/tex] = Volume = 0.6 m³
t = Time taken = 1 hour
P = Power = 4 W
Mass is given by
[tex]m=\rho V\\\Rightarrow m=1.2\times 0.6\\\Rightarrow m=0.72\ kg[/tex]
As there is no change in kinetic and potential energy, the specific volume of the tank will be unaffected
[tex]V_{s}=\dfrac{0.6}{0.72}\\\Rightarrow V_s=0.833\ m^3/kg[/tex]
The specific volume at the final state is 0.833 m³/kg
Energy is given by
[tex]E=Pt\\\Rightarrow E=-4\times 1\times 3600\\\Rightarrow E=-14400\ J[/tex]
The energy transfer by work is -14.4 kJ
Change in specific internal energy is given by
[tex]E=-m\Delta u\\\Rightarrow \Delta u=-\dfrac{E}{m}\\\Rightarrow \Delta u=-\dfrac{-14.4}{0.72}\\\Rightarrow \Delta u=20\ kJ/kg[/tex]
The change in specific internal energy is 20 kJ/kg
Answers:
0.833 m³/kg-14.4 kJ20 kJ/kgBy what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each object is 5.5 N
Answer:
Distance between both the object will be 404.51 m
Explanation:
We have given charge on two objects [tex]q_1=q_2=1C[/tex]
Coulomb force between the two objects is given F = 5.5 N
We have to fond the distance between both the object so that force between them is 5.5 N
According to coulomb law force between two charge particle is [tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}[/tex]
So [tex]5.5=\frac{9\times 10^9\times 1\times 1}{r^2}[/tex]
[tex]r^2=16.36\times 10^8[/tex]
r = 404.51 m
So distance between both object will be 404.51 m
A truck covers 45.0 m in 8.80 s while smoothly slowing down to final speed of 3.00 m/s. Find Its Original Speed.
Answer:
7.23 m/s
Explanation:
From Newton's equation of motion,
v = u + at ...................... Equation 1.
Where v = final velocity, u = initial velocity, a = acceleration, t = time.
Also,
s = ut+ 1/2at²........................ Equation 2
Where s = distance.
Given: t = 8.8 s, s = 45.0 m.
Substitute into equation 2
Note: we find the value of a in terms of u
45 = u(8.8)+1/2a(8.8)²
45 = 8.8u+38.72a
38.72a = 45 -8.8u
38.72a = (45-8.8u)
a = (45-8.8u)/38.72
also, v = 3.00 m/s
Substituting into equation 1
3 = u + 8.8[(45-8.8u)/38.72)]
3 = u + (45-8.8u)/4.4
3×4.4 = 4.4u + 45 - 8.8u
13.2 - 45 = 4.4u - 8.8u
-31.8 = -4.4u
u = -31.8/-4.4
u = 7.23 m/s.
Hence the initial velocity = 7.23 m/s
True / False:
O An ammeter must be placed in parallel with a resistor to measure the current through the resistor.
O A voltmeter is used to measure voltage.
O A voltmeter has a small internal resistance.
O An ammeter is used to measure current.
O A voltmeter must be placed in parallel with a resistor to measure the voltage across the resistor.
O An ammeter has a large internal resistance.
Answer: 1. False, an ammeter must be placed in a series connection to experience the same current which it is measuring
2. True, a voltmeter measures voltage (electric potential) in volts
3. False, a voltmeter should have high internal resistance such that it does not significantly alters current in a circuit.
4. True, a meters measure current
5. True, voltmeter would measure electric potential across a resistor
6. False, since they tend to influence the current in a circuit, ideally it should have very low resistance near zero
Explanation:
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of 20C
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
[tex]Q =\frac{KA(\delta T)}{L}[/tex]
[tex]Q =\frac{25.87*1*20}{1}[/tex]
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Answer: Rate of heat transfer q = 517.4W
Explanation:
Thermal conductivity is a material property that describes its ability to conduct heat. Thermal conductivity is the quantity of heat transmitted through a unit thickness of a material, in a direction normal to a surface of unit area,due to a unit temperature gradient under steady state conditions. It can be shown mathematically using the equation below;
Q/t = kA(∆T)/d
q = kA(∆T)/d
q = Q/t = rate of heat transfer
Q = total amount of heat transfer
t = time
k = thermal conductivity of material
A = Area
d= distance
For the case above, the material used is still air.
