Answer:
Empathy
Explanation:
Empathy: to understand the thoughts, feelings, or emotional state of another person
Appreciation: to show gratitude
Cooperation: is the association for mutual benefit
Commitment: is the agreement to do something.
From the definitions above, empathy is the term that describes one’s ability to understand how another person is feeling.For every _______ over 50 mph, your chances of being seriously injured, disfigured, or killed are doubled.
Answer:
10 mph
Explanation:
Exceeding speed limit is one of the major causes of road accident.Crash severity increases with the speed of the vehicle at impact.The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.The above consequences double for every 10 mph over
50 mph that a vehicle travels.
Answer:
10 mph
Explanation:
Exceeding speed limit is one of the major causes of road accident.
Crash severity increases with the speed of the vehicle at impact.
The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.
The above consequences double for every 10 mph over
50 mph that a vehicle travels
Explanation:
An 899-kg (1987 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s. (a) Find the average acceleration of the dragster during this time interval. 44.58 Correct: Your answer is correct. m/s2 (b) What is the magnitude of the average net force on the dragster during this time
Explanation:
Mass, m = 899 kg
Initial velocity, u = 0 m/s
Final velocity, v = 26.3 m/s
Time, t = 0.59 s
a) We have equation of motion v = u + at
Substituting
v = u + at
26.3 = 0 + a x 0.59
a = 44.58 m/s²
Acceleration of dragster = 44.58 m/s²
b) Force = Mass x Acceleration
F = ma
F = 899 x 44.58 = 40074.07 N
F = 40.07 kN
Average net force on the dragster during this time is 40.07 kN
If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
Final answer:
The question is related to determining the critical value of x, or xcritical, that ensures a bar with a heavy block remains stable by balancing torques around the pivot point.
Explanation:
The question asks about stability conditions in a physical system involving a block on a bar and requires the application of concepts such as the center of mass, force equilibrium, and tension. When the mass of the block is too large and positioned too close to the left end of the horizontal bar, the system may become unstable, potentially causing the bar to tilt. To determine the smallest value of x, denoted as xcritical, we must set up equations considering the torques about the pivot point, ensuring that the sum of torques must equal zero for stability. This involves calculating the torques due to the weight of the bar, the weight of the block, and any other forces acting on the system. The position of the center of mass of the bar is crucial as it influences the bar's propensity to rotate and the distribution of weight. Hence, xcritical is that specific distance at which the torques balance each other out, and the system remains in a stable, horizontal state.
Help Please ! A wave of wavelength 52 cm travels with frequency 400 Hz. What is the speed of the wave? Show all work.
The speed of the wave is 208 m/s
Explanation:
The wave equation states that ; speed of wave= wavelength *frequency
Mathematically, v=λ*f
Given;
λ=52 cm= 52/100 =0.52 m
f= 400 Hz
v=?
v=0.52*400 =208 m/s
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When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials A) conduct electricity, while others do not. B) can accept protons more easily than others. C) can give up electrons more readily than others. correct D) repel negative objects, and others attract them.
When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials can give up electrons more readily than others.
Answer: Option C
Explanation:
We know that charges can neither be created nor be destroyed by law of conservation of charges. So when we rub two objects, it is natural to have a transfer of charges. But the charges which get transferred may be negligible in most of the cases leading to no significant observations.
But for some materials, like when we rubbed a balloon with human hair, we observed clouding of excess static charge on the balloon surface. This indicates that hair can easily give up electrons to balloon leading to clouding of excess static charge on it.
A series combination of two resistors of resistance 12 Ω and 8 Ω is connected across a source of emf 24 V. What is the potential difference across 8 Ω resistors?
Answer:
The Potential difference across the 8 Ω resistor = 9.6 V
Explanation:
In a series combination of resistor, The total resistance
Rt = R₁ + R₂............... Equation 1
Where Rt = total resistance, R₁ = Resistance of the first resistor, R₂ = resistance of the second resistor.
Given: R₁ = 12Ω, R₂ = 8Ω.
Substituting these vales into equation 1
Rt = 12 + 8
Rt = 20 Ω.
