Answer:
force on the proton = 5.96 × [tex]10^{-14}[/tex] N
Explanation:
given data
kinetic energy = 1.16 ×[tex]10^{5}[/tex] eV
charge density of σ = +6.60 μC/m²
solution
we get here force on the proton that is express as
F = qE ....................1
here q is charge on proton i.e = 1.6 × [tex]10^{-19}[/tex] C
and E is electric field due to charge i.e E = [tex]\frac{\sigma }{2*\epsilon_o }[/tex]
so put the value in equation 1 we get
force on the proton = 1.6 × [tex]10^{-19}[/tex] × [tex]\frac{6.60*10^{-6}}{2*8.85*10^{-12}}[/tex]
force on the proton = 5.96 × [tex]10^{-14}[/tex] N
Final answer:
The magnitude of the force on the proton is 1.19 × 10-13 N.
Explanation:
In order to find the magnitude of the force on the proton, we can use the formula F = qE, where F is the force, q is the charge of the proton, and E is the electric field. First, we need to convert the charge density from μC/m² to C/m², which gives us σ = 6.60 × 10-6 C/m². Since the electric field is uniform, we can use the equation E = σ/ε₀, where ε₀ is the electric constant.
Plugging in the values, we have E = (6.60 × 10-6 C/m²) / (8.85 × 10-12 C²/N·m²), which gives us E = 7.46 × 10^5 N/C. Now, we can find the force using F = qE.
F = (1.60 × 10-19 C)(7.46 × 10^5 N/C) = 1.19 × 10-13 N.
A fireworks rocket explodes at a height of 120 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of 7.20 10⁻² W/m² for 0.205 s.
(a) What is the total amount of energy transferred away from the explosion by sound?
(b) What is the sound level in decibels heard by the observer?
Answer:
2670.90667586 J
108.573324964 dB
Explanation:
r = Distance = 120 m
A = Area = [tex]4\pi r^2[/tex]
I = Intensity of sound = [tex]7.2\times 10^{-2}\ W/m^2[/tex]
t = Time taken = 0.205 s
[tex]I_0[/tex] = Threshold intensity = [tex]10^{-12}\ W/m^2[/tex]
Power is given by
[tex]P=IA\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2[/tex]
Energy is given by
[tex]E=Pt\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2\times 0.205\\\Rightarrow E=2670.90667586\ J[/tex]
The total amount of energy is 2670.90667586 J
Sound intensity level is given by
[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow \beta=10log\dfrac{7.2\times 10^{-2}}{10^{-12}}\\\Rightarrow \beta=108.573324964\ dB[/tex]
The sound level is 108.573324964 dB
The total amount of energy transferred away from the explosion by sound is 1.476 x 10^-2 Joules. The sound level in decibels heard by the observer is 97.50 dB.
Explanation:(a) To find the total amount of energy transferred away from the explosion by sound, we can use the formula:
Energy = Intensity x Time
Substituting the given values:
Energy = (7.20 x 10-2 W/m2) x (0.205 s) = 1.476 x 10-2 Joules
(b) The sound level in decibels heard by the observer can be calculated using the formula:
Sound Level (dB) = 10 x log10(Intensity / Threshold Intensity)
Substituting the given values:
Sound Level (dB) = 10 x log10((7.20 x 10-2 W/m2) / (10-12 W/m2)) = 97.50 dB
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A 58.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of 5.00 m. What is the work done on the man by (a) the gravitational force and (b) the escalator
Answer
a) -2842 J
b) 2842 J
Explanation:
The escalator is moving the man of mass 58 kg till the height of 5 meters
So here;
m=58 kg
distance = height = h= 5 m
gravitational acceleration = g= 9.8 m/sec2
a) We know work is scalar product of force and displacement
i-e Work = F.S= FS cosθ, where θ is the angle between force vector and displacement vector
When escalator is moving up - the gravitational force is acting downward i- opposite to displcacement i-e angle between force and displacement is 180 degrees.
we have data
m= 58 kg
F=mg = 58×9.8=568.4 N
S=5 m
Work = F.S=FS cos 180°=568.4×5×(-1)= -2842 J
So the work done by gravitational force is -2842 J and -ve sign here indicates that distance is traveled in opposite direction of force.
b) When escalator is moving up the, force is exerted by the escalator equal to gravitational force and the displacement is in the same direction so the angle between force and displacement is 0 degrees
Work = F.S=FS cos (0)=568.4×5×1=2842 J
So work done by the gravitational force then will be 2842 J
So in both cases work is same but in opposite directions.
