Answer:
hard line, soft line
Explanation:
A cylinder which is in a horizontal position contains an unknown noble gas at 54700 Pa 54700 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 0.150 m 0.150 m , while 16800 J 16800 J of heat is removed from the gas. If the piston has a radius of 0.272 m 0.272 m , calculate the change in internal energy Δ U ΔU of the system.
Answer:
-14892.93 J
Explanation:
given,
Pressure, P = 54700 Pa
heat removed, Q = 16800 J
radius, r = 0.272 m
distance, d = 0.150 m
internal energy ΔU of the system = ?
We know,
Force = Pressure x area
F = P x A
F = 54700 x π x r²
F = 54700 x π x 0.272²
F = 12713.79 N
Work done = F.d
W = 12713.79 x 0.15
W = 1907.07 J
Change in internal energy ΔU is
ΔU = W + Q
= 1907.07 + (-16800)
= -14892.93 J
Hence, the change in internal energy is equal to -14892.93 J
Which of the following would decrease the resistance in a wire?
Increase the thickness of the wire
Increase the mass of the wire
Decrease the thickness of the wire
Decrease the mass of the wire
Increase the thickness of the wire would decrease the resistance in a wire
Explanation:
Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.
Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.
Answer:
Increase the thickness of the wire
Explanation:
I just took the test.
Your textbook discusses the cosmic calendar, a model of the history of the universe scaled to a single year. The length of time represented by one month on this cosmic calendar is therefore closest to__________
Answer:
1.15 Billion Years
Explanation:
If the average age of the universe is 13.772 billion years and that equals to a year in cosmic calendar, then the length of time in a month will be 13.772 / 12 = 1.14766666667 ~ 1.15 Billion Years
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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While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38 degrees with the vertical. While driving back home moments later at the same speed but in the opposite direction, you see that the rain is falling straight down. From these observations, determine the speed of the raindrops relative to the ground. From these observations, determine the angle of the raindrops relative to the ground.
Final answer:
To determine the speed of the raindrops relative to the ground and their angle, we analyze the observations of rain angle in two car speeds using trigonometry and the Pythagoras theorem, concluding with the application of the tangent function and arctan.
Explanation:
When driving north at 25 m/s, rain appears to make an angle of 38 degrees with the vertical due to the combination of the rain's vertical speed and the horizontal speed of the car. However, when driving south at the same speed, the rain appears to fall vertically, indicating the horizontal component of the rain's velocity is equal to the car's speed. To determine the speed of the raindrops relative to the ground and the angle of the raindrops relative to the ground, we can use trigonometry.
Given the observation that the rain appears vertical when driving south at 25 m/s, it implies the horizontal velocity of the rain is 25 m/s (equal but opposite to the car's velocity, thereby canceling it out). From the 38-degree angle observation, we can use tan(38 degrees) = vertical component / horizontal component to find the vertical speed. The answer is derived from tan(38 degrees) = vertical component / 25 m/s, which allows us to calculate the vertical component.
The speed of the raindrops relative to the ground is then found by calculating the resultant of the horizontal and vertical components using Pythagoras theorem, and the angle of the raindrops is found by taking the arctan of the vertical component over the horizontal component.
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
Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop. B speed up. C keep moving at the same speed. D follow a circular path.
Answer:
A slow down and stop
Explanation:
When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.
A 75.3 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 8.1 m starting from rest, its speed is 7.1 m/s. Find the magnitude of the net force on the bobsled. Answer in units of N
Answer:
233.43 N
Explanation:
Force: This is the product of mass and acceleration of a body.
The formula for force is given as,
F = ma .............. Equation 1
Where F = Net force on the bobsled, m = mass of the bobsled, a = acceleration of the bobsled
We can look for a using the equation of motion
v² = u² + 2as.............. Equation 2
Where V = final velocity, u = initial velocity, a = acceleration, s = distance.
making a the subject of the equation,
a = (v²-u²)/2s................... Equation 3
Given: v = 7.1 m/s, u = 0 m/s ( from rest), s = 8.1 m.