k for still air at 20°C and 1bara = 0.02587W/mK
A = 1m^2
d = 1mm = 0.001m
∆T = 20°C = 20K
q = 0.02587×1×20/0.001
q = 517.4W
Therefore, the rate of heat conduction through the still air is 517.4W
In testing thousands of different materials for use as lightbulb filaments, Thomas Edison best illustrated a problem-solving approach known as:
Group of answer choices
a. fixation.
b. belief perseverance.
c. trial and error.
d. the confirmation bias.
e. the representativeness heuristic.
Answer:
Trial and error.
Explanation:
This approach of Thomas Edison where in testing of thousand of different material for light bulb filament are used to find the most suitable material is called trial and error approach of problem solving.
Trial and error pursue a method, seeing if it performs, and not trying a new mechanism. This mechanism will be repeated until a solution or success is attained. Assume moving a large object like a couch into your house for example.
Starting with an initial value of P(0)equals30, the population of a prairie dog community grows at a rate of Pprime(t)equals30minusStartFraction t Over 2 EndFraction (in units of prairie dogs/month), for 0less than or equalstless than or equals60.
a. What is the population 9 months later?
b. Find the population P(t) for 0less than or equalstless than or equals60.
Answer:
a) The population of prairie dogs after nine months is 280.
b) P(t) = 30 + 30 · t - t²/4 for 0 ≤ t ≤ 60
Explanation:
Hi there!
We have the following information:
The initial population is P(0) = 30.
The rate of growth of the population is the following:
P´(t) = 30 - t/2 where
a) Let´s find the function of the population of prairie dogs P(t). For that, let´s integrate the P´(t) function between t = 0 and t and between P = 30 and P
P(t) = ∫P´(t)
P´(t) = dP/dt = 30 - t/2
Separating variables:
dP = (30 - t/2) dt
∫dP = ∫(30 - t/2) dt
P - 30 = 30 · t - t²/4
P(t) = 30 + 30 · t - t²/4
The population of prairie dogs at t = 9 months will be equal to P(9):
P(9) = 30 + 30(9) - (9)²/ 4
P(9) = 280 prairie dogs
The population of prairie dogs after nine months is 280.
b) P(t) = 30 + 30 · t - t²/4 (it was obtained in part a).
The population of the prairie dog community 9 months later would be 294 prairie dogs, and the population equation for the entire prairie dog community for t months is P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60.
Explanation:The equation given is P'(t) = 30 - t / 2. This is a differential equation, and to get P(t), we need to find the anti-derivative or integrate P'(t) with respect to time, t. The integral of P'(t) = ∫(30 - t / 2) dt = 30t - t^2/4.
Adding the initial condition, P(0) = 30, we find P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60. Therefore, a. The population of the prairie dog community 9 months later would be P(9) = 30*9 - (9^2) / 4 + 30 = 294 prairie dogs. b. P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60 is the population equation for the entire prairie dog community for t months.
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A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change the period to 2.04 s?
The mass that must be added is 0.628 kg
Explanation:
The period of a mass-spring system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
For the initial mass-spring system in the problem, we have
m = 0.500 kg
T = 1.36 s
Solving for k, we find the spring constant:
[tex]k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m[/tex]
In the second part, we want the period of the same system to be
T = 2.04 s
Therefore, the mass on the spring in this case must be
[tex]m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg[/tex]
Therefore, the mass that must be added is
[tex]\Delta m = 1.128 - 0.500 = 0.628 kg[/tex]
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A pulse waveform with a frequency of 10 kHz is applied to the input of a counter. During 100 ms, how many pulses are counted?
Answer:
n = 1000 pulses
Explanation:
Given that,
The frequency of a pulse waveform, [tex]f=10\ kHz=10^4\ Hz[/tex]
To find,
The number of pulses counted during 100 ms.