Beginning about 55 seconds into the video, you'll see an animation of a photographer looking through her camera at a man, a set of trees, and distant mountains. Notice that, as viewed through the camera, the positions of the man and the trees change (relative to distant mountains) as the photographer moves. Which of the following statements correctly describes what is really happening in this situation?
Answer and Explanation
The concept of parallax and relative motion is responsible for this.
That is, the photographer's motion prompts her to see parallax for the man and the trees, because their positions appear to keep shifting even though they are not really moving.
The apparent movement of the man and trees relative to distant mountains as the photographer moves is a result of a physics concept called parallax. Parallax causes nearer objects to appear to move more than farther ones as the observer's position changes.
Explanation:The phenomenon you're describing occurred as a result of a concept in physics known as parallax. Parallax refers to the apparent movement of objects when the observer's position changes. So in this case, as the photographer moves, the position of the man and the trees appear to change relative to the distant mountains. This is because nearer objects appear to move more than farther objects as the observer moves. So, the man and trees (which are closer) seem to shift their positions more than the mountains (which are farther away).
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Consider a three-phase generator rated 300 MVA, 23 kV, supplying a system load of 240 MVA and 0.9 power factor lagging at 230 kV through a 330 MVA, 23 D/ 230 Y-kV step-up transformer with a leakage reactance of 0.11 per unit a) Neglecting the exciting current and choosing base values at the load of 100 MVA and 230 kV, find the phasor currents IA, IB, and IC supplied to the load in per unit b) By choosing the load terminal voltage VA as reference, specify the proper base for the generator circuit and determine the generator voltage V as well as the phasor currents IA, IB, and IC, from the generator. (Note: Take into account the phase shift of the transformer.) c) Find the generator terminal voltage in kV and the real power supplied by the generator in MW d) By omitting the transformer phase shift altogether, check to see whether you get the same magnitude of generator terminal voltage and real power delivered by the generator.no-load speeds possible with this motor?
The phasor currents for the load are 0.573 per unit. With the transformer phase shift, the generator supplies a voltage of 24.3 KV and 54.6 MW. Ignoring the transformer phase shift, the magnitudes remain the same, but the angles differ.
Explanation:The complex power supplied to the load is P = 240 MVA * 0.9 = 216 MW and reactive power Q = 240 MVA * sqrt (1-0.9^2) = 106.3 MVAR. From this, you can compute the phasor currents IA, IB, and IC using the formula: I = S/V * conjugate of the power factor angle, giving IA = IB = IC = 0.573 per unit.
For b), taking leakage reactance into account, the generator voltage V = 1 + j0.11 * I, giving a voltage of 1.063 per unit. Because the generator is in Delta and the load is in Wye, there is a -30-degree shift in currents from the generator. So, IA, IB, and IC are 0.573 angle -30 degrees per unit.
For c), multiply the generator voltage by base voltage to determine the terminal voltage (1.063*23kV = 24.3 kV). The total real power supplied by the generator PG = VI*cos(theta) = 0.573*1.063*0.9 = 0.546 per unit, or 54.6 MW.
For d), if the transformer phase shift is ignored, we will still obtain the same magnitude of generator terminal voltage and real power, but the angles will be different, i.e., the currents will be in phase with the voltage, and not lagging by 30 degrees, because we didn't take into account the Y-D transformation.
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f the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would __________ and the wavelength of peak emission would shift toward __________ wavelengths.
Answer
IF the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would _increase_ and the wavelength of peak emission would shift toward _shorter_ wavelengths.
Explanation:
The Energy of radiation emitted by the earth varies directly with the average surface temperature of the earth and inversely with the wavelength of emissions.
E = hv/λ
That is, E = k/λ
Therefore the most peak emissions (highest energies) would have shorter wavelengths.
Complete question:
if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would __________ and the wavelength of peak emission would shift toward __________ wavelengths.
Answer:
if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would Increase and the wavelength of peak emission would shift towards Shorter wavelengths.
Explanation:
Stefan-Boltzmann law, a fundamental law of physics, explains the relationship between an object's temperature and the amount of radiation that it emits. This law states that all objects with temperatures above absolute zero (0K) emit radiation at a rate proportional to the fourth power of their absolute temperature.
Expressed mathematically as; E = σT⁴
From this formula above, temperature is directly proportional to amount of radiation emitted.
Thus, if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would Increase.Also, Energy of emitted radiation can be related to wavelength in the expression below
E =hc/λ
Where;
E is the energy of the emitted radiation
h is Planck's constant
c is the speed of light
λ is the wavelength of the emitted radiation
From the formula above, Energy of the emitted radiation is inversely proportional to the wavelength of the emitted rays.
