Since there's no answer choices available, I would say "reward mechanisms in the brain", "pleasure", "Schizophrenia" or "appetite".
Answer:
A neurotransmitter used in the parts of the brain involved in regulating movement and experiencing pleasure.
Explanation:
Dopamine is one of the chemicals in our brain which controls mood Interacting with the pleasure and reward center of our brain, dopamine, along with other chemicals like oxytocin and endorphins plays a very crucial role in how we feel
The total power consumption by all humans on earth is approximately 1013 W. Let’s compare this to the power of incoming solar radiation. The intensity of radiation from the sun at the top of the atmosphere is 1380 W/m2. The earth's radius is 6.37×106 m.
Answer:
Power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.
Explanation:
Intensity of sunlight = I = 1380 w/m^2
Area of earth = A = 4*pi*r^2 = 4*pi*(6.37*10^6)^2 = 5.09*10^14 m^2
he intensity is defined as the total power spread over the area of earth (Area of Sphere with radius equal to distance between earth and sun) and given by the following formula:
Intenity of sunlight = Power/Area of earth
I = P/A
P = IA
P = (1380)(5.09*10^14)
P = 7.036*10^17 W
if we take ratio:
7.036*10^17/1013 = 6.94 * 10^14
Hence, power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.
A block of mass 2 kg is traveling in the positive direction at 3 m/s. Another block of mass 1.5 kg, traveling in the same direction at 4 m/s, collides elastically with the first block. Find the final velocities of the blocks. How much kinetic energy did the system lose
Answer:
a. The final velocity of the block of mass 2 kg is 3 m/s or 3.86 m/s. The final velocity of the block of mass 1.5 kg is 4 m/s or 2.86 m/s b. The kinetic energy change is 0 J or -12.235 J. Since the collision is elastic, we choose ΔK = 0
Explanation:
From principle of conservation of momentum,
momentum before impact = momentum after impact
Let m₁ = 2 kg, m₂ = 1.5 kg and v₁ = 3 m/s, v₂ = 4 m/s represent the masses and initial velocities of the first and second blocks of mass respectively. Let v₃ and v₄ be the final velocities of the blocks. So,
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
(2 × 3 + 1.5 × 4) = 2v₃ + 1.5v₄
6 + 6 = 2v₃ + 1.5v₄
12 = 2v₃ + 1.5v₄
2v₃ + 1.5v₄ = 12 (1)
Since the collision is elastic, kinetic energy is conserved. So
1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v₃² + 1/2m₂v₄²
1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 1/2 ×2v₃² + 1/2 × 1.5v₄²
9 + 12 = v₃² + 0.75v₄²
21 = v₃² + 0.75v₄²
v₃² + 0.75v₄² = 21 (2)
From (1) v₃ = 6 - 0.75v₄ (3) . Substituting v₃ into (2)
(6 - 0.75v₄)² + 0.75v₄² = 21
36 - 9v₄ + 0.5625v₄² + 0.75v₄² = 21
36 - 9v₄ + 1.3125v₄² - 21 = 0
1.3125v₄² - 9v₄ + 15 = 0
Using the quadratic formula,
v₄ = [-(-9) ± √[(-9)² - 4 × 1.3125 × 15]]/(2 × 1.3125)
= [9 ± √[81 - 78.75]]/2.625
= [9 ± √2.25]/2.625
= [9 ± 1.5]/2.625
= [9 + 1.5]/2.625 or [9 - 1.5]/2.625
= 10.5/2.625 or 7.5/2.625
= 4 m/s or 2.86 m/s
Substititing v₄ into (3)
v₃ = 6 - 0.75v₄ = 6 - 0.75 × 4 = 6 - 3 = 3 m/s
or
v₃ = 6 - 0.75v₄ = 6 - 0.75 × 2.86 = 6 - 2.145 = 3.855 m/s ≅ 3.86 m/s
b. The kinetic energy change ΔK = K₂ - K₁
K₁ = initial kinetic energy of the two blocks = 1/2m₁v₁² + 1/2m₂v₂²
= 1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 9 + 12 = 21 J
K₂ = final kinetic energy of the two blocks = 1/2m₁v₃² + 1/2m₂v₄². Using v = 3 m/s and v = 4 m/s
= 1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 9 + 12 = 21 J.