Since both resistors are connected in series, The same current flows through both resistors.
from ohms law,
V = IR..................... Equation 3
I = V/R....................... Equation 2
where I = current flowing through both resistors, V= Emf, R = total resistance of the circuit
Also given : V = 24 V, R = 20 Ω,
Substituting these values into equation 3,
I = 24/20
I = 1.2 A.
Note: Since both resistors are connected in series, The same current flows through both resistors.
The Potential difference across the 8 Ω resistor
V = 1.2×8
V = 9.6 V
The Potential difference across the 8 Ω resistor = 9.6 V
The closure temperature represents the point when ________. A. a magma cools to the point where minerals begin to crystallize B. radioactive isotopes begin to decay C. isotopes are no longer free to move out of a crystal lattice D. the last remaining magma crystallizes
The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice.
Answer: Option C
Explanation:
The closure temperature can also be termed as blocking temperature. It is mostly used in radiometric dating. As the temperature decreases, below a certain point the isotopes may get freeze in their lattice positions. And there may be slowing of diffusion.
At the closure temperature, that rate of diffusion will be zero as the isotopes will be no longer free to move out of crystal lattice. So, this is termed as closure or blocking temperature. As the isotopes loose their ability to move, their concentration will remain fixed in their position leading to measurement of radiation dating.
Final answer:
Closure temperature is the point where isotopes are trapped within a crystal lattice, marking the mineral as a closed system, which is crucial for isotopic dating.
Explanation:
The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice. This is a point in the cooling process of a rock or mineral, after crystallization, at which its isotopic constituents are no longer able to escape from its crystal lattice. Essentially, it is the temperature below which a mineral becomes a closed system for certain isotopes. Until reaching this temperature, isotopic parent and daughter products can freely exchange with the environment. As the temperature drops below the closure temperature, the system locks in place, preserving the isotopic date that will be analyzed to determine the history of the rock.
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Problem-Solving Strategy: Simple Harmonic Motion II: Energy
The energy equation, E=12mvx2+12kx2=12kA2, is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time, it is usually easier to use the equation for simple harmonic motion, ax=d2xdt2=−kmx (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or of vx; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then x is positive and vx is positive.
IDENTIFY the relevant concepts
Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.
SET UP the problem using the following steps
Part A
The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
Select all that apply.
maximum velocity vmax
amplitude A
force constant k
mass m
total energy E
potential energy U at x
kinetic energy K at x
position x from equilibrium
Part B
What is the kinetic energy of the object on the spring when the spring is compressed 5.1 cm from its equilibrium position?
Part C
What is the potential energy U of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Answer:
Part A
Mass = 50g
Vmax = 3.2m/s
Amplitude= 6cm
Position x from the equilibrium= 5.1cm
Part B
Kinetic energy = 0.185J
Part C
Potential energy = 0.185J
Explanation:
Kinetic energy = 1/2mv×2
Vmax = wa
w = angular velocity= 53.33rad/s
Kinetic energy = 1/2mv^2×r^2 = 0.185J
Part c
Total energy = 1/2m×Vmax^2= 0.256J
1/2KA^2= 0.256J
K= 142.22N/m (force constant)
Potential energy = 1/2kx^2
=1/2×142.22×0.051^2
= 0.185J
To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.
Explanation:In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.
To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.
Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.
Substituting the known values into the equations, we can calculate the kinetic energy of the toy.
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Large, heavy vehicles have slow acceleration, long__________distances and respond poorly through sharp turns.
Answer:braking
Explanation:
Large vehicles as compared to small vehicles require long braking distance, otherwise, it could topple the heavy vehicles. Heavy vehicles provide high torque thus it is used to carry heavy loads.
They run at relatively low speed as compared to the light vehicles as they are slow to accelerate and thus require long braking distance as Momentum associated with them is very high.
If sudden brakes are applied it may cause the vehicle to skid and flip over it due to the presence of large momentum.
Worker A works at a constant rate and, working alone, can complete a job in 6 hours. Worker B works at a constant rate and, working alone, can complete the same job in 5 hours. Worker C works at a constant rate and, working alone, can complete the same job in 3 hours.
Answer:
[tex]R= \frac{1job}{1.429 hours}[/tex]
So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.
Explanation:
If we want to express the situation in math terms and find the number of hours that takes to complete 1 job with the 3 at the same time, we can do this.
