In a water molecule, the dipole's negative portion is located on the oxygen atom and points towards the hydrogen atoms due to the difference in electronegativity between these elements.
Explanation:The water molecule, H2O, is a polar molecule, meaning it has a net dipole due to the presence of polar bonds, which result from a significant difference in electronegativities of the atoms involved. In a water molecule, the oxygen atom is more electronegative than hydrogen and therefore pulls the shared electrons toward itself. This creates a charge separation where the oxygen side of the molecule becomes partially negative, and the hydrogen side becomes partially positive. Looking at the provided options, the most accurate statement would be that the negative portion of the dipole is D) pointing from the oxygen through the hydrogen atoms
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The outside of a picture frame has a length which is 3 cm more than width. The area enclosed by the outside of the picture frame is 154 square cm. Find the width of the outside of the picture frame.
Answer:
To me it would only make sense for it to be 51.33
Explanation:
w≈51.33
Length 3
Area 154
The width of the outside of the picture frame is 11 cm.
What is width?The distinction between a figure's length and width is that length denotes a figure's longer side, while width denotes its shorter side.
Given
L = w + 3
L w = 154
(w+3)w = 154
w² + 3 w - 154 = 0
(w+14)(w-11) = 0
w = 11 cm
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The rate of change of speed of the belt is given by 0.06(10 - t) m/s^2, where t is in seconds. The speed of the belt is 0.8 m/s at t = 0. When the normal acceleration of a point in contact with the pulley is 40 m/s^2, determine (a) the speed of the belt; (b) the time required to reach that speed; and (c) the distance traveled by the belt.
Answer:
a) speed of belt = 0.8114m/s
b) time required = 0.02secs
c) distance traveled = 0.016m
Explanation:
The detailed and step by step calculation is as shown in the attached files.
Which answer most completely describes the difference/s between the gravitational force and the electrostatic force? Question 6 options:
The gravitational force is much weaker than the electrostatic force
The gravitational force is weaker and can only attract objects but the electrostatic force is strong and can both attract and repel
The gravitational force occurs throughout the universe but the electrostatic force only occurs on Earth
The gravitational force can only attract but the electrostatic force can attract or repel
I believe the answer would be D
The gravitational force is weaker and can only attract objects but the electrostatic force is strong and can both attract and repel. - This is the most completely describes the differences between the gravitational force and the electrostatic force.
What are the differences between gravitational force and electrostatic force?The main differences are:
Electrostatic force acts on particles with charge, whereas gravitational force acts on particles with mass.While the electrostatic force can be either attracting or repulsive, the gravitational force is always attractive.While the electric potential can be either negative or positive, the gravitational potential is always negative.Gravitational force is weak but has low range whereas electrostatic force is strong bur short ranged.Learn more about electrostatic force here:
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A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 4.68 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?
Answer:
0.8m/s
Explanation:
Weight of mas,F=763 N
Mass of man=[tex]\frac{F}{g}=\frac{763}{9.8}=77.86 kg[/tex]
By using [tex]g=9.8m/s^2[/tex]
Weight of flatcar=F'=3513 N
Mass of flatcar=[tex]\frac{3513}{9.8}=358.5 Kg[/tex]
Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg
Velocity of system=19.8m/s
Let v be the velocity of flatcar with respect to ground
Velocity of man relative to the flatcar=[tex]-4.68m/s[/tex]
Final velocity of man with respect to ground=v-4.68
By using law of conservation of momentum
Initial momentum=Momentum of car+momentum of flatcar
[tex]436.36(19.8)=77.86(v-4.68)+358.5v[/tex]
[tex]8639.928=77.86v-364.3848+358.5v[/tex]
[tex]8639.928+364.3848=436.36 v[/tex]
[tex]9004.3128=436.36v[/tex]
[tex]v=\frac{9004.3128}{436.36}[/tex]
[tex]v=20.6 m/s[/tex]
Initial speed of flatcar=Speed of system
Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s
A 16 V battery does 1705 J of work transferring charge. How much charge is transferred? Answer in units of C.
Answer:
Explanation:
Given
Voltage [tex]V=16\ V[/tex]
Work done [tex]W=1705\ J[/tex]
Work is Equivalent to energy
We know that Charge is given by
[tex]Q=I\cdot t[/tex]
where I=current
t=time
Energy [tex]E=P\times t[/tex]
P=power
P=V\times I
V=Voltage
I=current
Also Energy [tex]E=V\cdot I\cdot t[/tex]
[tex]I=\frac{E}{V\cdot t}[/tex]
Substitute the value of I in charge
[tex]Q=\frac{E}{V\cdot t}\times t[/tex]
[tex]Q=\frac{E}{V}[/tex]
[tex]Q=\frac{1705}{16}[/tex]
[tex]Q=106.56\ C[/tex]
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is:_______
a) g/.
b) g/4.
c) 2g.
d) g/2.
e) g.