Thus, there would be a shift towards shorter wavelengths.when electrons are excited to differnt energy levels, the average radii from the nucleus also changes. Rank the following electron energy states according to the average distance of the electron from the nucleus. Rank from largest to smallest distances..
Answer:
The problem can be solved using the equation below
Explanation:
ΔE = E1 − E2
ΔE = 2.179*〖10〗^(-18)*(〖1/〖n_1〗^2 〗-〖1/〖n_2〗^2 〗 )
A 10-kg sled carrying a 30-kg child glides on a horizontal, frictionless surface at a speed of 6.0 m/s toward the east. The child jumps off the back of the sled, propelling it forward at 20 m/s. What was the child’s velocity in the horizontal direction relative to the ground at the instant she left the sled?
Answer:
- 1.33m/s
Explanation:
We choose the system to be the child and the sled. The surface is friction less, which means that there are no forces exerted on the system horizontally. This means that the horizontal momentum component of the system is constant and conserved.
So we can use the conservation momentum principle to find the velocity of the child just after he leaves the sled.
This is shown in the attached file.
The child’s velocity in the horizontal direction relative to the ground at the instant she left the sled is 1.33 m/s.
How to calculate the velocity?It should be noted that velocity simply means the directional speed of an object in motion.
In this case, the velocity will be:
= ((30 + 10) × 6) - (10 × 20)/30
= (40 × 6) - (200)/30
= (240 - 200)/30
= 40/30
= 1.33 m/s
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A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate?
The dimensions of the rectangular certificate with a perimeter of 32 inches and an area of 63 square inches can be 7 inches by 9 inches or 9 inches by 7 inches.
Explanation:The subject of this question is Mathematics, specifically an application of algebra to solve for the dimensions of a rectangle when given its perimeter and area. Let’s denote length as L and width as W. The formulas for area and perimeter of a rectangle are given by Area = L * W and Perimeter = 2 * (L + W), respectively. Given that the area is 63 square inches and the perimeter is 32 inches, we can set up two equations.
From the perimeter, we have 2L + 2W = 32, simplifying gives us L = 16 - W.
We can then substitute this into the area equation, so (16 - W) * W = 63.
This simplifies and solved that gives us W = 7 or W = 9.
To find L, we substitute W = 7 / W = 9 into L = 16 - W. We get the pairs (L, W) as (9, 7) or (7, 9).
So the dimensions of the certificate can be 7 inches by 9 inches or 9 inches by 7 inches.
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An oblique rectangular prism with a square base has a volume of 539 cubic units. The edges of the prism measure 7 by 7 by 14 units. How many units longer is the slanted edge length of the prism, 14, compared to its perpendicular height?
Answer:
3 units
Solution:
V=539 cubic units
Square base, with edge a=7 units
Slanted edge length: s=14 units
V=Ab h
Ab=49 square units
539 cubic units = (49 square units) h
h= 11 units
s-h=14 units-11 units
s-h=3 units
Answer:
3 units
Explanation :
Volume of an oblique rectangular prism with a square base = A×B×h
Where h = perpendicular height
From the question, Volume of an oblique rectangular prism with a square base = 539 cubic units
We were asked from the question to find how many units longer the slanted edge length of the prism, 14 is compared to its perpendicular height.
The first step is : Find the perpendicular height
Edges A = 7 units
Edges B = 7 units
Perpendicular height ?
Hence,
539 = 7 × 7 × h
539 = 49h
h = 539 ÷ 49
h = 11 units
Therefore, perpendicular height of the prism = 11 units
To find how many units longer, we would subtract the perpendicular height of the prism from the slanted edge length of the prism
= 14 units - 11 units
= 3 units .
Therefore the slanted edge length of the prism , 14, is 3 units longer compared to its perpendicular height.
Perception is defined as the ____. a.conversion of electromagnetic energy into electrochemical energyb.process of encoding experiencesc.transduction of environmental stimulid.interpretation of sensory information
Answer:
d.interpretation of sensory information
Explanation:
The organizing, recognition, and understanding of sensory information is perception(the Latin perceptio). It is the experience in order to represent and comprehend the knowledge as well as the world.It comprises data collection from sensory organs to brain interpretation
Suppose you had one circuit with a 9V battery connected to two 100 ohm light bulbs in series. and a second circuit with a 9V battery connected to two 100 ohm light bulbs in parallel. How does the brightness of the two bulbs in series compare to two bulbs connected in parallel? Which set is dimmer? Explain why.