Substitute into equation 3
a = (7.1²-0²)/(2×8.1)
a = 50.41/16.2
a = 3.1 m/s²
Also given: m = 75.3 kg
Substitute into equation 1
F = 75.3×3.1
F = 233.43 N
Hence the net force on the bobsled = 233.43 N
In a transverse wave, the motion of the disturbance is in what direction relative to the wave motion? opposite parallel perpendicular in the same direction
Answer:
[tex]\displaystyle Perpendicular[/tex]
Explanation:
Longitudinal waves are parallel to the direction of the motion of the disturbance, while transverse waves are perpendicular to the direction of the motion of the disturbance.
I am joyous to assist you anytime.
In a transverse wave, the motion of the disturbance is perpendicular to the wave motion.
A transverse wave is a type of wave where movements oscillate along paths at a right angle to the advance of the wave.Examples of this type of wave include seismic waves and electromagnetic waves.The electromagnetic waves can be both radio waves and light waves.In conclusion, in a transverse wave, the motion of the disturbance is perpendicular to the wave motion.
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A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions:
a. when the ropes are horizontal
b. when the ropes make a 30.0° angle with the vertical
c. at the bottom of the circular arc
Answer:
A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.
Gpe @ A=mgh=40*9.81*2=784.8J
B. At 30degree vertical angle the vertical displacement from lowest position is given by
2-2cos(30)=2-1.73=0.27m
Gpe @B= 40*9.81*0.27=106 J
C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.
(a) The gravitational potential energy when the ropes are horizontal is 784.8 J.
(b) The gravitational potential energy when the ropes make a 30 degree angle with vertical is 106 J.
(c) At the bottom of circular arc the gravitational potential energy is zero.
Given data:
The mass of child is, m = 40.0 kg.
The length of ropes is, L = 2 m.
The energy possessed by any object under the influence of gravity and by the virtue of position of object is known as gravitational potential energy.
(a)
As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position. Then the gravitational potential energy is,
[tex]PE = mgL\\\\PE = 40 \times 9.8 \times 2\\\\PE=784.8 \;\rm J[/tex]
Thus, the gravitational potential energy when the ropes are horizontal is 784.8 J.
(b)
At 30 degree vertical angle the vertical displacement from lowest position, the gravitational potential energy is given by,
[tex]PE = mgL(1-cos30^{\circ})\\\\PE = 40 \times 9.8 \times 2 \times (1-cos30^{\circ})\\\\PE = 106 \;\rm J[/tex]
Thus, the gravitational potential energy when the ropes make a 30 degree angle with vertical is 106 J.
(c)
At the bottom of circular arc it's gravitational potential energy is zero relative to lowest position as bottom of arc itself is lowest position.
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According to Coulomb’s Law, the force between two charged objects is related to _____.
a. the inverse of the square of the distance separating them
b. the distance separating them
c. the inverse of the charges of the objects
d. the mass of the objects
Answer:
A.) the inverse of the square of the distance separating them
Explanation:
Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."
Mathematically, F = kq1q2/r²
Where q1 and q2 are the charges
r is the distance between the charges.
According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.
Coulomb’s Law states that the force between two charged objects is directly related to the inverse of the square of the distance between them.
Explanation:According to Coulomb’s Law, the force between two charged objects is related to (a) the inverse of the square of the distance separating them. This law states that the force (F) between two charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. It's written as F = k * q1 * q2 / r^2, where k is Coulomb's constant. Thus, if the distance between the objects is doubled, the force between them is reduced to one fourth of its original value.
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person walks in the following pattern: 2.3 km north, then 2.5 km west, and finally 5.4 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point
Answer:
a.3.86 Km
b.Direction=233.3 degree
Explanation:
We are given that
Person wale in North direction,DE=2.3 Km
Person walk in West direction==BD=2.5 Km
Person walk in South direction,EC=5.4 Km
a.We have to find the distance between initial and final position.
Vector AB=-2.5 Km
BD=-EA=-2.3 Km
CA=y=-(EC-EA)=-(5.4-2.3))=-3.1 m
AB=-DE=x=-2.5 Km
Magnitude of resultant vector=[tex]\sqrt{AB^2+CA^2}=\sqrt{(-2.5)^2+(-3.1)^2}[/tex]
Magnitude of resultant vector=3.86 Km
Hence, the distance between initial and final position=3.86 Km
b.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
Direction=[tex]\theta=tan^{-1}(\frac{-3.1}{-2.3})\approx 53.3^{\circ}[/tex] S of W
x-coordinate and y-coordinate are negative therefore, the angle lies in third quadrant.