Answer:
d) g/2
Explanation:
We need to use one of Newton's equations of motion to find the position of the stone at any time t.
x(t) = x₀(t) + ut - ¹/₂at²
Where
x₀(t) = initial position of the stone.
x(t) - x₀(t) = distance traveled by the stone at any time.
u = initial velocity of the stone
a = acceleration of the stone
t = time taken
On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.
=> 0 = x₀(t)
=> x₀(t) = 0
On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.
=> 0 = 0 + uT - ¹/₂gT² (a = g)
=> uT = ¹/₂gT²
=> g = 2u/T
On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.
=> 0 = 0 + u(2T) - ¹/₂a(2T)²
=> 2uT = 2aT²
=> a = u/T
By comparing we see that a = g/2.
Final answer:
The acceleration due to gravity on planet X is determined by comparing the time it takes for a stone thrown vertically upward to return to the thrower's hand. Using kinematic equations and given that it takes twice the time to return on planet X as it does on Earth, we deduce the acceleration on planet X is half that of Earth's gravity (g/2).
Explanation:
The question concerns finding the acceleration due to gravity on an unknown planet, which is an essential concept in physics. To determine the gravitational acceleration (g) on planet X, given that a stone thrown with the same initial upward velocity as on Earth returns in 2T instead of T, we can use the kinematic equations for motion under constant acceleration.
On Earth, the total time for the stone to return to the thrower's hand is T, which implies that the time to reach the maximum height is T/2. Applying the equation v = u + at, where v is the final velocity (0 m/s at maximum height), u is the initial upward velocity, a is the acceleration due to gravity (which is -g on Earth), and t is the time (T/2), we have 0 = u - g(T/2). Solving for u gives us u = g(T/2).
On Planet X, the total time for the stone to return to the thrower's hand is 2T, so the time to reach maximum height is T. Using the same equation with the acceleration on Planet X as aX, we get 0 = u - aX(T). Since u = g(T/2), we have g(T/2) = aX(T), leading to aX = g/2.
Therefore, the correct answer is (d) g/2.
You find a bag labeled "10 kg of lead-210". Its contents now weigh 1.25 kg. How many half-life periods have there been since the bag was weighed and labeled? Lead-210 is a radioisotope with a half-life of about 22 years.
Answer : The number of half-life periods will be, 3
Explanation : Given,
Initial amount of lead = 10 kg
Amount of lead after decay = 1.25 kg
Half-life = 22 years
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives
[tex]a_o[/tex] = Initial amount of the reactant
n = number of half lives
[tex]t_{1/2}[/tex] = half-life
Now put all the given values in the above formula, we get:
[tex]1.25=\frac{10}{2^n}[/tex]
[tex]2^n=8[/tex]
[tex]2^n=2^3[/tex]
[tex]n=3[/tex]
Thus, the number of half-life periods will be, 3
A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseball travels a horizontal distance of 16.0 m and rotates through an angle of 44.5 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?
Answer:
The tangential speed of a point on the "equator" of the baseball is 4.33 m/s.
Explanation:
Given that,
Linear speed of base ball = 42.5 m/s
Distance = 16.0 m
Angle = 44.5 rad
Radius of baseball = 3.67 cm
We need to calculate the flight time
Using formula of time
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{16.0}{42.5}[/tex]
[tex]t=0.376\ sec[/tex]
We need to calculate the number of rotation
Using formula of number of rotation
[tex]n=\theta\time 2\pi[/tex]
[tex]n=\dfrac{44.5}{2\pi}[/tex]
[tex]n=7.08[/tex]
We need to calculate the time for one rotation
Using formula of time
[tex]T=\dfrac{t}{n}[/tex]
Put the value into the formula
[tex]T=\dfrac{0.376}{7.08}[/tex]
[tex]T=0.053\ sec[/tex]
We need to calculate the circumference
Using formula of circumference
[tex]C=2\pi\times r[/tex]
Put the value into the formula
[tex]C=2\pi\times3.67\times10^{-2}[/tex]
[tex]C=0.23\ m[/tex]
The tangential speed is equal to the circumference divided by the time. it takes to complete one rotation.
We need to calculate the tangential speed
Using formula of tangential speed
[tex]v=\dfrac{C}{T}[/tex]
Put the value into the formula
[tex]v=\dfrac{0.23}{0.053}[/tex]
[tex]v=4.33\ m/s[/tex]
Hence, The tangential speed of a point on the "equator" of the baseball is 4.33 m/s.