ΔK = K₂ - K₁ = 21 - 21 = 0
Using v = 3.86 m/s and v = 2.86 m/s
K₂ = 1/2 × 2 × 3.86² + 1/2 × 1.5 × 2.86² = 14.8996 - 6.1347 = 8.7649 J ≅ 8.765 J
ΔK = K₂ - K₁ = 8.765 - 21 = -12.235 J
Since the collision is elastic, we choose ΔK = 0
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is 0.53 kg/m3 at 7.5 km and 0.74 kg/m3 at 3.8 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.
Answer:
[tex]\frac{D_{jet}}{D_{prop}}=2.865[/tex]
Explanation:
Given data
Speed of jet Vjet=1190 km/h
Speed of prop driven Vprop=595 km/h
Height of jet 7.5 km
Height of prop driven transport 3.8 km
Density of Air at height 10 km p7.8=0.53 kg/m³
Density of air at height 3.8 km p3.8=0.74 kg/m³
The drag force is given by:
[tex]D=\frac{1}{2}CpAv^2\\[/tex]
The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:
[tex]\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865[/tex]
What is the name of the german scientist that proposed the theory of continental drift
Answer:
Alfred Wegener
Explanation:
Alfred Wegener was a German scientist who first discovered the idea of continental drift.
The continental drift hypothesis says that the large continents drift from one location to another over the broad ocean water bodies with respect to the fixed poles. This was the first step in understanding the interior of the earth and how the tectonic activities take place on earth.
He contributed many pieces of evidence in order to support this hypothesis, but it was initially not accepted as he was not able to explain the main mechanism for the continental motion.
Solar energy heats the surface of the earth including the ground rocks and even roadways as the temperatures of the surfaces increase heat energy is released back
Answer: into the atmosphere
Explanation:
As this energy is released into the atmosphere, bubbles of warm air is formed which is released and it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere, and is known as....
convection
While convection is a form of heat transfer from one place to the other involving the movement of fluids. So in the case narrated above the fluid which serves as a medium is Air.
Suppose you watch a leaf bobbing up and down as ripples pass it by in a pond. You notice that it does two full up and down bobs each second. Which statement is true of the ripples on the pond?
They have a frequency of 2 hertz. (hertz = cycles per second)
Answer:
They have a frequency of 2 hertz
Explanation:
Frequency is the number of occurrences of a repeating event per unit of time and Wavelength is the distance from one crest to another, or from one trough to another.
In the question, it is stated that the leaf does two full up and down bobs, this means that it completes 2 full cycles in one second. Therefore, its frequency is 2/s
where, s⁻¹ is hertz
so, They have a frequency of 2 hertz
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies.Express your answer in terms of Kinitial and Kfinal.W=
Answer:
[tex]W=K_f-K_i[/tex]
Explanation:
The work done on a particle by external forces is defined as:
[tex]W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,[/tex]
According to Newton's second law [tex]F=ma[/tex]. Thus:
[tex]W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\[/tex]
Acceleration is defined as the derivative of the speed with respect to time:
[tex]W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,[/tex]
Speed is defined as the derivative of the position with respect to time:
[tex]W=m\int\limits^{v_f}_{v_i} v \cdot dv \,[/tex]
Kinetic energy is defined as [tex]K=\frac{mv^2}{2}[/tex]:
[tex]W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i[/tex]
(a) Calculate the magnitude of the gravitational force exerted by the Moon on a 75 kg human standing on the surface of the Moon. (The mass of the Moon is 7.41022 kg and its radius is 1.7106 m.)
Answer:
128 N
Explanation:
Using
F = Gm'm/r²....................... Equation 1
Where F = Force, G = Universal constant, m = mass of the human, m' = mass of the moon, r = radius of the moon
Given: m = 75 kg, m' = 7.4×10²² kg, r = 1.7×10⁶ m
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 1
F = (6.67×10⁻¹¹ )(75)(7.4×10²²)/(1.7×10⁶)²
F = (3.7×10¹⁴)/(2.89×10¹²)
F = 1.28×10²
F = 128 N
Explanation:
Below is an attachment containing the solution.
A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density of 6.37 10 6 C m2 A charge of 0.370 µC is now introduced into the cavity inside the sphere a What is the new charge density on the outside of the sphere b Calculate the strength of the electric field just outside the sphere
Answer:
a. [tex]6.032\times10^{-6}C/m^2[/tex]
b.[tex]6.816\times10^5N/C[/tex]
Explanation:
#Apply surface charge density, electric field, and Gauss law to solve:
a. Surface charge density is defined as charge per area denoted as [tex]\sigma[/tex]
[tex]\sigma=\frac{Q}{4\pi r_{out}^2}[/tex], and the strength of the electric field outside the sphere [tex]E=\frac{\sigma _{new}}{\epsilon _o}[/tex]
Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.