Answer:
Bulbs in series will be dim
Explanation:
The voltage remain constant in a series circuit, thus bulbs in series circuit will get equally distributed voltage which is not equal to the voltage across the circuit or the voltage supplied
In case of parallel circuit, the voltage across the bulb is the same as the voltage across the circuit due to which bulbs will be brighter in case of parallel circuit
A small object moves along the xx-axis with acceleration ax(t)ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At tt = 0 the object is at xx = -14.0 mm and has velocity v0xv0x = 8.70 m/sm/s.
What is the xx-coordinate of the object when tt = 10.0 ss?
Answer:
x = 54.3m (on the +ve x axis)
Explanation:
This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.
Given that acceleration, a, is:
ax(t) = - 0.032(15.0 - t)
And the initial values are:
x(t = 0) = - 14.0m
v(t = 0) = 8.7m/s
Hence,
a = - 0.032(15 - t)
a = - 0.48 + 0.032t
a = dv/dt = -0.48 + 0.032t
To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:
v = ∫dv/dt = ∫(-0.48 + 0.032t)
∫dv = ∫(-0.48 + 0.032t)dt
(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)
Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:
=> v - 8.7 = -0.48t + 0.032t²/2
v = 8.7 - 0.48t + 0.016t²
To obtain distance, x, integrate the velocity and apply the initial values:
v = dx/dt = 8.7 - 0.48t + 0.016t²
=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)
∫dx= ∫(8.7 - 0.48t + 0.016t²)dt
(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)
Inputting the initial values t₀ = 0s, x₀ = - 14.0m:
(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3
x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0
Now that the distance, x, has been obtained, when t = 10s:
x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0
x = 87 - 24 + 5.3 - 14.0
x = 54.3m
Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).
Acceleration of a body is the change in velocity with respect to time.
The x-coordinate of the object when value of time is 10 second is 54.3 m.
What is acceleration?Acceleration of a body is the change in velocity with respect to time.
Given information-
A small object moves along the x-axis with acceleration,
[tex]ax(t)=-0.0320(15-t)[/tex]
Initial position and initial velocity of the object is,
[tex]x_0=-14\rm m\\v_0=8.7\rm m/s[/tex]
For the velocity integrate the given equation as,
[tex]ax(t)=-0.0320(15-t)\\\dfrac{dv}{dt} =\int ({-0.48+0.032t} )\, dt\\v-v_0=-0.48t+0.032\times\dfrac{t^2}{2} \\v-8.7=-0.48t+0.016t^2\\v=0.016t^2-0.48t+8.7[/tex]
Integrate it again to find the distance of the object.
[tex]v=0.016t^2-0.48t+8.7\\\int\dfrac{dx}{dt}=\int(0.016t^2-0.48t+8.7)dt\\x-x_0=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\x-14=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\\\x=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t-14\\[/tex]
Put the value of [tex]t[/tex] as 10 seconds as,
[tex]x=0.016\times\dfrac{10^3}{3}-0.48\times\dfrac{10^2}{2}+8.7\times10-14\\[/tex]
[tex]x=54.3\rm m[/tex]
Hence the x-coordinate of the object when value of time is 10 second is 54.3 m.
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3. If you start at the equator and travel to 100 N latitude, approximately how many kilometers (or miles) north of the equator will you be? Take the circumference of the Earth to be 40,000 kilometers (24,900 miles). Show your calculations.
Answer:
travel = 1111.11 km or ( 691.66 miles ) north of the equator
Explanation:
given data
travel = 10° N latitude
circumference of the Earth = 40,000 kilometers
solution
when we start at equator and travel whole world and finally reach that same point
so we cover 40,000 km and that is circumference of Earth
so that we cover = 360° latitude
and here distance travel in each degree of latitude will be
distance travel in each degree = [tex]\frac{40000}{360}[/tex]
distance travel in each degree = 111.11 km
so here for travel to 10 degrees N
so travel is = 111.11 × 10 =
travel = 1111.11 km or ( 691.66 miles ) north of the equator
A physical change
A) occurs when iron rusts.
B) occurs when sugar is heated into caramel.
C) occurs when glucose is converted into energy within your cells.
D) occurs when water is evaporated.
E) occurs when propane is burned for heat.
Answer: D) occurs when water is evaporated.
Explanation:
A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.
A chemical change is defined as a change in which a change in chemical composition takes place. A new substance is formed in these reactions.
a). iron rusts :the chemical reaction occurs by combination of iron with oxygen , thus a chemical change.
b) sugar is heated into caramel. the chemical reaction occurs by decomposition , thus a chemical change.
c) glucose is converted into energy within your cells: the chemical reaction occurs by decomposition , thus a chemical change.
d) when water is evaporated : Only the state changes and thus a physical change and cannot be reversed.
e) when propane is burned for heat: the chemical reaction as propane combines with oxygen to form carbon dioxide and water , thus a chemical change.