A swimmer is determined to cross a river that flows due south with a strong current. Initially, the swimmer is on the west bank desiring to reach a camp directly across the river on the opposite bank. In which direction should the swimmer head?
Answer:
Swim in a strong north easterly direction so the the river carries him right across
Explanation:
The swimmer is on the west river bank with intentions of reaching the east river bank directly across him. If he just swims towards the east, the current of the river will carry him downstream landing him in a south easterly position, below where he intended to go. So the right thing to do would be to swim in a strong north easterly direction so the the river carries him right across.
To reach directly across the river from the west bank, the swimmer should head northeast to counteract the southward current of the river. This balancing act of velocities ensures they reach their desired location.
Explanation:This question deals with the concept of Relative Velocity in Physics. When a swimmer attempts to cross a river, they must take into account the current of the river pushing them downstream. The velocity of the swimmer relative to the water needs to be added to the velocity of the river to get the resultant velocity, which describes the direction the swimmer actually travels relative to the shore.
In this scenario, if the current of the river is flowing due south, the swimmer on the west bank should not aim directly for the point across the river. They should aim slightly upstream to the north to counteract the force of the current. Hence, the swimmer should head in a northeasterly direction, which when combined with the river current, will get them directly across to the eastern shore.
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A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s . The child is 500 m from the shore and 1100 m upstream of the boat dock from which the rescue team sets out.
If their boat speed is 7.30 m/s with respect to the water, at what angle from the shore must the boat travel in order to reach the child?
Final answer:
To reach a child being carried by a river current, the rescue boat, with a speed of 7.30 m/s versus water, must aim at a specific angle upstream accounting for the current's speed of 2.20 m/s. The calculation involves vector addition and trigonometry to counteract the current and ensure a direct path to the child.
Explanation:
The question involves calculating the angle at which a rescue boat must travel to reach a child being carried away by a river current. The river flows at a speed of 2.20 m/s, while the boat's speed in still water is 7.30 m/s. Given the distances from the shore to the child (500 m) and from the boat dock upstream to the child (1100 m), we need to determine the angle relative to the shore for the boat to make a direct rescue.
To solve this, we need to use the concept of vector addition, where the boat's velocity vector and the river's current vector combine to create a resultant vector that points directly to the child's location. The angle can be calculated using trigonometry, particularly the tangent function. However, since the specific calculations and steps to obtain the angle are not provided in the question or the previous examples, it is crucial to understand the principles of vectors and relative motion to approach this problem.
Essentially, the rescue team needs to aim the boat at an angle upstream to counteract the river's flow. The effect of the river's current will adjust the boat's path towards the child. It's a classic application of relative motion in two dimensions, requiring careful consideration of both the magnitude and direction of the vectors involved.
A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. If the driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?
Answer:
For 24 seconds force exerted is 5092 N towards opposite direction of motion of bus.For 3.90 seconds force exerted is 31333 N towards opposite direction of motion of bus.Explanation:
We have equation of motion v = u + at
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Case 1:-
Time, t = 24 s
Substituting
v = u + at
0 = 20 + a x 24
a = -0.8333 m/s²
Force = Mass x Acceleration = 6110 x -0.8333 = -5092 N
Force exerted is 5092 N towards opposite direction of motion of bus.
Case 2:-
Time, t = 3.90 s
Substituting
v = u + at
0 = 20 + a x 3.90
a = -5.13 m/s²
Force = Mass x Acceleration = 6110 x -5.13 = -31333 N
Force exerted is 31333 N towards opposite direction of motion of bus.
The average force exerted on the bus with gentle braking is approximately -5092 N, and with slamming the brakes, it is approximately -31342 N. Negative values indicate the force direction is opposite to the motion.
Explanation:To calculate the average force exerted on the bus during both stops, we will use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by the acceleration (F = ma). Here, the acceleration can be calculated using the formula for deceleration since the bus is coming to a stop:
For gentle braking: Deceleration = (Final velocity - Initial velocity) / Time = (0 - 20.0 m/s) / 24.0 s = -0.8333 m/s2.For slamming the brakes: Deceleration = (Final velocity - Initial velocity) / Time = (0 - 20.0 m/s) / 3.90 s = -5.1282 m/s2.Now, we can calculate the average force (F) exerted in both cases by multiplying the mass of the bus (m = 6110 kg) by the deceleration (a).
Gentle braking average force: F = m × a = 6110 kg × -0.8333 m/s2 = -5092 N (approximately).Slamming the brakes average force: F = m × a = 6110 kg × -5.1282 m/s2 = -31342 N (approximately).Since the bus is stopping, the negative sign indicates the direction of the force is opposite to the initial direction of motion.