Solution,
The frequency of a pulse waveform is equal to the number of pulses per unit time. It is given by :
[tex]f=\dfrac{n}{t}[/tex]
[tex]n=f\times t[/tex]
[tex]n=10^4\ Hz\times 100\times 10^{-3}\ s[/tex]
n = 1000 pulses
So, there are 1000 pulses counted in a pulse waveform.
the earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. where did this energy come from>
Explanation:
Earth or any planet are actually born from huge clouds of gas and dust. Their stellar mass are fairly distributed at a radius from the axis of rotation. Gravitational force cause the cloud to come together. Now the whole gathered in smaller area. Now, individual particles come close to the roational axis. Thus, decreasing the moment of inertia of the planet.
As
I=mr^2
reducing r reduces I. However, the angular moment of the system remains always conserved. So, to conserve the angular momentum the angular velocity of the planet increases and so did the otational kinetic energy
The additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.
The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed because of the conservation of angular momentum. As the cloud collapsed under its own gravity, the rotation rate increased to conserve angular momentum, which is the product of moment of inertia and rotational velocity. Since the moment of inertia decreases as the mass moves closer to the axis of rotation (due to the shrinking size of the collapsing cloud), the rotational velocity must increase to keep the angular momentum constant. This results in an increase in rotational kinetic energy, which is given by the equation [tex]\( KE_{rot} = \frac{1}{2} I \omega^2 \)[/tex], where [tex]\( I \)[/tex] is the moment of inertia and [tex]\( \omega \)[/tex] is the angular velocity.
The gravitational potential energy of the cloud is converted into kinetic energy as the cloud collapses. The initial potential energy is high because the particles in the cloud are far from the center of mass. As the cloud contracts, this potential energy is transformed into kinetic energy, both linear and rotational, due to the conservation of energy. The increase in rotational kinetic energy is a direct consequence of this conversion and the conservation of angular momentum.
In summary, the additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.
The complete question is:
The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?
A large truck breaks down out on the road and receives a push back to town by a small compact car.
Pick one of the choices A through F below which correctly describes the forces that the car exerts on the truck and the truck exerts on the car for each of the questions. You may use an answer more than once or not at all.
A. The force exerted by the car pushing against the truck is equal to that exerted by the truck pushing back against the car.
B. The force exerted by the car pushing against the truck is less than that exerted by the truck pushing back against the car.
C. The force exerted by the car pushing against the truck is greater than that exerted by the truck pushing back against the car.
D. The car's engine is running so it exerts a force as it pushes against the truck, but the truck's engine isn't running so it can't exert a force back against the car.
E. Neither the car nor the truck exert any force on each other. The truck is pushed forward simply because it is in the way of the car.
F. None of these descriptions is correct.
If the car and truck are moving at cruising speed, which choice(s) below are true?
1. The car is pushing on the truck, but not hard enough to make the truck move.
2. The car, still pushing the truck, is speeding up to get to cruising speed.
3. The car, still pushing the truck, is at cruising speed and continues to travel at the same speed.
4. The car, still pushing the truck, is at cruising speed when the truck puts on its brakes and causes the car to slow.
Newton's Third Law states the forces between the car and the truck are equal and opposite. Thus, the force the car exerts on the truck is equal to the force the truck exerts on the car, making Choice A correct for all situations described, including when the car and truck are at cruising speed or if the truck puts on its brakes.