Sam, whose mass is 79 kg, stands at the top of an 11-m-high, 120-m-long snow-covered slope. His skis have a coefficient of kinetic friction on the snow of 0.07. If he uses his poles to get started, then glides down, what is his speed at the bottom?
Answer:
The speed of Sam at the bottom is 7.19 m/s.
Explanation:
Given that,
Mass of Sam = 79 kg
Height = 11 m
Length = 120 m
Coefficient of kinetic friction = 0.07
Suppose, an object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, micro-kilometer , is small enough that the object will slide down the slope if given a very small push to get it started.
We need to calculate the speed at the bottom
Using conservation of energy
[tex]P.E=K.E+\text{energy lost of friction}[/tex]
[tex]mgh=\dfrac{1}{2}mv^2+\mu mg(\sqrt{L^2-h^2})[/tex]
[tex]v^2=2gh-2\mu g(\sqrt{L^2-h^2}[/tex]
[tex]v=\sqrt{2gh-2\mu g(\sqrt{L^2-h^2}}[/tex]
Where, m = mass
h = height
L= length
v = speed
g = acceleration due to gravity
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times11-2\times0.07\times9.8(\sqrt{120^2-11^2})}[/tex]
[tex]v=7.19\ m/s[/tex]
Hence, The speed of Sam at the bottom is 7.19 m/s.
Velocity is the rate of change of position. Sam's velocity at the bottom of the slope with a height of 11 m is 13.8844 m/s.
What is velocity?Velocity is the rate of change of position of an object with respect to time.
[tex]v = \dfrac{ds}{dt}[/tex]
We know that at the topmost height, the weight of Sam will act as potential energy, while during skiing down this potential energy will be partially converted to kinetic energy and part will be converted to heat or can say will be lost due to the friction, therefore,
Potential Energy = Kinetic energy + Energy loss due to the friction,
[tex]mgh = \frac{1}{2}mv^2 + \mu gh\sqrt{L^2+H^2}\\\\mgh - \mu gh\sqrt{L^2+H^2} = \frac{1}{2}mv^2\\\\2mgh - 2\mu gh\sqrt{L^2+H^2} = mv^2\\\\2gh(m - \mu \sqrt{L^2+H^2}) = mv^2\\\\v^2 = \dfrac{2gh}{m}(m - \mu \sqrt{L^2+H^2})[/tex]
Substitute the values,
Mass, m = 79 kg
Height, H = 11 m
Length, L = 120 m
Coefficient of kinetic friction, μ = 0.07
Acceleration due to gravity, g = 9.81 m/s²
[tex]v^2 = \dfrac{2 \times 9.81 \times 11}{79}[79-0.07\sqrt{120^2+11^2}]\\\\v^2 = 215.82 - 23.0442\\\\v = 13.8844\rm\ m/s[/tex]
Hence, Sam's velocity at the bottom of the slope with a height of 11 m is 13.8844 m/s.
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The electric force between objects A and B is F. If the charge of object A were twice as large as it is, but everything else was kept the same, what would be the new electric force between objects A and B?
Answer:
[tex]F'=2F[/tex]
Explanation:
According to Coulomb's law and assuming the objects as point charges, the magnitude of the electric force that each object exerts on the other is defined as:
[tex]F=\frac{kq_Aq_B}{d^2}[/tex]
Here k is thee Coulomb constant, [tex]q_A[/tex] and [tex]q_B[/tex] are the charges of the objects and d is the distance of separation between them. We have [tex]q'_A=2q_A[/tex]:
[tex]F'=\frac{kq'_Aq_B}{d^2}\\F'=\frac{k2q_Aq_B}{d^2}\\F'=2\frac{kq_Aq_B}{d^2}\\F'=2F[/tex]
A porcelain cup of mass 303 g and specific heat 0.260 cal/g-°C contains 161 cm³ of coffee, which has a specific heat of 1.00 cal/g-°C. If the coffee and cup are initially at 71.0 °C, how much ice at 0.00 °C must be added to lower the temperature to 49.0 °C?
Answer:
[tex] m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr[/tex]
So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C
Explanation:
For this case we need to use the fact that the sum for all the heats involved in the system are 0, since we assume an equilibrium state.
Data given
[tex]m_p = 303 gr[/tex] mass of the porcelain cup
[tex] cp_p = 0.260 cal/g C[/tex] the specific heat for the porcelain cup
[tex] T_{ip} = T_{ic}= 71 C[/tex] initial temperature for the coffee and the porcelain cup.
[tex] V_{c}= 161 cm^3[/tex] Volume of the coffee.