For this case we have the following rates:
[tex]R_A =\frac{1job}{6hours}[/tex]
[tex]R_B =\frac{1 job}{5 hours}[/tex]
[tex]R_C=\frac{1job}{3 hours}[/tex]
And we know that working together the rate would be the addition of the rates like this:
[tex]R=R_A +R_B +R_C = \frac{1job}{6hours}+\frac{1job}{5hours}+\frac{1job}{3hours} =\frac{7 jobs}{10hours}[/tex]
And if we divide the numerator and denominator by 7 we got:
[tex]R= \frac{1job}{1.429 hours}[/tex]
So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.
A bullet with a mass of 8.0 g and a speed of 650 m/s is fired at a block of wood with a mass of 0.250 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 12 m/s. What is the speed of the bullet when it exits the block
The speed of the block after the bullet exits is 7.5 m/s. This velocity is found by using the conservation of momentum principle and plugging in the known values into the momentum equation to solve for the unknown variable.
To solve for the speed of the wooden block after the bullet exits, we will use the principle of conservation of momentum. This principle states that in an isolated system, the total momentum before an event is equal to the total momentum after the event. The formula to calculate the final velocity of the block is:
[tex]m_{bullet} \times u_{bullet} + m_{block} \times u_{block} = m_{bullet} \times v_{bullet} + m_{block} \times v_{block}[/tex]
Where:
[tex]m_{bullet}[/tex] = mass of the bullet
[tex]u_{bullet}[/tex] = initial velocity of the bullet
[tex]m_{block}[/tex] = mass of the block
[tex]u_{block}[/tex] = initial velocity of the block (0 m/s, since it's stationary)
[tex]v_{bullet}[/tex] = final velocity of the bullet after passing through the block
[tex]v_{block}[/tex] = final velocity of the block
Given:
[tex]m_{bullet}[/tex] = 0.050 kg (50 g)
[tex]u_{bullet}[/tex] = 500 m/s
[tex]m_{block}[/tex] = 2 kg
[tex]u_{block}[/tex] = 0 m/s
[tex]v_{bullet}[/tex] = 200 m/s
Plugging in the values:
0.050 × 500 + 2 × 0 = 0.050 × 200 + 2 × [tex]v_{block}[/tex]
Solving for [tex]v_{block}[/tex]:
[tex]v_{block}[/tex] = (0.050 × 500 - 0.050 × 200) / 2
[tex]v_{block}[/tex] = (25 - 10) / 2
[tex]v_{block}[/tex] = 15 / 2
[tex]v_{block}[/tex] = 7.5 m/s
Therefore, the velocity of the block after the bullet exits is 7.5 m/s.
Some substances have the same chemical composition, but their atoms are arranged differently. A classic example is carbon. A pencil lead is made of graphite, which contains only carbon atoms. A diamond also contains only carbon atoms. However, the carbon atoms in a diamond are packed more closely together. Would you expect the density of graphite and a diamond to be the same? Explain your answer.
Answer: No, because the atoms are arranged differently. Looking at 100 atoms in each sample, the volume would be smaller in a diamond, because the atoms are packed more closely together. The density of a diamond would be higher.
Explanation:
Final answer:
Graphite and diamond have different densities due to the varied arrangements of carbon atoms in their structures, with diamonds being denser and harder due to a three-dimensional bond network.
Explanation:
The difference in density between graphite and diamond can be attributed to the variation in how carbon atoms are arranged in these two allotropes of carbon. Graphite has a layered structure with weak bonds between each layer, allowing the sheets to easily slip past each other, leading to its softer structure and lower density.
In contrast, diamonds have carbon atoms bonded together in all three dimensions, creating a dense, strong lattice, which is why diamonds are very hard and have a higher density compared to graphite.
Why do astronauts (and cans of soft drink) float around in the shuttle instead of falling?
Answer: Astronauts only float around in the shuttle when they are outside the gravitational pull of the earth
Explanation: when astronauts takes off from the earth, they get to a point (space) where the earth's gravity can no longer pull them. At this state, they experience weightlessness because there is no gravity. Since there is no gravity to pull them down, hence they start floating.
Astronauts and objects float in the shuttle due to the microgravity state from continual free-fall around the Earth, creating the feeling of weightlessness.