Explanation:According to Newton's Third Law of Motion, every action has an equal and opposite reaction. This law explains the forces between two objects—the car and the truck—when one object exerts force on another. In this scenario, when the compact car pushes the large truck:
Choice A is correct. The force exerted by the car on the truck is equal to the force exerted by the truck on the car.Choices B, C, and D are incorrect because they misrepresent Newton's Third Law.Choice E is incorrect as it suggests no force interaction, which is not possible according to Newton's laws.Choice F is incorrect because there is a correct description provided in Choice A.Therefore, for the situations described:
Choice A is the correct answer.Choice A remains correct as the action-reaction forces are still present.When the car and truck are moving at cruising speed (Choice A), the forces are equal, even at a constant speed.If the truck puts on its brakes, the car will experience a change in motion, but the forces between the car and truck will remain equal (Choice A).A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It is to deliver a mass flow rate of 0.01 kg/s air to a pipeline. Assuming a constant-pressure specific heat of Cp = 1.004 kJ/kg-K for the air, determine the maximum possible exit pressure of the compressor
Answer:
[tex]P_2=4091\ KPa[/tex]
Explanation:
Given that
T₁ = 290 K
P₁ = 100 KPa
Power P =5.5 KW
mass flow rate
[tex]\dot{m}= 0.01\ kg/s[/tex]
Lets take the exit temperature = T₂
We know that
[tex]P=\dot{m}\ C_p (T_2-T_1)[/tex]
[tex]5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K[/tex]
If we assume that process inside the compressor is adiabatic then we can say that
[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}[/tex]
[tex]\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\[/tex]
[tex]2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}[/tex]
[tex]P_2=40.91\times 100 \ KPa[/tex]
[tex]P_2=4091\ KPa[/tex]
That is why the exit pressure will be 4091 KPa.
Waves on a string are described by the following general equation
y(x,t)=Acos(kx−ωt).
A transverse wave on a string is traveling in the +xdirection with a wave speed of 8.75 m/s , an amplitude of 6.50×10−2 m , and a wavelength of 0.540 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.59 m and t = 0.150 s .
Q. In general, the cosine function has maximum displacements, either positive or negative, when its argument is equal to an integer multiple of π. When t = 0.150 s , k = 11.6 rad/m , and ω = 102 rad/s use the wave equation to select all of the x positions that correspond to points of maximum displacement.
Check all that apply considering only positive arguments of the cosine function.
0.795 m
1.59 m
1.73 m
1.86 m
2.00 m
2.13 m
Answer:
The displacement is at x=1.59m and t=0.150s is [tex]-6.50\cdot10^{-2}\text{ m}.[/tex]
Out of the given points, the argument of the cosine is an integer multiple of [tex]\pi[/tex] for x=1.59m, 1.86m, 2.13m.
Explanation:
The displacement at x and t is given by y(x,t). We now need to find [tex]k[/tex] and [tex]\omega[/tex]. The speed of the wave is given by
[tex]c=\frac{\omega}{k}[/tex]
while the wavelength satisfies
[tex]k=\frac{2\pi}{\lambda}=\frac{2\pi}{0.540\text{ m}}=11.6\text{ m}^{-1}.[/tex]
Substituting this into the previous equation we find
[tex]\omega=ck=8.75\text{ m/s}\cdot 11.6\text{ m}^{-1}=102\text{ rad/s}.[/tex]
Now we have
[tex]y(1.59\text{ m},0.150\text{ s})=6.50\cdot10^{-2}\text{ m}\cos(11.6\cdot1.59-102\cdot0.150)=-6.50\cdot10^{-2}\text{ m}.[/tex]
Now we calculate [tex]kx-\omega t[/tex] at [tex]t=0.150\text{ s}[/tex] at each given x and check whether it is integer multiple of [tex]\pi[/tex].
[tex]11.6\text{ m}^{-1}\cdot 0.795\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=-6.08=-1.93\pi\text{ not an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 1.59\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=3.14=\pi\text{ it is an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 1.73\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=4.77=1.52\pi\text{ not an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 1.86\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=6.28=2\pi\text{ it is an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 2.00\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=7.90=2.51\pi\text{ not an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 2.13\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=9.41=3\pi\text{ it is an integer multiple of }\pi;[/tex]
Final answer:
The wave function y(x, t) = Acos(kx−ωt) can be used to find x positions of maximum displacement for a transverse wave on a string.
Explanation:
The wave function y(x, t) = Acos(kx−ωt) represents a transverse wave on a string. Given the values of k = 11.6 rad/m and ω = 102 rad/s, we can find the x positions that correspond to points of maximum displacement by setting the argument of the cosine function to be an integer multiple of π.