__________ refers to the spreading of the signal that different parts of the signal arrive at different times at the destination. a. turnaround time b. propagation delay c. dispersion d. insulation e. attenuation
Final answer:
Dispersion refers to the spreading of a signal where different parts of the signal reach the recipient at varying times, affecting the wave by causing signal broadening due to the speed differential of its frequency components.
Explanation:
The term that refers to the spreading of a signal where different parts arrive at the receiver at different times is known as dispersion. Dispersion affects waves by causing a broadening of the signal, as different frequency components of the wave travel at different speeds. This is especially important in optical fibers, where dispersion can cause pulse broadening, potentially limiting the bandwidth and impairing the integrity of the data being transmitted. Option C is correct .
Factors such as scattering and refraction directly relate to the phenomenon of dispersion. For instance, scattering causes the redirection of light within a medium and refraction is the bending of a wave when it enters a medium where its speed is different. In the context of signal transmission, residence time is the average time a molecule spends in its reservoir, though this concept is less directly related to the question at hand.
A car is traveling at 22 m/s and slows to 4.9 m/s in 4.0 seconds. What is the acceleration? Give the answer to one decimal place.
Answer:
a= - 4.2 m/s².
Explanation:
Given that
u = 22 m/s
v= 4.9 m/s
t= 4 s
The average acceleration = a
We know v = u +at
v=final velocity
u=initial velocity
Now by putting the values in the above equation
4.9= 22 + a x 4
[tex]a=\dfrac{4.9-22}{4}\ m/s^2[/tex]
a= - 4.2 m/s².
Therefore the acceleration will be - 4.2 m/s².
a= - 4.2 m/s².
Negative indicates that velocity and acceleration is is opposite direction.
A gas occupies 97 mL at 130 kPa. Find its volume at 225 kPa You must show all of your work to receive credit. .
Answer:
The answer to your question is 56 ml
Explanation:
Data
V1 = 97 ml
P1 = 130 kPa
V2 = ?
P2 = 225 kPa
Formula
Use Boyle's law to solve this problem because it relates to the volume and the pressure.
V1P1 = V2P2
Solve for V2
V2 = (V1P1) / P2
Substitution
V2 = (97 x 130) / 225
Simplification
V2 = 12610 / 225
Result
V2 = 56 ml
One of the moons of a planet is at an average distance of 0.7 AU from the planet and its period of revolution is 27.8 earth days. What is the period of revolution of another moon of the same planet that is at 1.2 AU from in?
a. 62.3 earth days
b. 128.4 earth days
c. 224 earth days
d. 27.8 earth days
Answer:
option A
Explanation:
given,
distance of the moon 1 from planet = 0.7 AU
Time period of moon 1 = 27.8 earth days
distance of the moon 2 from the planet = 1.2 AU
time period of the moon 2 = ?
using Kepler's formula
[tex]T^2 = \dfrac{4\pi^2}{GM}\ R^3[/tex]
now,
only variable in the formula is time period and distance.
so,
[tex]\dfrac{T_2^2}{T_1^2}=\dfrac{R_2^3}{R_1^3}[/tex]
[tex]\dfrac{T_2^2}{27.8^2}=\dfrac{1.2^3}{0.7^3}[/tex]
T₂² = 3893.49
T₂ = 62.3 earth days
Hence, the correct answer is option A
A bicyclist is moving toward a sheer wall while holding a tuning fork rated at 488 Hz. 1) If the bicyclist detects a beat frequency of 8 Hz (between the waves coming directly from the tuning fork and the echo waves coming from the sheer wall), calculate the speed of the bicycle. Assume the speed of sound is 343 m/s.
Answer:
speed of the bicycle = 2.81 m/s
Explanation:
given data
tuning fork rated f = 488 Hz
beat frequency fb = 8 Hz
speed of sound vs = 343 m/s
solution
we get here frequency that rider is listening that is
rider hearing frequency = 488 Hz + 8 Hz
rider hearing frequency = 496 Hz
and when it bouncing from sheer wall at frequency of related to stationary object
frequency = 488 Hz + 4 Hz
frequency = 492 Hz
so now we get speed of the bicycle that is
speed of the bicycle = ( [tex]\frac{492}{488}[/tex] × 343 ) - 343 Hz
speed of the bicycle = 2.81 m/s
The speed of the bicycle is 2.81 m/s.
Based on the given information,
• The tuning fork rated at frequency (f) 488 Hz.
• The beat frequency (fb) given is 8Hz.
• The speed of the sound is 343 m/s.