Explanation:Astronauts and objects like cans of soda float in the shuttle because of the lack of gravity in space, a state known as microgravity. When the shuttle is orbiting the earth, it's actually falling towards the earth but also moving forward. This forward motion allows the shuttle and everything in it to keep missing the Earth, so they keep falling towards it but never hitting it. This continual state of free-fall creates the feeling of weightlessness and is why astronauts and objects inside the shuttle appear to float.
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Which of the following is characteristic of proficient catching?
A. pointing the fingers upward to catch a high ball
B. pointing the fingers upward to catch a low ball
C. letting the shoulders and elbows move to give with contact of the ball
D. A and C E. B and C
Answer:
The correct answer is option D i.e. A and C
Explanation:
The correct answer is option D i.e. A and C
for proficient catching player must
- learn to absorbed the ball force
- moves the hang according to ball direction to hold the ball
- to catch ball at high height move the finger at higher position
- to catch ball at low height move the finger at lower position
A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A transit train left from station A 1 hour later than the passenger train, but it arrived at the station B at the same time with the passenger train. What was the average speed of the transit train?
Explanation:
Average speed of passenger train = 45 mph
Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours
Distance between station A to station B = 45 x 4 = 180 miles.
Time taken from station A to station B for transit train = 4 - 1 = 3 hours
Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train
180 = Average speed of transit train x 3
Average speed of transit train = 60 mph
Average speed of transit train is 60 mph
a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?
Answer:
Elevation =31.85[m]
Explanation:
We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.
We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.
The point where the ball is thrown shall be taken as a reference point of potential energy.
[tex]E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J][/tex]
m = mass of the ball = 300 [gr] = 0.3 [kg]
v = initial velocity = 25 [m/s]
[tex]E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J][/tex]
[tex]93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m][/tex]
The current through a 10-ohm resistor connected to a 120-v power supply is
Answer:
12 A
Explanation:
Voltage, V = 120 V
Resistance, R 10 ohm
By using ohm's law
V = i x R
where, i is the current
i = V / R
i = 120 / 10
i = 12 A
thus, the current is 12 A.
Which of the following forms an interference pattern when directed toward two suitably-spaced slits?
a. electrons
b. Sound
c. light
d. all of these
enone of these
Answer:
Option d
Explanation:
Interference is the phenomenon in which two or more waves having the same frequency combines or cancel out each other to form a resulting wave with the amplitude equals the sum of the amplitudes of the waves that are combined.
Thus it is clear that the particle with the wave nature can produce interference pattern. Hence, electron, light, sound, all of these produces interference pattern when directed towards the slits that are suitably spaced.
Kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
Answer:
Explanation:
Speed: Speed can be defined as the ratio of distance to time. The S. I unit of speed is m/s. And it is expressed mathematically as,
speed = distance/time
S = d/t............................. Equation 1
Conversion: (i) If 1 miles = 1609.344 m,
then, 26.2 miles = 1609.344× 26.2
= 42164.813 m.
(i) if 1 hours = 3600 seconds,
then, 5.5 hours = 5.5×3600
=19800 seconds.
Given: d = 26.7 miles= 42164.813 m, t = 5.5 hours = 19800 seconds
Substituting these values into equation 1
S = 42164.813/19800
S = 2.1 m/s.
Therefore Kenneth's average speed = 2.1 m/s
Answer:4.8 mph on edge
Explanation: have a great day :)
Find the rate of change for x3. You need to work out the change in f(x)=x3 when x is increased by a small number h to x+h. So you will work out f(x+h)-f(x). Then do some algebra to simplify this. Then divide this by h to get the average rate of change of f(x) between x and x+h.
Answer:
Explanation:
Given
[tex]F(x)=x^3[/tex]
Rate of change of F(x) is given by
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{F(x+h)-F(x)}{x+h-x}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{(x+h)^3-x^3}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{h^3+3x^2h+3xh^2}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\h^2+3x^2+3xh[/tex]
Putting limits
[tex]F'(x)=3x^2[/tex]
The average rate of change will be "3x²".