Using the wave equation, the x positions that correspond to points of maximum displacement are:
0.795 m
1.59 m
2.00 m
2.13 m
A tuned mass damper for a skyscraper consists of a mass–spring system with spring constant 0.288 M/m. What should be its mass if it’s to oscillate with a period of 5.91 s
Answer:
m = 0.255kg
Explanation:
from the formular of a mass - spring system
T = 2π√m/k
making m as the subject of formular
m = T² k/4π²
T =5.91s
k = 0.288 N/m
m = 10.059/39.489
m = 0.255kg
Wilma I. Ball walks at a constant speed of 5.93 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.
(a)
What is Wilma's average speed over the entire trip?Wilma I. Ball walks at a constant speed of 4.51 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.
(a)
What is Wilma's average speed over the entire trip?
(b)
What is Wilma's average velocity over the entire trip?
Wilma's average speed over the entire trip is approximately 3.71 m/s, calculated by dividing the total distance by the total time. However, as she returned to her starting point, her total displacement is zero, and thus, her average velocity is 0 m/s.
Explanation:The average speed is given by the total distance divided by the total time. Since Wilma I. Ball travels the same distance twice (A to B and back), we can say the total distance is 2d. Suppose the distance between A to B is 'd' m. The time taken for the first trip is d/4.51 s, and the time taken for the return trip is d/3.15 s. Therefore, the average speed is given by (2d) / ((d/4.51) +(d/3.15)). This simplifies to approximately 3.71 m/s.
However, average velocity is the total displacement divided by total time. Since she ends at her starting point, her total displacement is zero. Thus, her average velocity over her round trip is 0 m/s.
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The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long, 710 yards wide, and 112 ft high. what is the cubic volume in feet, convert your result from part a to cubic meters.
Answer with Explanation:
We are given that
a.Length building=632 m
1 m=3.2808 feet
632 m=[tex]3.2808\times 632=2073.47 feet[/tex]
Length of building,l=2073.47 feet
Width of building,b=710 yards=[tex]710\times 3=2130 ft[/tex]
1 yard=3 feet
Height of building,h=112 ft
Volume of building=[tex]l\times b\times h[/tex]
Volume of building=[tex]2073.47\times 2130\times 112=494647003.2ft^3[/tex]
Volume of building=494647003.2 cubic feet
b. 1 cubic feet=0.028 cubic meter
Volume of building= [tex]494647003.2\times 0.028[/tex]cubic meters
Volume of building=13850116.0896 cubic meters
A rigid container equipped with a stirring device contains 1.5 kg of motor oil. Determine the rate of specific energy increase when heat is transferred to the oil at a rate of 1 W and 1.5 W of power is applied to the stirring device.
To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.
Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,
[tex]\dot{U} = \dot{Q}-\dot{W}[/tex]
Substitute 1W for [tex]\dot{Q}[/tex] and 1.5 W for [tex]\dot{W}[/tex]
[tex]\dot{U} = 1-1(1.5)[/tex]
[tex]\dot{U} = 2.5W[/tex]
Now calculcate the rate of specific internal energy increase,
[tex]\dot{u} = \frac{\dot{U}}{m}[/tex]
[tex]\dot{u} = \frac{2.5}{1.5}[/tex]
[tex]\do{u} = 1.6667W/kg[/tex]
The rate of specific internal energy increase is 1.6667W/kg
To answer the student's question, combine the total power input from heat and stirring, then calculate the temperature increase of the oil using its specific heat capacity with the energy conservation principle.
Explanation:The student is asking about the rate of specific energy increase when heat is transferred to motor oil while work is being done on it by stirring. In thermodynamics, this relates to the conversion of work into thermal energy. The first step is to calculate the total energy input per second (power), which is the sum of heat transfer and the work done by stirring. The next step is to use the specific heat capacity to determine the rate of temperature increase for the oil.
Given the specific heat capacity of water and a weight lifting scenario, the student needs to apply the energy conservation principle, converting the work done into the thermal energy, which will result in raising the temperature of the water. To find this temperature increase, one should use the formula Q = mcΔT, where Q is the energy transferred as heat, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.