Now with the help of the given information, the rider is hearing a frequency of,
[tex]=(f+fb)\\=(488+8)\\=496 Hz[/tex]
However, it is bounding from sheer wall at the frequency of,
[tex]=(488+\frac{8}{2})\\= 492 Hz[/tex]
Now the speed of the bicycle will be,
[tex]=(\frac{492}{488} * 343)-343\\=2.811 m/s\\= 2.81 m/s[/tex]
Thus, the speed of the bicycle is 2.81 m/s.
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an electron aquires 3.45 e-16j of kinetic energy when it is accelerated by an electric field from plate a to b in a computer monitor what is the potential difference between the plates and which plate has the higher potential?
Answer:
Potential difference will be equal to 2156.25 volt
Explanation:
We have given kinetic energy of electron [tex]KE=3.45\times 10^{-16}J[/tex]
Charge on electron [tex]e=1.6\times 10^{-19}C[/tex]
Let the potential difference is V
So energy electron will be equal to [tex]E=qV[/tex], here q is charge on electron and V is potential difference
This energy will be equal to kinetic energy of the electron
So [tex]1.6\times 10^{-19}\times V=3.45\times 10^{-16}[/tex]
V = 2156.25 volt
So potential difference will be equal to 2156.25 volt
If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was, how many rotations per second will result?
Answer:
There are finally 4 rotations per second.
Explanation:
If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was. We need to find the final angular velocity. It is a case of conservation of angular momentum such that :
[tex]I_1\omega_1=I_2\omega_2[/tex]
Let [tex]I_1=I[/tex] , [tex]I_2=\dfrac{I}{4}[/tex] and [tex]\omega_1=1[/tex]
So,
[tex]\omega_2=\dfrac{I_1\omega_1}{I_2}[/tex]
[tex]\omega_2=\dfrac{I\times 1}{(I/4)}[/tex]
[tex]\omega_2=4\ rev/sec[/tex]
So, there are finally 4 rotations per second. Hence, this is the required solution.
Final answer:
When a trapeze artist reduces their rotational inertia to one fourth while rotating, their rotational speed increases to four times its original rate to conserve angular momentum, resulting in 4 rotations per second.
Explanation:
The question pertains to the conservation of angular momentum, which is a principle in physics stating that if no external torque acts on an object, the total angular momentum will remain constant. For a trapeze artist or any rotating body, if the moment of inertia decreases, the rotational speed (number of rotations per second) must increase proportionally to conserve angular momentum.
If a trapeze artist is rotating once each second and contracts to reduce the rotational inertia to one fourth, her rotational speed must increase to four times what it was to conserve angular momentum. Therefore, the new rotation rate will be 4 rotations per second.
One of the many steps in the production of toothpaste is to screw the caps on the tubes, which is still a manual process, performed by on man, Mr. Bucket. Which statement about this situation is BEST?
A) This is most appropriate for an output measure of capacity.
B) In this case, the capacity of this step is not the maximum rate of output.
C) This is most appropriate for an input measure of capacity.
D) Utilization of the worker at this process step cannot be measures as it is a manual process.
Answer:A) This is most appropriate for an output measure of capacity.
Explanation: Toothpaste is a paste used to promote oral health and hygiene, Most toothpaste contain an active agent(floride) and it acts as an abrasive agent which helps to remove dental plaques or unwanted attachments to the teeth.
Among the options the Option(A) which says that THIS IS MOST APPROPRIATE FOR THE MEASURE OF CAPACITY is the statement that best Describes the situation. Toothpastes have a long history as it can be traced to about 4000years ago.
The distance between two successive crests of a certain transverse wave is 1.40 m. 25 crests pass a given point along the direction of travel every 12.4 s. Calculate the wave speed. Answer in units of m/s.
Answer: 2.82m/s
Explanation:
Wave speed(v) is the product of the frequency (f) of a wave and its wavelength (¶)
V = f¶
Since wavelength is the distance between two successive crests or trough, the wavelength given is 1.40m i.e ¶ = 1.40m
If the total crest is 25, this means that 25 oscillations occurs in 12.4s.
Since the number of oscillation made in one second is the frequency, frequency = 25/12.4 = 2.01cycle/seconds
Speed of the wave v = 2.01×1.4
V =
Wave speed is 2.82m/s
What does the slope of gravitational potential energy vs velocity^2 graph represent?
Explanation:
Mechanical energy (the sum of potential and kinetic energy) is constant:
ME = PE + KE
ME = PE + ½ mv²
PE = ME − ½ mv²
So the slope of the line is -½ of the mass.
In a gravitational potential energy vs velocity^2 graph, the slope symbolizes the mass of the object. The graph represents the conversion of energies - kinetic and gravitational throughout a motion course. Peaks correspond to max potential energy and min kinetic energy, and vice-versa.