Average rate of changeAccording to the question,
The function, f(x) = x³
then,
f(x + h) = (x + h)³
Now,
→ f(x + h) - f(x) = (x + h)³ - x³
= x³ + h³ + 3x²h + 3xh² - x³
= h³ + 3x²h + 3xh²
and,
→ [tex]\frac{f(x+h) -f(x)}{h}[/tex] = [tex]\frac{h^3+3x^2h+3xh^2}{h}[/tex]
= [tex]\frac{h[h^2+3x^2+3xh]}{h}[/tex]
= h² + 3x² + 3xh
By applying the limit, we get
→ [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex] = [tex]\lim_{h \to 0}[/tex] h² + 3x² + 3xh
By substituting the values,
= 0² + 3x² + 3x(0)
= 3x²
Thus the above approach is correct.
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to obtain a multimeter reading of 1 v between b and c in the circuit the value of r2 would have to be.
Complete question:
In the circuit shown in the figure below (See image attached), suppose that the value of R1 is [tex] 500\,k\Omega [/tex]. To obtain a multimeter reading of 1 V between points B and C in the circuit, the value of R2 would have to be.
Answer:
[tex]R2=0.1\Omega [/tex]
Explanation:
First, we are going to the find current trough the circuit, because the resistors are on series the current is the same on each resistor so I=I1=I2. The Ohm's law for the circuit is:
[tex] V=R_{T}*I [/tex] (1) , with V the voltage of the battery (6V), I the current trough the circuit and [tex]R_{T} [/tex] the total resistance of the circuit, but for resistors on series the total resistance is the sum of the individual resistance so [tex] R_{T} = R1+R2[/tex] (2).
Using (2) on (1) and solving for I:
[tex]I=\frac{V}{R1+R2} [/tex] (3)
Ohm's law is true for the individual resistors too so we're going to apply that on R2:
[tex]V2=R2*I2 [/tex], but remember I2=I
[tex]V2=R2*I [/tex] (4), using (3) on (4)
[tex]V2=R2* \frac{V}{R1+R2} [/tex], solving for R2:
[tex] R1*V2+R2*V2=R2V[/tex]
[tex] R1*V2=R2(V-V2)[/tex]
[tex] R2=\frac{R1V2}{V-V2}=\frac{500\times10^{3}\Omega*1V}{6V-1V}[/tex], V2= 1V because we want that reading on the multimeter.
[tex]R2=100\,k\Omega [/tex]
Answer:
10k
Explanation:
A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25 % of heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27^{\circ}C. [For lead, melting point = 327^{\circ}C, specific heat = 0.03 cal/g-^{\circ}C, latent heat of fusion = 6 cal/g and J = 4.2 J/cal.]
Answer:
409.87803 m/s
Explanation:
v = Velocity of bullet
L = Latent heat of fusion = 6 cal/g
c = Specific heat of lead = 0.03 cal/g°C
[tex]\Delta T[/tex] = Change in temperature = (327-27)
m = Mass of bullet
[tex]1\ J=4.2\ J/cal[/tex]
The heat will be given by the kinetic energy of the bullet
[tex]Q=\dfrac{1}{2}mv^2[/tex]
According to the question
[tex]Q=0.75\dfrac{1}{2}mv^2[/tex]
This heat will balance the heat going into the obstacle
[tex]Q=mc\Delta T+mL\\\Rightarrow 0.75\dfrac{1}{2}mv^2=m(c\Delta T+L)\\\Rightarrow v^2=\dfrac{2}{0.75}\times (0.03\times (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40\times 4.2\times 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=\sqrt{168000}\\\Rightarrow v=409.87803\ m/s[/tex]
The speed of the bullet is 409.87803 m/s
if five joules were required to move a crate in 3.7 seconds, what power was applied?
Answer:
The answer to your question is 1.35 Watts
Explanation:
Data
Work = W = 5 J
time = t = 3.7 s
Power = P = ?
Formula
Power is a rate in which work is done or energy is transferred over time
P = [tex]\frac{W}{t}[/tex]
Substitution
[tex]P = \frac{5}{3.7}[/tex]
Result
P = 1.35 W
An insulating, solid sphere has a uniform, positive charge density of [tex]\rho[/tex] =4.40*10⁻⁷C/m³. The sphere has a radius R of 0.370m.
What is the potential [tex]V_r[/tex] at a point located at r = 0.160 m from the center of the sphere?
The electric potential at a point within a uniformly charged solid sphere can be calculated using the given formula. By substituting the given values of charge density, radius of the sphere, and distance from the center of the sphere, the electric potential can be computed.