[tex]\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2[/tex] #surface charge outside sphere.
[tex]\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2[/tex]
Hence, the new charge density on the outside of the sphere is [tex]6.032\times10^{-6}C/m^2[/tex]
b. The strength of the electric field just outside the sphere is calculated as:
From a above, we know the new surface charge to be [tex]6.032\times10^{-6}C/m^2[/tex],
[tex]E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C[/tex]
Hence, the strength of the electric field just outside the sphere is [tex]6.816\times10^5N/C[/tex]
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.
Answer:
The lowest possible frequency of sound is 971.4 Hz.
Explanation:
Given that,
Distance between loudspeakers = 2.00 m
Height = 5.50 m
Sound speed = 340 m/s
We need to calculate the distance
Using Pythagorean theorem
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=2.00^2+5.50^2[/tex]
[tex]AC=\sqrt{(2.00^2+5.50^2)}[/tex]
[tex]AC=5.85\ m[/tex]
We need to calculate the path difference
Using formula of path difference
[tex]\Delta x=AC-BC[/tex]
Put the value into the formula
[tex]\Delta x=5.85-5.50[/tex]
[tex]\Delta x=0.35\ m[/tex]
We need to calculate the lowest possible frequency of sound
Using formula of frequency
[tex]f=\dfrac{nv}{\Delta x}[/tex]
Put the value into the formula
[tex]f=\dfrac{1\times340}{0.35}[/tex]
[tex]f=971.4\ Hz[/tex]
Hence, The lowest possible frequency of sound is 971.4 Hz.
In a Millikan oil-drop experiment, a uniform electric field of 5.71 x 10^5 N/C is maintained in the region between two plates separated by 6.49 cm. Find the potential difference (in V) between the plates.
Answer:
37057.9V
Explanation:
Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.
Electric potential (E) = Potential Difference (V)/distance between plates(d)
Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m
Since E = V/d
V = Ed
V = 5.71×10^5×0.0649
V = 37057.9Volts
The potential difference (in V) between the plates is 37057.9V
Answer:
V = 3.71×10⁴ V
Explanation:
Potential difference: This can be defined as the work done in moving a positive charge from infinity to any point in an electric field.
The S.I unit of potential difference is Volt (V).
The expression for potential difference is
V = E×d.............................. Equation 1
Where V = potential difference between the plates, E = Electric field , d = distance of separation between the plates
Given: E = 5.71×10⁵ N/C, d = 6.49 cm = 0.0649 m.
Substitute into equation 1
V = 5.71×10⁵×0.0649
V = 3.71×10⁴ V
When a 70 kg man sits on the stool, by what percent does the length of the legs decrease? Assume, for simplicity, that the stool's legs are vertical and that each bears the same load.
The diameter of one leg of the stool is missing and it's 2cm.
Answer:
(ΔL/L) = 0.00729%
Explanation:
If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.
Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.
Now, weight of man is 70kg.
Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m
Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m
Now as stated earlier, the force on one leg is; F/3.
Now F = mg = 70 x 9.81 = 686.7N
So, force on one leg = 686.7/3 = 228. 9N
Now we know youngs modulus(Y) = Stress/Strain.
Stress = F/A while Strain = ΔL/L
Therefore Y = (F/A) / (ΔL/L)
And therefore, (ΔL/L) = F/(AY)
So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)
When expressed in percentage, it becomes 0.00729%
An accurate calculation for the percent decrease in the stool legs' length under a 70 kg man's weight requires knowledge of the materials properties and dimensions of the stool. General concepts of stress, strain, and deformation in materials are discussed, including their relationship to applied forces and material properties as described by Hooke's Law.
Explanation:To determine the percent decrease in the length of the stool legs when a 70 kg man sits on it, we need additional information such as the material properties of the stool legs and any relevant physical dimensions. However, with the given scenario, we can talk generally about stress and strain in materials and how they are calculated in relation to force and deformation. Stress is the force per unit area applied to a material, while strain is the deformation experienced by the material relative to its original length. For an object like a stool leg, the deformation (change in length) due to a person sitting on it could be calculated using Hooke's Law if the deformation remains within the linear elastic range of the material.