Explanation:The slope of the gravitational potential energy vs the square of the velocity graph represents the object's mass in specific physical circumstances. This is because the equation for kinetic energy (which this graph is showing an indirect relationship with) is (1/2)mv^2, where m is mass and v is velocity. The slope can be interpreted as mass due to the rearrangement of this formula into the form used for this graph (Potential energy = 1/2 * m * v^2), wherein the slope (m) symbolizes the mass of the object.
Moreover, the information encapsulated in the potential energy graph as a whole represents energy conversion from kinetic to gravitational and vice versa. At the peak of a slope/motion course, potential energy is at its maximum and kinetic energy at its minimum. The opposite occurs at the bottom of the slope: both kinetic energy and potential energy reach their minimum values.
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A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100 rad/s2. After making 2844 revolutions, its angular speed is 140 rad/s.
Complete answer
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100rad/s2. After making 2844 revolutions, its angular speed is 140rad/s
(a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
Answer:
a) 126.59 radians per second
b) 134.1 seconds
Explanation:
We can use the rotational kinematic equations for constant angular acceleration.
a) For a) let’s use:
[tex]\omega^{2}=\omega_{0}^{2}+2\alpha\varDelta\theta [/tex] (1)
with [tex] \omega_{0}[/tex] the initial angular velocity, [tex] \omega[/tex] the final angular velocity, [tex] \alpha[/tex] the angular acceleration and [tex] \Delta \theta [/tex]the revolutions on radians (2844 revolutions = 17869.38 radians). Solving (1) for initial velocity:
[tex]\sqrt{\omega^{2}-2\alpha\varDelta\theta}=\omega_{0} [/tex]
[tex]\omega_{0}^2=\sqrt{(140)^2 -(2)(0.100)(17869.38)=126.59 \frac{rad}{s}}[/tex]
b) Knowing those values, we can use now the kinematic equation
[tex] \omega=\omega_{0}+\alpha t[/tex]
with t the time, solving for t:
[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{140-126.59}{0.1} [/tex]
[tex] t=134.1 s[/tex]
If the L-shaped rod has a moment of inertia I=9kgm2, F1=12N, F2=27N, and again F3=0, how long a time t would it take for the object to move through 45∘ ( π/4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures.
Answer:
Time, t = 2.80seconds
Explanation:
Calculating net Torque, T
Torque(net)= F2(0.08m) - F1(0.03)
Torque(net) = (27×0.08)-(12×0.03)
Torque(net)= 2.16 - 0.36 =1.8N/m
Pi/4= Torque(net)/ Inertia
3.142/4= 1.8/9 = 0.2rad/s^2=a
Pi/4= 1/2 at^2
3.142/4=1/2×0.2 ×t^2
0.7855= 0.1 t^2
t = sqrt(0.7856/0.1)
t= sqrt(7.855)
t=2.80 seconds
By Assuming that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. then the time in seconds is 2.80seconds
What are the types of force ?Force can be denoted as the parameter obtained by pushing or pulling of any object which result in object’s interaction or movement, without force the objects can not move, it can be stopped or change the direction.
Force is a quantitative property and it is an interaction between two physical bodies, means an object and its environment, there are different types of forces in nature.
If an object is in its moving state then it will be either static or motion, if it is pushed or pulled the object will change its position and The external push or pull upon the object called as Force.
There are two types of forces such as contact force which are the force occurs when we apply some effort on the object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force
Non-Contact forces can be defined as the force occur from a distance such as Electromagnetic Force, Gravitational Force, Nuclear Force
Calculating net Torque, T, Torque(net)= F2(0.08m) - F1(0.03)
Torque(net) = (27×0.08)-(12×0.03)
Torque(net)= 2.16 - 0.36 =1.8N/m
Pi/4= Torque(net)/ Inertia
3.142/4= 1.8/9 = 0.2rad/s^2=a
Pi/4= 1/2 at^2
3.142/4=1/2×0.2 ×t^2
0.7855= 0.1 t^2
t = sqrt(0.7856/0.1)
t= sqrt(7.855)
t=2.80 seconds
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Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.95 m/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 39.6 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.50 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
Answer:
Explanation:
Given
Speed of Henrietta is [tex]v=2.95\ m/s[/tex]
Height of tower [tex]h=39.6\ m[/tex]
Bruce throws the bangle after 7.5 s
During 7.5 s Henrietta travels
[tex]x=2.95\times 7.5=22.125\ m[/tex]
Suppose bangle hit the ground after t sec so bangle will has to cover a distance of x and distance traveled by Henrietta during this time t
Range of bangle when thrown with speed u
[tex]R=u\times t[/tex]
[tex]R=x+2.95\times t-----1[/tex]
bangle will also cover a vertical distance of 39.6 m
so using equation of motion
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
here initial vertical velocity is zero
[tex]39.6=0+\frac{1}{2}\cdot 9.81\cdot t^2[/tex]
[tex]t=\sqrt{8.0816}[/tex]
[tex]t=2.84\ s[/tex]
Substitute the value of t in
[tex]u\times 2.84=22.125+2.95\times 2.84[/tex]
[tex]u=10.743\ m/s[/tex]
An elevator packed with passengers has a mass of 1950 kg.