Explanation:The potential Vr within a uniformly charged solid sphere of charge density ρ, radius R, at a distance r from the center is given by the formula:
Vr = ρr²/6ε₀ + (ρR²/2ε₀) [ 1 - 3r²/R² ] for r < R
Where ρ is the charge density (4.40*10⁻⁷ C/m³), r is the distance from the center of the sphere (0.160 m), R is the radius of the sphere (0.370 m) and ε₀ is the permittivity of free space (8.854 * 10⁻¹² C²/N.m²).
Substituting the given values into the formula, we can calculate the electric potential at a point located at r = 0.160 m from the center of the sphere.
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The potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]
The potential at a point located at a distance \(r\) from the center of a uniformly charged solid sphere can be calculated using the formula:
[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]
where \(k\) is the Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(Q_{enc}[/tex]\) is the charge enclosed within the radius \(r\), and \(r\) is the distance from the center of the sphere to the point where the potential is being calculated.
For a uniformly charged sphere, the charge enclosed within a radius \(r\) is given by:
[tex]\[ Q_{enc} = \frac{4}{3}\pi r^3 \rho \][/tex]
where [tex]\(\rho\)[/tex] is the charge density of the sphere.
Given that [tex]\(\rho = 4.40 \times 10^{-7} \, \text{C/m}^3\)[/tex] and [tex]\(r = 0.160 \, \text{m}\),[/tex] we can calculate \(Q_{enc}\) as follows:
[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.160 \, \text{m})^3 (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]
[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.004096 \, \text{m}^3) (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]
[tex]\[ Q_{enc} = \frac{4}{3}\pi (1.8192 \times 10^{-9} \, \text{C}) \][/tex]
[tex]\[ Q_{enc} = 2.42528 \times 10^{-9} \, \text{C} \][/tex]
Now, we can calculate the potential [tex]\(V_r\):[/tex]
[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]
[tex]\[ V_r = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) (2.42528 \times 10^{-9} \, \text{C})}{0.160 \, \text{m}} \][/tex]
[tex]\[ V_r = \frac{2.17836 \times 10^{-8} \, \text{Nm}^2/\text{C}}{0.160 \, \text{m}} \][/tex]
[tex]\[ V_r = 1.36148 \times 10^{-7} \, \text{V} \][/tex]
Therefore, the potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]
In an experiment to determine ΔH for the reaction between HBr and NaOH , 43.6 mL of 1.08 M HBr at 19.95 °C is placed in a coffee cup calorimeter which has a heat capacity of 7.99 J/°C. 43.6 mL of 1.08 M NaOH at 19.95 °C is added to the acid solution in the calorimeter and quickly mixed. A final temperature of 26.88 °C was recorded. What is ΔH for this reaction, in kJ? Assume that the specific heats of all solutions are the same as for pure water (4.18 J g-1 °C-1), and that all solution densities are 1.00 g/mL. Don't forget to put the proper sign on your ΔH value.
Answer:
ΔH = -57.78kj/mol
Explanation:
Assumptions
These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.1984 J/goC.
the heat evolved will be
q = mcΔt
first of all convert ml to grams
(43.6 mL + 43.6 mL ) = 87.2mL of solution.
density of water=1g/ml
87.2 mL X 1 g/ml = 87.2 grams of solution.
(mass = Volume X Density)
Find the temperature change.
Δt =tfinal - tinitial = 26.88oC - 19.95oC = 6.93oC
q = mcΔt
= 87.2 grams X 4.184 J/goC X 6.9oC
= 2.52 X 10^3 J
This is the heat gained by the water, but in fact it is the heat lost by the reacting HBr and NaOH, therefore q = -2.52 x 10^3 J.
i.e. it is an exothermic reaction, heat was lost to the water and it got warmer
to find the how much of HBr that was used in mol
43.6 mL of HBr X 1.00 mol HBr/ 1000 mL HBr = 0.0436 mol HBr
same quantity of base NaoH was used
. molar enthalpy = J/mol = -2.52 x 10^3 J / 0.0436 mol
-57.78kj/mol
Therefore, for the neutralization of HBrl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -57.78kj/mol
If Mars were the same size as Mercury (instead of its actual size), which surface features would it have?