The velocity of P-waves in the crust is ~ 7 km/s. Of the epicenter of an M 7.6 earthquake occurred 280 km from the closest seismic station, how long does it take the P-wave to arrive at the station
Answer:
t = 40 s
Explanation:
given,
Speed of the P-wave = 7 km/s
distance of the seismic station = 280 Km
time taken by the P-wave = ?
we know,
distance = speed x time
[tex]t = \dfrac{d}{s}[/tex]
[tex]t = \dfrac{280}{7}[/tex]
t = 40 s
time taken by the P-wave to arrive at the station is equal to 40 s.
Final answer:
The P-wave takes approximately 40 seconds to arrive at the closest seismic station.
Explanation:
The P-waves in the crust travel at a velocity of approximately 7 km/s. To determine the time it takes for the P-wave to arrive at a seismic station, we can use the equation:
Time = Distance / Velocity
In this case, the distance to the epicenter of the earthquake is given as 280 km. Plugging this value into the equation, we get:
Time = 280 km / 7 km/s = 40 seconds
So, it takes approximately 40 seconds for the P-wave to arrive at the closest seismic station.
Estimate the acceleration you subject yourself to if you walk into a brick wall at normal walking speed. Make a reasonable estimate of your speed and the time it takes you to come to a stop. Explain your answer!
Answer:
Walking into a brick wall at normal walking speed (1.4 m/s), you will come to a complete stop in a short time (0.1 s) and experience much more acceleration (14 m/s²) back the way you came, but because you are softer than the wall, the inelastic collision will probably cause you to bounce back off the wall, changing the actual experienced acceleration you feel.
Explanation:
Assuming normal human walking speeds and time it takes to come to rest.
Normal human walking speed = 5km/h = 1.4 m/s
Time it takes you to come to a complete stop = 0.1 seconds
acceleration = Δ Velocity/ Time
Δ Velocity = final velocity - initial velocity
Δ Velocity = (1.4 - 0)m/s = 1.4 m/s
acceleration = 1.4/0.1
acceleration = 14 m/s²
Walking into a brick wall at normal walking speed, you will experience much more acceleration back the way you came, but because you are softer than the wall, the inelastic collision will probably cause you to bounce back off the wall, changing the actual experienced acceleration you feel.
Acceleration of the subject.
The acceleration is the rate at which the body velocity changes with time. It can be referred to the body the speed and direction. The point of the object moves in the straight line as acceleration is both a magnitude and a direction. Estimated acceleration of the subject and yourself
thus the answer is 1.4 m/s and stops at 0.1s
As per the question of you walking into the brick wall at a normal walking speed than then walking towards the wall at normal walking at speed is 1.4 m, you will completely stop at a short time of 0.1 sec. The experience much more acceleration (14 m/s²) back the way you came, but due to the inelastic collision that probably causes you to bounce back by the wall, which will change the initial experience you felt.Hence answer is 1.4 and 01 sec.Learn more about the acceleration.
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Tests reveal that a normal driver takes about 0.75s before he or she can react to a situation to avoid a collision. It takes about 3s for a driver having 0.1% alcohol in his system to do the same.
If such drivers are traveling on a straight road at 30mph (44 )ft/s and their cars can decelerate at 24ft/s^2 , determine the shortest stopping distance d for normal driver from the moment he or she see the pedestrians.
Also, Determine the shortest stopping distance for drunk driver from the moment he or she see the pedestrians.
Answer:
NORMAL DRIVER: d = 73.3 ft
DRUNK DRIVER: d = 172.3
Explanation:
NORMAL DRIVER:
Distance covered in initial 0.75s = 0.75s *44 = 33ft
USING THE THIRD EQUATION OF MOTION
V^2-U^2 = 2as
0-(44)^2 = 2 (-24) s
s = 1936/48 =40.3 ft
d = 33 + 40.3 = 73.3 ft
DRUNK DRIVER:
Distance covered in initial 3s = 3s *44 = 132 ft
USING THE THIRD EQUATION OF MOTION
V^2-U^2 = 2as
0-(44)^2 = 2 (-24) s
s = 1936/48 =40.3 ft
d = 132 + 40.3 = 172.3 ft
A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 94 cm. Calculate the focal length of the mirror.