Without specific data on the stool's material, cross-sectional area, and elastic modulus, we cannot calculate the exact percent decrease in length of the stool's legs. If such information were provided, we could use the formula σ = F/A, where σ is the stress, F is the force, and A is the cross-sectional area; and the formula ε = δL/L, where ε is the strain, δL is the change in length, and L is the original length. Combined with Hooke's Law (σ = Eε, E being the elastic modulus), we could find the deformation and thus the percent decrease in length.
What is the coefficient of static friction between the coin and the turntable?
Answer:The coefficient of static friction between the turntable and the coin is 0.1
Explanation:
The coefficient of static friction is the friction force between two objects when neither of the objects is moving. ... A value of 1 means the frictional force is equal to the normal force. It is a misconception that the coefficient of friction is limited to values between zero and one.
Which is true about the spacing of the streamlines in a wire?
The statement " Close spacing represents greater current densities" is true about the spacing of the streamlines in a wire (option F)
Why is this correct?
In a scenario where a conductor is wider on the left and narrower on the right, the electric field lines and current density are depicted by streamlines. With current flowing from the wider end to the narrower end, the total charge and current remain constant. However, the current density fluctuates, being higher at the narrower end.
Hence, when observing streamlines, closer spacing within the wire signifies a higher current density, specifically in the narrower sections compared to the wider ones.
Complete question:
Which is true about the spacing of the streamlines in a wire?
A. Wide spacing represents faster random-motion velocities.
B. Wide spacing represents greater electric field vectors.
C. Wide spacing represents greater current densities.
D. Close spacing represents faster random-motion velocities.
E. Close spacing represents greater electric field vectors.
F. Close spacing represents greater current densities.
The wave function of a particle in a one-dimensional box of width L is Ψ(x) = Asin(πx/L). If we know the particle must be somewhere in the box, what must be the value of A? Express your answer in terms of L.
Answer: A = square root (2/L)
Explanation: find the attached file for explanation
Final answer:
The wave function of a particle in a one-dimensional box needs to be normalized. The normalization condition requires that the value of A for the wave function Ψ(x) = Asin(πx/L) is found to be A = √(2/L) after solving the normalization integral.
Explanation:
The wave function Ψ(x) of a particle in a one-dimensional box of width L must be normalized so that the total probability of finding the particle within the box is 1. This normalization condition implies that the integral of the square of the absolute value of the wave function over the interval from 0 to L should equal 1.
The normalization integral for the given wave function Ψ(x) = Asin(πx/L) is:
∫ |Asin(πx/L)|² dx = A² ∫ sin²(πx/L) dx = 1
When you solve the integral from 0 to L, the result is:
A² * L/2 = 1
Therefore, solving for A, we get:
A = √(2/L)
So the value of A in terms of L is √(2/L).
Explain Rutherford's experiment?
Answer:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
Explanation:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.
The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.
Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.
C2 = 20.0 μF, and C3 = 25.0 μF. If no capacitor can withstand a potential difference of more than 100 V without failure, what are (a) the magnitude of the maximum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?
The maximum potential difference between points A and B is 100 V, and the maximum energy stored in the three-capacitor arrangement is determined by the capacitor with the smallest capacitance.
The maximum potential difference (voltage) between points A and B in a series combination of capacitors is limited by the capacitor with the lowest breakdown voltage. Given C2 = 20.0 μF and C3 = 25.0 μF, and no capacitor can exceed 100 V, the maximum potential difference that can exist between points A and B is 100 V.
The maximum energy that can be stored in the three-capacitor arrangement is determined by the capacitor with the least energy storage capability. In this case, the capacitors are in series, so the total energy stored in the arrangement is the energy stored by the capacitor with the smallest capacitance, which is 20 μF, at the maximum allowed voltage of 100 V.
How many times did thomas edison fail before inventing the lightbulb
Answer:
he failed thousands of times
Explanation:
There is no known number for his failings. Edison may have failed in many of his experiments and in his schooling, but he had something better working in his favor. He had great determination and persistence.
He failed thousands of times in an attempt to develop an electric light, the great Edison simply viewed each unsuccessful experiment as the elimination of a solution that wouldn’t work, thereby moving him that much closer to a successful solution.
A block with mass m = 4.90 kg is placed against a spring on a frictionless incline with angle θ = 45° (The block is not attached to the spring). The spring, with spring constant k = 35.0 N/cm, is compressed 16.0 cm and then released.a.) What is the elastic potential energy of the compressed spring?b.) What is the change inn the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline?c.) How far along the inline is the highest point from the release point?