Answer:
A surface that is full of craters, very similar to that one seen in mercury.
Explanation:
Mercury is the smaller planet in the solar system whit a very thin atmosphere. If Mars were the same size of mercury it will certainly have a surface full of craters.
When meteoroid (fragments of rock) reach the atmosphere of a planet, they get incinerate as a consequence of the friction between the object and the atmosphere. However, if the meteoroid has the necessary size it can reach the ground (at this point is known as meteorite).
In the case propose, it is most likely that bigger meteorite reach the surface of the planet, which leads to the result of a surface full of craters since the size of mars is not enough to maintain a very dense atmosphere due to the weak gravitational field.
Technician A says that the waveform produced by an ac generator after rectification is called a sine wave. Technician B Says that the waveform produced by the ac generator after the output moves through the diodes is a straight line because it is a constant dc voltage. who is correct?
A.Tech A
B.Tech b
C.A and B
D. neither a nor b
Neither Technician A nor B is correct.
Answer: Option D
Explanation:
In AC generator, diodes are usually used to rectify the alternator current. This is the transformation into direct current from alternating current. Sinusoidal voltage is generated by generators when it rotates in both directions. The output voltage [tex]V_{i n}[/tex] is converted to DC with diode help.
This process is called rectification. The output voltage [tex]V_{i n}[/tex] is treated as the input signal of the diode circuit. So, after rectification, the sine wave is not produced by an AC generator. And, the sine wave resulted by AC generator when the output passes via diodes and not a straight line.
The waveform produced by an AC generator after rectification is a pulsating DC waveform, not a sine wave. Technician A is correct in this case.
Explanation:Technician A is correct. The waveform produced by an AC generator after rectification is not a sine wave, but a pulsating DC waveform. After the output of the generator moves through the diodes, the negative half-cycles are clipped, resulting in a waveform that resembles a series of straight lines instead of a smooth sine wave. This pulsating DC waveform still contains alternating current components, but it is not a constant DC voltage.
The maximum force a pilot can stand is about seven times his weight. What is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250 m/s?
Answer:
Radius of curvature of the path is 1063 meters
Explanation:
It is given that,
Force acting on the pilot is about seven times of his weight. Speed with which pilot moves, v = 250 m/s.
As per Newton's second law of motion, the net force acting on the pilot at the bottom is given by :
[tex]N-mg=\dfrac{mv^2}{r}[/tex]
Where
N is the normal force
r is the radius of curvature
According to given condition,
[tex]7mg-mg=\dfrac{mv^2}{r}[/tex]
[tex]6mg=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv^2}{6mg}[/tex]
[tex]r=\dfrac{mv^2}{6mg}[/tex]
[tex]r=\dfrac{v^2}{6g}[/tex]
[tex]r=\dfrac{(250)^2}{6\times 9.8}[/tex]
r = 1062.92 meters
or
r = 1063 meters
So, the radius of curvature of the path is 1063 meters. Hence, this is the required solution.
A 57 kg skier starts from rest at a height of H = 27 m above the end of the ski-jump ramp. As the skier leaves the ramp, his velocity makes an angle of 28° with the horizontal. Neglect the effects of air resistance and assume the ramp is frictionless.
(a) What is the maximum height h of his jump above the end of the ramp?
(b) If he increased his weight by putting on a backpack, would h then be greater, less or, the same?
Answer:
(a)[tex]h=5.95m[/tex]
(b) h is the same
Explanation:
According to the law of conservation of energy:
[tex]E_i=E_f\\U_i+K_i=U_f+K_f[/tex]
The skier starts from rest, so [tex]K_i=0[/tex] and we choose the zero point of potential energy in the end of the ramp, so [tex]U_f=0[/tex]. We calculate the final speed, that is, the speed when the skier leaves the ramp:
[tex]mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}[/tex]
Finally, we calculate the maximum height h above the end of the ramp:
[tex]v_f^2=v_i^2-2gh\\[/tex]
The initial vertical speed is given by:
[tex]v_i=vsin\theta[/tex]
and the final speed is zero, solving for h:
[tex]h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m[/tex]
(b) We can observe that the height reached does not depend on the mass of the skier
The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?