Answer:
The focal length of the concave mirror is -15.5 cm
Explanation:
Given that,
Height of the object, h = 20 cm
Radius of curvature of the mirror, R = -31 cm (direction is opposite)
Object distance, u = -94 cm
We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :
R = 2f
f is the focal length
[tex]f=\dfrac{R}{2}[/tex]
[tex]f=\dfrac{-31}{2}[/tex]
f = -15.5 cm
So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.
List and explain briefly similarities and differences between the electric force between two charges and the gravitational force between two masses.
Answer:
Please see below as the answers are self-explanatory.
Explanation:
Similarities1) The resultant force is along the line that joins both charges or both masses (assuming both objects can be represented as points)
2) Both type of forces obey Newton's 3rd law.
3) Both are proportional to the product of the property that is affected by the force (charges and masses)
4) Both obey an inverse - square law (consequence of our universe being three-dimensional)
Differences1) Main difference, is that while the gravitational force is always attractive, the electrostatic force can be attractive or repulsive, as there are two types of charges, which attract each other being of different type, and repel each other if they are of the same type.
2) It is possible, artificially, to block the influence of the electrostatic force, shielding a room, for instance, which is not possible for the gravitational force.
A strong man is compressing a lightweight spring between two weights. One weight has a mass of 2.3 kg , the other a mass of 5.3 kg . He is holding the weights stationary, but then he loses his grip and the weights fly off in opposite directions. The lighter of the two is shot out at a speed of 6.0 m/s .
What is the speed of the heavier weight?
To solve this problem we will apply the concepts related to the conservation of momentum. For this purpose we have that the initial momentum must be equivalent to the final momentum of the system. Mathematically this can be expressed as
[tex]m_1v_1 = m_2v_2[/tex]
Here,
[tex]m_{1,2}[/tex] = Mass of each object
[tex]v_{1,2}[/tex]= Velocity of each object
Rearranging to find the speed of the heavier weight,
[tex]v_2 = \frac{m_1v_1}{m_2}[/tex]
[tex]v_2 = \frac{(2.3)(6)}{5.3}[/tex]
[tex]v_2 = 2.6m/s[/tex]
Therefore the speed of the heavier weight is 2.6m/s
What equation gives the position at a specific time for an object with constant acceleration? a. x=x0+v0t+1/2}at^2b. x=v0t+at^2c. vf=v0+atd. v^2f=v0^2+2aΔx
The first option is mathematically described as
[tex]x = x_0 +v_0 t +\frac{1}{2} at^2[/tex]
Here,
[tex]x_0 =[/tex]Initial position
[tex]v_0 =[/tex] Initial velocity
t = Time
a = Acceleration
As we can see in this equation, the position of a body is described taking into account its initial point with respect to the reference system, the initial velocity of this body and the acceleration when it is constant. All this depending on time.
The second option despises the initial position, so it does not allow the exact calculation of the position.
The third option does not consider the position, only the speed and acceleration with respect to time
The fourth option considers acceleration and distance but does not take into account the time of the object.
Therefore the correct answer is A.
A girl delivering newspapers covers her route by traveling 5.00 blocks west, 5.00 blocks north, and then 7.00 blocks east. What is her final position relative to her starting location?
Answer:
P = 2i + 5j
Therefore she is 2 blocks east and 5 blocks north.
Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks
Angle = taninverse (5/2)
Angle = 68.2°
Explanation:
Given:
Let west be negative and east be positive x axis.
Let north be the positive y axis.
5.00blocks west = -5.00 i
5.00 blocks north = 5.00 j
7.00 blocks east = 7.00i
Addition of the vector form of hee position is;
P = -5i +7i -5j
P = 2i + 5j
Therefore she is 2 blocks east and 5 blocks north.
Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks
Angle = taninverse (5/2)
Angle = 68.2°
A gadget of mass 21.85 kg floats in space without motion. Because of some internal malfunction, the gadget violently breaks up into 3 fragments flying away from each other. The first fragment has mass m1 = 6.42 kg and speed v1 = 6.8 m/s while the second fragment has mass m2 = 8.26 kg and speed v2 = 3.54 m/s. The angle between the velocity vectors ~v1 and ~v2 is θ12 = 64 ◦ . What is the speed v3 of the third fragment? Answer in units of m/s.