Answer:
a) 44.8J
b) 44.8J
c) 127cm
Explanation:
a) The elastic potential energy of a compressed spring is given by the formula:
[tex]U_e=\frac{1}{2} kx^{2}[/tex]
Where U_e is the elastic potential energy, k is the spring constant and x is the distance the spring is compressed. In this case, we have:
[tex]U_e=\frac{1}{2} (35.0N/cm)(16.0cm)^{2} = 4480Ncm[/tex]
To express the result in Joules, we have to use the fact that 1cm=0.01m. Then:
[tex]U_e=4480Ncm=4480N(0.01m)=44.8J[/tex]
In words, the elastic potential energy of the compressed spring is 44.8J.
b) Using the law of conservation of mechanical energy, we have that:
[tex]E_o=E_f\\\\U_{eo}+U_{go}+K_o=U_{ef}+U_{gf}+K_f[/tex]
Taking t=0 the moment in which the block is released, and t=t_f the point of its maximum height, we have that [tex]U_{g0}=0;K_0=0;U_{ef}=0;K_f=0[/tex] because in t=0 the block has no speed and is in tis lowest point; and in t=t_f the block has stopped and isn't in contact with the spring. So, our equation is reduced to:
[tex]U_{gf}=U_{e0}\\\\U_{gf}=44.8J[/tex]
So, the gravitational potential energy of the block in its highest point is 44.8J.
c) Using the gravitational potential energy formula, we have:
[tex]U_g=mgh\\\\\implies h=\frac{U_g}{mg} \\\\h=\frac{44.8J}{(4.90kg)(9.8m/s^{2}) }=0.9m=90cm[/tex]
Using trigonometry, we can compute the distance between the release point and its highest point:
[tex]d=\frac{h}{sin\theta}=\frac{90cm}{sin45\°}=127cm[/tex]
A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate the force (in newtons) that would need to be applied to the rope to just start the wagon moving.
Answer:
The force that would be applied on the rope just to start moving the wagon is 122 N
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/[tex]s^{2}[/tex])
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/[tex]s^{2}[/tex]
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/[tex]s^{2}[/tex]
F = 122.63 N
Therefore a force of 122 N must be applied to the rope just to start the wagon.
Explanation:
Below is an attachment containing the solution.
A municipal water supply is provided by a tall water tower. Water from this tower flows to a building. How does the water flow out of a faucet on the ground floor of a building compare with the water flow out of an identical faucet on the second floor of the building
Answer:
THE ANSWER IS: Water flows more rapidly out of the ground-floor faucet.
Explanation:
Water flows more rapidly out of the ground-floor faucet.
Why is the water flow more rapid out of a faucet on the first floor of a building than in an apartment on a higher floor?The first floor of a building has the biggest pressure differential, which is why water flows more quickly out of a faucet there than in an apartment on a higher floor. As we move up the structure, however, the pressure difference reduces.How much water does a water tower hold?An average water tower is usually about 165 feet (50 meters) tall, and its tank can hold about a million gallons of water or more.Why does water flow on the flour easily?Proteins: the higher protein content the higher water absorption.Pentosans: the higher the pentosans content the higher the water absorption.Is water pressure lower on higher floors?In actuality, the idea that a building's age affects water pressure is a fiction. However, it is true that in buildings where the roof tank serves as the supply of water, the water pressure at fixtures is lower in upper floor apartments than in lower level apartments.Learn more about municipal water supply here:
https://brainly.com/question/11961382
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If the temperature is held constant during this process and the final pressure is 703 torr , what is the volume of the bulb that was originally filled with gas
Answer:
[tex]2.42liter[/tex]
Explanation:
The question is incomplete; you need some additional data.
I will assume the missing data from a similar question and keep the final pressure of 703 torr given in your question.
An ideal gas at a pressure of 1.20 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.720 L as shown here. When the stopcock is opened the gas expands into the empty bulb
If the temperature is held constant during this process and the final pressure is 703 torr , what is the volume of the bulb that was originally filled with gas?