We can convert this to m^3 and we got 0.000161m^3 and assuming the density fot the coffee equal to the water 1 Kg/m^3 the mass would be:
[tex] m_c = 1 kg/m^3 *0.000161 m^3 = 0.000161 kg=0.161 Kg[/tex]
[tex] Cp_{c} = 1 cal/g C[/tex] Specific heat for the coffee
[tex] m_i =?[/tex] mass of ice required
[tex] T_e= 49C[/tex] equilibrium temperature
[tex]L_f = 80 cal/g [/tex] represent the latent heat of fusin since the ice change the state to liquid.
Solution to the problem
Using this formula:
[tex] \sum_{i=1}^n Q_i = 0[/tex]
We have this:
[tex] m_p cp_p (T_e -T_{ip}) + m_{c} cp_c (T_e -T_{ic}) +m_i L_f + m_i cp_w (T_e -0) =0[/tex]
Now we can replace and we have this:
[tex] 303 gr *(0.260 cal/g C) (49-71)C + 0.161 gr*(1 cal/g C)(49-71)C +m_i [80 cal/gr+(1cal/g C)(49-0)C]=0[/tex]
And now we can solve for [tex] m_i[/tex] and we have:
[tex]-1733.16cal -3.542cal +m_i [129 cal/g]=0[/tex]
[tex]m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr[/tex]
So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C
The motor winds in the cable with a constant acceleration, such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. Determine the tension developed in the cable. The coefficient of kinetic friction between the crate and the plane is mk = 0.3.
Answer:
85 N
Explanation:
Given that crate mass = 20kg
Distance = 6m
Time = 3 seconds
Coefficient of kinetic friction = 0.3
We begin by calculating for acceleration
Which was gotten as 1.33 m/s sq
SEE THE ATTACHEMENT FOR DETAILS
To determine the tension developed in the cable, we consider forces acting on the crate, including the tension in the cable and the force of kinetic friction. We use Newton's second law of motion and equations for tension and force of kinetic friction to calculate the tension. By substituting values into the equations, we can solve for the tension.
Explanation:To determine the tension developed in the cable, we need to consider the forces acting on the crate. Since the crate is moving with constant acceleration, we can use Newton's second law of motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the forces acting on the crate are the tension in the cable and the force of kinetic friction. The tension in the cable can be calculated using the equation:
Tension = mass of the crate * acceleration + force of kinetic friction
Given that the mass of the crate is 20 kg and the acceleration is the change in velocity divided by the time taken (6 m / 3 s = 2 m/s), we can calculate the tension using the equation:
Tension = 20 kg * 2 m/s^2 + force of kinetic friction
We also need to determine the force of kinetic friction. The force of kinetic friction can be calculated using the equation:
Force of kinetic friction = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the crate, which can be calculated using the equation:
Normal force = mass of the crate * gravitational acceleration
Given that the coefficient of kinetic friction is 0.3 and the gravitational acceleration is 9.8 m/s^2, we can calculate the force of kinetic friction using the equation:
Force of kinetic friction = 0.3 * (20 kg * 9.8 m/s^2)
Now we can substitute this value back into the equation for tension:
Tension = 20 kg * 2 m/s^2 + (0.3 * (20 kg * 9.8 m/s^2))
Solving this equation will give us the tension developed in the cable.
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008 (part 1 of 3) 10.0 points A 0.338 kg particle has a speed of 3.8 m/s at point A and kinetic energy of 10.1 J at point B. What is its kinetic energy at A? Answer in units of J. 009 (part 2 of 3) 10.0 points What is the particle’s speed at B? Answer in units of m/s. 010 (part 3 of 3) 10.0 points What is the total work done on the particle as it moves from point A to B?
Answer:
1) 2.44 joules
2) 7.73 m/s
3) 7.6 joules
Explanation:
Kinetic energy (K) of a particle is:
[tex] K=\frac{mv^{2}}{2} [/tex] (1)
with m the mass, and v the velocity
1) Because we already now velocity on A (va) and the mass of the object we can calculate its kinetic energy:
[tex]K_{a}=\frac{mv_{a}^{2}}{2}=\frac{(0.338kg)(3.8\frac{m}{s})^{2}}{2}=2.44J [/tex]
2) Because on B we know mass and kinetic energy we should solve (1) for v and use our values to find the velocity on B:
[tex]v_{b}=\sqrt{\frac{2K_{b}}{m}}=\sqrt{\frac{2(10.1J)}{(0.338kg)}}=7.73\frac{m}{s} [/tex]
3) Work-energy theorem states that the change of kinetic energy of an object is equal to the total work done on it, so:
[tex]W=K_b-K_a=10.1J-2.44J= 7.6J [/tex]
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v?
Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:
[tex]F_{B}=-5755N[/tex]
Explanation:
Set up force equation
∑F=ma
∑F=W+FB
[tex]\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N[/tex]
The minus sign for downward direction
Which provides evidence that humans are causing an increase in the carbon dioxide concentration in the atmosphere?
A. The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is positive.
B. The net flux of carbon out of the atmosphere without including the contribution from burning fossil fuels is zero.
C. The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is negative.
D. The net flux of carbon out of the atmosphere without including the contribution from burning fossil fuels is negative.
Answer: The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is positive.
Explanation:
The human activity of burning fossil fuels inevitably contributes to the increase in carbon dioxide and other gases in the Earth's atmosphere. As a result of this occurrence, a natural greenhouse appears. The greenhouse effect is the process of warming the Earth's surface and the lower layers of the atmosphere, resulting from the leakage of heat radiation.
Answer:
C
Explanation:
earth science climate change unit test
When the person is skating from the bottom of the track back up to the top, her potential energy _______ . When the person is skating from the top of the track down to the bottom, his kinetic energy _______ . When the person is skating from the top of the track down to the bottom, her velocity _______ .
Answer:
Explanation:
Potential Energy of an object is a function of elevation from datum thus assuming datum to be ground, the potential energy of the person increases as he moves up to the top.
When a person is skating from the top of the track to the bottom then his potential energy is getting converted into the kinetic energy so his kinetic energy increases as he moves down.
When a person is moving from top to bottom his kinetic energy increases which is a function of mass and velocity. As mass is fixed therefore his velocity increases as he moves down
Final answer:
When the person is skating from the bottom of the track back up to the top, potential energy increases. When the person is skating from the top of the track down to the bottom, kinetic energy increases. When the person is skating from the top of the track down to the bottom, velocity increases.
Explanation:
When the person is skating from the bottom of the track back up to the top, her potential energy increases. When the person is skating from the top of the track down to the bottom, his kinetic energy increases. When the person is skating from the top of the track down to the bottom, her velocity increases.
A 30kg rock is swung in a circular path and in a vertical plane on a .25m length string at the top of the path, the angular speed 12.0 rad/s. what is the tension in the string at that point?
A. 7.9 N.
B. 16 N.
C. 18 N.
D. 83 N.
Answer
given,
mass of the rock, m = 30 Kg
Length of the string, r = 0.25 m
angular speed of the rock, ω = 12 rad/s
Tension in the string at the top of the vertical circle = ?
At the top of the string the tension is
[tex]T_{top}=\dfrac{mv^2}{r} - m g[/tex]
we know v = r ω
[tex]T_{top}= mr\omega^2 - m g[/tex]
[tex]T_{top}= 30\times 0.25\times 12^2-30\times 9.8[/tex]
[tex] T_{top} = 786 N[/tex]
if the given mass is 0.3 Kg then tension in the rope
[tex]T_{top}= 0.30\times 0.25\times 12^2-0.30\times 9.8[/tex]
[tex]T_{top} = 7.86 N[/tex]
If the mass is 0.3 then the correct answer is option A.
Question: Suppose you see a crescent Moon; how much of the Moon's entire surface (the full globe of the Moon) is in daylight?
Final answer:
When you see a crescent Moon, only a small portion of the Moon's surface is in daylight due to the angle of sunlight. The rest is faintly illuminated by Earthshine.
Explanation:
Crescent Moon: When you see a crescent Moon, only a small portion of the Moon's entire surface is in daylight. The illuminated portion of the Moon's surface is determined by the angle at which sunlight strikes the Moon. The rest of the Moon's surface not in direct sunlight is faintly illuminated by Earthshine, caused by sunlight reflecting off the Earth.
Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Just as they reach the lake below
A) the speed of Alice is larger than that of Tom.
B) the splashdown speed of Alice is larger than that of Tom.
C) they will both have the same speed.
D) the speed of Tom will always be 9.8 m/s larger than that of Alice.
E) the speed of Alice will always be 25 m/s larger than that of Tom.
Final answer:
Alice and Tom will both have the same speed upon reaching the water because gravity accelerates them both at the same rate for their vertical descent. Alice's initial horizontal velocity does not affect her vertical downward acceleration.