Answer:
-5.985*10^7Ns=momentum
ds=31.578m
Explanation:
The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?
from newton's second law of motion , which states that the rate of change in momentum is directly proportional to the force applied
f=kdv/dt
k=1
ft=dv/dt
dp=dv/dt
f=negative because its moving in the opposite direction
2.85× 10^6*21=
-5.985*10^7Ns=momentum
-dp=mv2-mv1
-dp+mv1=mv2
-5.985*10^7+2.0× 10^7 kg*3.0 m/s=2.0× 10^7kgV2
v2=150000/(2.0× 10^7)
v2=0.0075m/s to the right
ds=1/2(v2+v1)dt
ds=0.5*(3.0075)*21
ds=31.578m
the barge will travel 31.578m before it is brought to rest.
Answer:
∆p = –5.985×10⁷ kg×(m/s)
∆x = 31.57875 m
The boat's change in momentum is –5.985×10⁷ kg×(m/s), and it will travel 31.57875 meters before it stops.
Explanation:
First off, let's take a quick look at the formula for momentum: momentum equals the force times the time the force is applied for. This means:
p = FΔt
Why is momentum represented by the letter p? That's actually a good question, to be honest—I guess they were running out of letters because the letter m is already used for "mass". But I digress.
Let's go back to the equation. We have both those values—the force applied and the time it is applied for—and so we can substitute those into the equation. But be careful: here, the force is applied in the opposite direction as the actual motion. This means the force must be negative. We get:
p = (–2.85×10⁶ N)(21 s)
p = –5.985×10⁷ kg·(m/s)
That's our value for initial momentum, not the final momentum. Why? Momentum and impulse are equivalent and their units are the same. If we calculated this as the impulse (J = mΔv), we'd be using the initial velocity, not the final velocity (the velocity after the force is applied).
But this problem asked us to solve for two things: the change in momentum (Δp) and the stopping distance (Δx). We can't find either value if we don't know the velocity of the block after the force is applied. This means we need to solve for it.
This is where knowing impulse and momentum are equal comes in handy: if the two values are equal, their formulas should be equal, too. It's like solving a system of equations with the substitution method. With the impulse and momentum formulas, we make them equivalent and get:
mΔv = FΔt
The change in something is like the difference in subtraction: it's the final value minus the initial value. We can then rewrite the equation as:
m(v₂ – v₁) = FΔt
Here, I used (v₂) for the final (second) velocity and (v₁) for the initial (first) velocity. Now, since we need to find the final velocity, (v₂), we need to isolate it by solving the equation for this value.
m(v₂ – v₁) = FΔt
[m(v₂ – v₁)] ÷ m = (FΔt) ÷ m
v₂ – v₁ = (FΔt) ÷ m
v₂ – v₁ + v₁ = [(FΔt) ÷ m] + v₁
v₂ = [(FΔt) ÷ m] + v₁
That's our formula to find the final velocity. The problem already gave us all the values we need to solve this equation: the force applied, the time the force is applied for, the mass of the object being stopped, and the object's initial velocity. When we substitute those into the equation, we get:
v₂ = [(FΔt) ÷ m] + v₁
v₂ = [(–2.85×10⁶ N)(21 s) ÷ 2.0×10⁷ kg] + 3.0 (m/s)
v₂ = [(–5.985×10⁷ kg·(m/s)) ÷ 2.0×10⁷ kg] + 3.0 (m/s)
v₂ = 3.0 (m/s) – 2.9925 (m/s)
v₂ = 0.0075 (m/s)
That gives us the final velocity that we need to find (Δp) and (Δx). Now, using the equations for each value, we can substitute in and finish this up!
Δp = m(Δv) = m(v₂ – v₁)
Δp = (2.0×10⁷ kg)[0.0075 (m/s) – 3.0 (m/s)]
Δp = (2.0×10⁷ kg)[–2.9925 (m/s)]
Δp = –5.985×10⁷ kg·(m/s)
Δx = ¹/₂(v₂ + v₁)(Δt)
Δx = ¹/₂[0.0075 (m/s) + 3.0 (m/s)](21 s)
Δx = ¹/₂[3.0075 (m/s)](21 s)
Δx = [3.0075 (m/s)](10.5 s)
Δx = 3.157875 m
And there we go! Problem solved! I hope this helps you! Have a great day!