(a) The elevator accelerates upward (in the positive direction) from rest at a rate of 2 m/s2 for 2.05 s. Calculate the tension in the cable supporting the elevator in newtons.
(b) The elevator continues upward at constant velocity for 8 s. What is the tension in the cable, in Newtons, during this time?
(c) The elevator experiences a negative acceleration at a rate of 0.45 m/s2 for 2.6 s. What is the tension in the cable, in Newtons, during this period of negative accleration?
(d) How far, in meters, has the elevator moved above its original starting point?
Answer:
a) Fa= Tension in the cable= 23010N
b) 19110N
c) 18232.5N
d=41.28m ,final velocity=0
Explanation:
a) Newtons 2nd law is given by Fnet=EF=ma
Fa= m(a + g)
Fa= 1950 (2+9.8)
Fa= 1950×11.8= 23010N
b) Fnet=0
Therefore Fb= W= 1950×9.8= 19110N
c) Fc= m(g - a)
Fc= 1950(9.8 - 0.45)
Fc= 18232.5N
d) First distance
ya= vt + 0.5at^2 = 0 + (0.5)(2)(2.05)^2
ya= 4.203m
yb= vt= 4.203×8=33.62m
yc = vt - (0.5at^2)
yc= 4.203×2.6) - (0.5×0.45×8^2)
yc = 10.93-14.4
yc =-3.46m
Dtotal = -3.46+33.62+4.203
Dtotal=41.28m
Explanation:
Below is an attachment of the solution.
A stuntman with a mass of 82.5 kg swings across a pool of water from a rope that is 12.0 m. At the bottom of the swing the stuntman's speed is 8.65 m/s. The rope's breaking strength is 1,000 N. Will the stuntman make it across the pool without falling in?
Answer:
The stuntman will not make it
Explanation:
At the bottom of the swing, the equation of the forces acting on the stuntman is:
[tex]T-mg = m\frac{v^2}{r}[/tex]
where:
T is the tension in the rope (upward)
mg is the weight of the man (downward), where
m = 82.5 kg is his mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]m\frac{v^2}{r}[/tex] is the centripetal force, where
v = 8.65 m/s is the speed of the man
r = 12.0 m is the radius of the circule (the length of the rope)
Solving for T, we find the tension in the rope:
[tex]T=mg+m\frac{v^2}{r}=(82.5)(9.8)+(82.5)\frac{8.65^2}{12.0}=1322 N[/tex]
Since the rope's breaking strength is 1000 N, the stuntman will not make it.
in circuit A, two resistors R1 and R2 are connected in series. In circuit B, two identical resistors are connected in parallel. Both circuits are powered by identical energy sources and include no other resistors. In which circuit, if any, is the current greater across R1?
A. Circuit A
B. Neither circuit
C. Circuit B
The current across R1 in circuit A is the same as the current across the resistors in circuit B.
Explanation:In circuit A, two resistors R1 and R2 are connected in series. In circuit B, two identical resistors are connected in parallel. In a series circuit, the current is the same in each resistor. Therefore, the current across R1 in circuit A is the same as the current across the resistors in circuit B, since they are connected in parallel.
Why are biogeochemical cycles important? Earth is a closed system. That means the amount of rocks, metals, carbon, nitrogen, oxygen, and water on Earth.... So, it is essential that biogeochemical cycles..... these materials as they move through Earth's subsystems.
Answer:
The earth is considered to be a closed system, where the concentration of rocks, metals, C, N, O and H₂O on earth remains constant.
So, it is essential that the biogeochemical cycle renews these materials while moving through the earth subsystems.
Explanation:
The biogeochemical cycle is usually defined as the circulation of chemical components such as Carbon, Nitrogen, Phosphorous, Oxygen, rocks, and metals within the different spheres and interaction of biotic and non-biotic factors.
These components are continuously in motion from one sphere to another. For example, the organism takes up the elements through food and some of these components such as carbon is released into the atmosphere in the form of CO₂. After the death of the organisms, the decomposers feed on these elements, where some of it is consumed by them and the remaining is eliminated into the atmosphere.