Explanation:
The correct answer is C) they will both have the same speed. This is because their vertical descent is solely affected by gravity, which accelerates all objects at the same rate (approximately 9.8 m/s2) regardless of their horizontal velocity component. Alice's initial horizontal velocity of 25 m/s contributes only to her horizontal movement and does not affect the rate at which she falls vertically due to gravity. Consequently, both Alice and Tom will have the same vertical speed when they reach the lake. When you consider both vertical and horizontal components for Alice, her overall splashdown speed will be greater than Tom's; however, the question asks about just the speed. Since only the vertical descent is mentioned, we infer that it's the vertical component being considered. Therefore, their speeds, considering just their vertical motion under gravity, are the same.
A particular type of resistor has a tolerance of 3%. Technician A says this indicates that the resistor’s current value can be 3% above or below its stated specification. Technician B says this indicates the resistor’s resistance value can be 3% above or below its stated specification. Who is right?
Answer:
Technician B is correct.
Explanation:
The given value of resistor having 3% tolerance means that the given quantity of the physical parameter can vary by 3% of what is specified. This variation can be either more or less than the quantified value.
When the variation can occur on both sides of the stated value then it is called bilateral tolerance, usually represented as, [tex]\pm3\%[/tex].When the variation is permissible only in one direction then it is called unilateral tolerance, represented by the sign + for the impressibility on the higher side and [tex]-[/tex] sign for the impressibility on the lower side.A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 249 m/s. What would be the speed of the wave on the longer wire?
Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Final answer:
The speed of the wave on the longer wire would be the same as on the shorter wire, which is 249 m/s, because the tension and linear mass density are the same.
Explanation:
The speed of a transverse wave on a string is determined by the tension of the string and the linear mass density of the string, according to the formula v = √{Ft/mu}, where v is the wave speed, Ft is the tension in the string, and \\mu is the linear mass density. Since both wires have the same mass and tension, the linear mass density is the same. Therefore, the speed of the wave on the longer wire would be the same as that on the shorter wire, which is 249 m/s.
For Other Than One And Two Family Dwellings, When Building A New Electrical Service At Least One 125 Volt, Single Phase, 15 Or 20 Ampere Rated Receptacle Outlet Shall Be Located Within At LEAST ________ Feet Of The Electrical Service Equipment.
Answer and Explanation
It was initially specified that the receptacle outlet to be located within 50 ft of the electrical service equipment.
As well, instead of the receptacle being required within 50 ft., it is now required within 25 ft. of the service. This was to accommodate the typical 25 ft. cord used by many service electricians.
The rules apply to indoor service locations other than one-and two-family dwellings. For these locations, at least one 125-volt, single-phase, 15- or 20-ampere receptacle outlet must be installed in an accessible location within 25 ft. of the indoor electrical service equipment. The receptacle must be within the same room or area as the actual service equipment.
Having a maintenance receptacle near the electrical service allows for testing, servicing and connection of portable electrical data acquisition equipment for analyzing the electrical system.
A dog running in an open field has components of velocity vx = 3.1 m/s and vy = -1.3 m/s at time t1 = 11.3 s . For the time interval from t1 = 11.3 s to t2 = 22.5 s , the average acceleration of the dog has magnitude 0.52 m/s2 and direction 27.5 ∘ measured from the +x−axis toward the +y−axis. At time t2 = 23.6s , what are the x-component of the dog's velocity?
Answer:
x -component of dog's velocity = 8.265m/s
y- component of dog's velocity = 1.389m/s
Explanation:
The detailed and step by step calculation is as shown in the attached file.
The speed of a boat in still water is v. The boat is to make a round trip in a river whose current travels at speed u.
A) Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving upstream and back downstream.
Express your answer in terms of the variables D, v, and u.
B) Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving directly across the river and back.
Express your answer in terms of the variables D, v, and u.
Answer:
a) time needed to make a round trip = DV/v2 - u2
b) time needed to make a round trip = D/root ( v2 -u2)
Explanation:
The detailed explanation is as shown in the attachment
For a round trip moving upstream and downstream, the formula for the total time needed is D / (v - u) + D / (v + u). But, if the boat moves directly across the river and back, the total time needed is D / sqrt(v² - u²). This involves the concept of relative velocity.
Explanation:First, we need to understand the concept of relative velocity here. When the boat travels upstream, it moves against the current, so its net speed is (v - u). Downstream, it moves with the current at a net speed of (v + u).
A) The total time for a round trip moving upstream and downstream is the total distance divided by the speed in each direction. The time to travel a distance D upstream is D / (v - u), and the time to travel the same distance downstream is D / (v + u). So, the formula for the total time needed is D / (v - u) + D / (v + u).
B) If the boat is going directly across the river and back, it is moving perpendicular to the current. The boat makes the round trip at a constant speed of sqrt(v² - u²). Hence, the time needed for a round trip of total distance D is D / sqrt(v² - u²).
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