The speed of third fragment of the given gadget is 8.62 m/s.
The given parameters:
total mass of the gadget, Mt = 21.85 kgmass of first fragment, m₁ = 6.42 kgmass of the second fragment, m₂ = 8.26 kgspeed of the first fragment, v₁ = 6.8 m/sspeed of the second fragment, v₂ = 3.54 m/sangle between the first and second fragment, θ = 64⁰The mass of the third fragment is calculated as follows;
[tex]m_3 = 21.85-(6.42 + 8.26)\\\\m_3 = 7.17 \ kg[/tex]
Apply the principle of conservation of linear momentum to determine the speed of the third fragment as follows;
[tex]m_3v_3 = m_1v_1 cos(\frac{\theta}{2} ) \ + \ m_2v_2 cos(\frac{\theta}{2} )\\\\7.17v_3 = 6.42\times 6.8 \times cos(\frac{64}{2} ) \ + 8.26 \times 6.8 \times cos(\frac{64}{2} )\\\\7.17 v_3 = 61.82 \\\\v_3 = \frac{61.82}{7.17} \\\\v_3 = 8.62 \ m/s[/tex]
Thus, the speed of third fragment of the given gadget is 8.62 m/s.
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Using conservation of momentum and resolving in x and y directions, we find the speed of the third fragment to be approximately 8.69 m/s. Initial conditions ensure total momentum is zero.
To solve this, we use the conservation of momentum. The total initial momentum of the system is zero since the gadget was initially at rest. For the fragments, the equation can be set up as follows:
m₁ * v₁ + m₂ * v₂ + m₃ * v₃ = 0
First, we resolve this in the x and y directions.
In the x-direction:
6.42 * 6.8 * cos(0°) + 8.26 * 3.54 * cos(64°) + 7.17 * v₃x = 0
⇒ 6.42 * 6.8 * 1 + 8.26 * 3.54 * 0.438 + 7.17 * v₃x = 0
⇒ 43.656 + 12.807 + 7.17 * v₃x =0
⇒ 7.17 * v₃x = - 56.463
v₃x = -7.88 m/s
In the y-direction:
6.42 * 6.8 * sin(0°) + 8.26 * 3.54 * sin(64°) + 7.17 * v₃y = 0
⇒ 6.42 * 6.8 * 0 + 8.26 * 3.54 * 0.899 + 7.17 * v₃y = 0
⇒ 26.281 + 7.17 * v₃y =0
⇒ 7.17 * v₃y = - 26.281
v₃y = -3.67 m/s
Then, using Pythagorean theorem to find v3:
v₃ = √(v₃x² + v₃y²)
= √((-7.88)² + (-3.67)²)
≈ √(62.1+ 13.5)
≈ √75.6
≈ 8.69 m/s
Thus, the speed of the third fragment is approximately 8.69 m/s.
How strong is the electric field between two parallel plates 4.8 mm apart if the potential difference between them is 220 V?
Answer:
Electric field, E = 45833.33 N/C
Explanation:
Given that,
Separation between the plates, d = 4.8 mm = 0.0048 m
The potential difference between the plates, V = 220 volts
We need to find the electric field between two parallel plates. The relation between the electric field and electric potential is given by :
[tex]V=E\times d[/tex]
[tex]E=\dfrac{V}{d}[/tex]
[tex]E=\dfrac{220}{0.0048}[/tex]
E = 45833.33 N/C
So, the electric field between two parallel plates is 45833.33 N/C. Hence, this is the required solution.
The electric field between the plates is 45833.33 N/C.
Electric Field
Given that the separation d between the plates is 4.8 mm and the potential difference V between the plates is 220 V.
The electric field E between the plates can be calculated by the formula given below.
[tex]E =\dfrac {V}{d}[/tex]
Substituting the values in the above equation, we get
[tex]E = \dfrac {220}{0.0048}[/tex]
[tex]E = 45833.33 \;\rm N/C[/tex]
Hence we can conclude that the electric field between the plates is 45833.33 N/C.
To know more about the electric field, follow the link given below.
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