So, this is how the elements are renewed, and they are conserved. The total concentration of these elements remains constant considering the earth to be a closed system.
Thus, the correct answers are given above.
Answer:
remains constant and renews
Explanation:
Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from rest by the locomotive. Discuss the advisability of this loose coupling and slack between cars from the point of view of impulse and momentum.
Answer:
Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).
Explanation:
Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour. (Assume 1 second deceleration time)
Force is derived from the equation F = ma, resulting in units of Newtons (N). For a 10-ton truck decelerating over 1 second from 55 mph, we convert mass to kg and speed to m/s to calculate force, likely resulting in a force measured in kilonewtons (kN) or meganewtons (MN) due to the large mass and deceleration involved.
Explanation:The subject of this question is how to derive a unit for force starting with SI base units, and finding a convenient unit for the force resulting from a specific collision scenario. Starting from the base, force (F) can be defined using Newton's second law, F = ma (force equals mass times acceleration), where mass has units of kilograms (kg), and acceleration is measured in meters per second squared (m/s²). Thus, the unit of force in the SI system is the Newton (N), defined as the force needed to accelerate a 1-kg object by 1 m/s².
To find a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour and assuming a 1 second deceleration time, we first convert the truck's mass to kilograms (10 tons = 9,071.85 kg, assuming 1 ton = 907.185 kg) and its speed to meters per second (55 mph ≈ 24.5872 m/s, assuming 1 mile = 1.60934 km and 1 hour = 3600 seconds). The deceleration rate (a) can be calculated assuming the final velocity (v) is zero and the time (τ) is 1 second, giving us a = -24.5872 m/s² (negative sign denotes deceleration). Thus, applying F = ma, the force of the collision can be calculated, and due to the large numbers involved, units such as kilonewtons (kN) or meganewtons (MN) may be more convenient depending on the exact outcome of the calculation.
The force resulting from the collision with a 10-ton trailer truck moving at 55 miles per hour is approximately 245.87 kN.
Starting with the equation for force, F = ma, where:
- F represents force,
- m represents mass,
- a represents acceleration.
In the International System of Units (SI), the base units are:
- Mass ( m ) is measured in kilograms (kg).
- Acceleration ( a ) is measured in meters per second squared [tex](\( \mathrm{m/s^2} \)).[/tex]
Therefore, the unit of force ( F ) in the SI system is:
[tex]\[ \mathrm{kg \cdot m/s^2} \][/tex]
This unit is commonly known as the Newton (N).
Now, let's calculate the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour with a deceleration time of 1 second.
First, we need to convert the mass of the trailer truck from tons to kilograms:
[tex]\[ 1 \text{ ton} = 1000 \text{ kg} \][/tex]
So, [tex]\( 10 \text{ tons} = 10,000 \text{ kg} \).[/tex]
Next, let's convert the speed from miles per hour to meters per second:
[tex]\[ 1 \text{ mile} = 1609.34 \text{ meters} \][/tex]
[tex]\[ 1 \text{ hour} = 3600 \text{ seconds} \][/tex]
So, [tex]\( 55 \text{ miles per hour} \)[/tex] is equivalent to:
[tex]\[ \frac{55 \times 1609.34}{3600} \text{ meters per second} \approx 24.587 \text{ m/s} \][/tex]
Given that deceleration time ( t ) is [tex]\( 1 \text{ second} \)[/tex], we can use the formula for acceleration:
[tex]\[ a = \frac{v}{t} \][/tex]
where v is the change in velocity.
Since the truck is decelerating, the change in velocity is from [tex]\( 24.587 \text{ m/s} \) to \( 0 \text{ m/s} \).[/tex]
[tex]\[ a = \frac{0 - 24.587}{1} \text{ m/s}^2 = -24.587 \text{ m/s}^2 \][/tex]
Now, we can calculate the force:
[tex]\[ F = m \times a = 10,000 \text{ kg} \times (-24.587 \text{ m/s}^2) \][/tex]
[tex]\[ F \approx -245,870 \text{ N} \][/tex]
Since force is a vector quantity and its direction is opposite to the direction of motion, the negative sign indicates that the force is acting in the opposite direction of the truck's motion.
For convenience, we can express this force in kilonewtons ( kN ):
[tex]\[ -245,870 \text{ N} = -245.87 \text{ kN} \][/tex]
So, the suggested convenient unit for the force resulting from the collision is [tex]$\mathrm{kN}$[/tex].
Complete Question:
Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour. (Assume 1 second deceleration time)
a. MN
b. GN
c. mN
d. kN
e. TN