SolutionSince the amount of gas and the temperature remain constant, we may use Boyle's law:
[tex]PV=constant\\\\P_1V_1=P_2V_2[/tex]
Form the data:
[tex]P_1=1.20atm\\\\P_2=703torr\\\\V_1=unknown\\\\V_2=V_1+0.720liter[/tex]
1. Convert 703 torr to atm:
[tex]P_2=703torr\times 1atm/760torr=0.925atm[/tex]
2. Sustitute the data in the equation:
[tex]1.2atm\times V_1=0.925atm\times(V_1+0.720liter)[/tex]
3. Solve:
[tex]1.2V_1=0.925V_1+ 0.666liter\\\\0.275V_1=0.666liter\\\\V_1=2.4218liter\approx 2.42liter[/tex]
The answer is rounded to 3 significant figures, according to the data.
How does electric force depend on the amount of charge and the distance between charges
Explanation:
The attractive or repulsive forces which act between any two charged species is an electric force.The electric force depends on the distance between the charged species and the amount of charge which can be calculated by the formula given as followsF = k×[tex]\frac{q1q2}{r2}[/tex]
where, K is coulombs constant, which is equal to - 9 x10^9 [tex]Nm^2/C^2.[/tex]
The unit for K is newtons square meters per square coulombs.This is known as Coulomb's Law.A 2.0 ???????? capacitor and a4.0 ???????? capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitor?
Answer:
0.27J
Explanation:
[tex]C_eq= C_1 + C_2\\= 2+4\\= 6UF\\U = (1/2) CV^2\\= (1/2)(6 - 6)(300 * 300)\\= 0.27J[/tex]
A spectrophotometer measures the transmittance or the absorbance. True or False
Answer: FALSE
Explanation: Could you help me with a question?
Consider two uniform solid spheres where one has twice the mass and twice the diameter of the other. The ratio of the larger moment of inertia to that of the smaller moment of inertia is:_________.a) 2b) 8c) 4d) 10e) 6
Answer:
b) 8.
Explanation:
Below is an attachment containing the solution.
\ describes the size and distance relationship of our sun and the nearest star?
Answer: Two marbles separated by 300 kilometers
Explanation: Hope i helped have a great day and please mark brainliest i would appreciate it!
The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 hz. Which frequencies will not resonate in the same pipe?
Answer:
The frequency 400 hz is not possible .
Explanation:
Given that,
Frequency = 200 hz
Length = l
Suppose, The given frequencies are,
600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.
The possible resonance frequencies are
We need to calculate the fundamental frequency
Using formula of fundamental frequency for pipe
[tex]F=\dfrac{nv}{4L}[/tex]
Where, n = odd number
Put the value of frequency
[tex]200= \dfrac{nv}{4L}[/tex]
We need to calculate the first over tone
Using formula of fundamental frequency
n = 3,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{2}=3\times\dfrac{v}{4l}[/tex]
[tex]F_{2}=3\times200[/tex]
[tex]F_{2}=600\ Hz[/tex]
We need to calculate the second over tone
Using formula of fundamental frequency
n = 5,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{3}=5\times200[/tex]
[tex]F_{3}=1000\ Hz[/tex]
We need to calculate the third over tone
Using formula of fundamental frequency
n = 7,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{4}=7\times200[/tex]
[tex]F_{4}=1400\ Hz[/tex]
We need to calculate the fourth over tone
Using formula of fundamental frequency
n = 9,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{5}=9\times200[/tex]
[tex]F_{5}=1800\ Hz[/tex]
Hence, The frequency 400 hz is not possible .
Final answer:
In a pipe that is closed at one end and open at the other, only odd multiples of the fundamental frequency will resonate. Frequencies that are even multiples of the fundamental 200 Hz, such as 400 Hz, 600 Hz, 800 Hz, will not resonate.
Explanation:
The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 Hz. This pipe supports harmonic frequencies following the sequence fn = n(v/4L), where n = 1, 3, 5..., v is the speed of sound, and L is the length of the pipe. Therefore, frequencies that will not resonate in the same pipe are those that are even multiples of the fundamental frequency, such as 400 Hz, 600 Hz, 800 Hz, and so on.
How does the water table change around a pumping water well?
What is the significance of electron transport in the photochemical reactions of photosynthesis
Explanation:
Photosynthetic electron transport is helpful in the conversion of solar energy into chemical energy in the process of photosynthesis through transferring electrons sequentially from [tex]H_2O[/tex] through Photosystem II and Photosystem I to NADP+. Cyclically flowing electrons generate ATP molecules, because after passing down the first step of the electron transport chain protons are pumped into the thylakoid lumen, and establishes a gradient in between.However, cyclic electron flow does not involve in the formation of NADPH, nor does it involve in the splitting of water or production of oxygen.