When we measured from East then
[tex]\theta=53.3+180=233.3^{\circ}[/tex]
Therefore, the direction=233.3 degree
The Pentium 4 Prescott processor, released in 2004, had a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average, it consumed 10 W of static power and 90 W of dynamic power.The Core i5 Ivy Bridge, released in 2012, had a clock rate of 3.4 GHz and voltage of 0.9 V. Assume that, on average, it consumed 30 W of static power and 40 W of dynamic power.Find the percentage of the total dissipated power comprised by static power for the Pentium 4 Prescott. Round to a whole integer between 0-1
Answer:
For Pentium 4 Prescott:
% of Static Power = 10
For core i5 Ivy Bridge:
% of Static Power = 43
Given Information:
Static Power of P4 = 10 W
Dynamic Power of P4 = 90 W
Static Power of i5 = 30 W
Dynamic Power of i5 = 40 W
Required Information:
% of static power w.r.t total power dissipation = ?
Explanation:
For Pentium 4 Prescott:
% of static power = static power/total power * 100
% of static power = 10/(10 + 90) * 100
% of static power = 10/(100) * 100
% of static power = 10
For core i5 Ivy Bridge:
% of static power = static power/total power * 100
% of static power = 30/(30 + 40) * 100
% of static power = 30/(70) * 100
% of static power = 43 (rounded to nearest whole integer)
Final answer:
The static power comprises 10% of the total dissipated power for the Pentium 4 Prescott processor.
Explanation:
To calculate the percentage of the total dissipated power that is comprised of static power for the Pentium 4 Prescott, we need to add the static power to the dynamic power to get the total power and then find what percentage the static power is of the total power.
Total power for Pentium 4 Prescott = Static power + Dynamic power
Total power for Pentium 4 Prescott = 10 W + 90 W
Total power for Pentium 4 Prescott = 100 W
Percentage of static power = (Static power / Total power) imes 100
Percentage of static power = (10 W / 100 W) imes 100
Percentage of static power = 10%
So, the static power comprises 10% of the total dissipated power for the Pentium 4 Prescott.
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.
A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally rotating at 55 rpm. What was the initial rotation rate in radians per second?
Answer: 5.76 rads/s
Explanation:
The initial rotation is 55 rpm
1 rev = 2π radians
55 revs = 55 × 2π/1 = 345.58 radians/min
345.58 rads/min = 345.58rads/60s = 5.76 rads/s
What is the net charge on a sphere that has the following? (a) 5.87 106 electrons and 8.11 106 protons C
Answer:
3.6 * 10^-13 C
Explanation:
The net charge of the sphere will be the sum of the total electron charge and total proton charge. Mathematically,
Q = Qe + Qp
TOTAL ELECTRON CHARGE:
An electron has an electronic charge of -1.6022 * 10^-19C.
Hence, the charge of 5.87 * 10^6 electrons will be:
Qe = - 1.6022 * 10^-19 * 5.87 * 10^6
Qe = - 9.405 * 10^-13 C
TOTAL PROTON CHARGE:
A proton has an electronic charge of 1.6022 * 10^-19. Hence, 8.11 * 10^6 protons will have:
Qp = 1.6022 * 10^-19 * 8.11 * 10^6
Qp = 1.3 * 10^-12 C
=> Q = (-9.405 * 10^-13) + (1.3 * 10^-12)
Q = 3.6 * 10^-13 C
Answer:
3.58*10⁻¹³ C
Explanation:
When we have a net charge on a sphere, this means that there must be a difference in the number of electrons and protons on the sphere, otherwise, the sphere would be electrically neutral.
In this case, we can find this difference as follows:
Np -Ne = 8.11*10⁶ protons - 5.87*10⁶ electrons = 2.24*10⁶ protons.
The total charge carried by all these protons is just the product of the charge of only one proton (which is equal to the elementary charge e) times the number of excess protons, as follows:
Qnet = 2.24*10⁶ protons * 1.6*10⁻19 C/proton = 3.58*10⁻¹³ C
Kohlberg believed that_____________.A. moral development progressed through distinct, qualitatively different stages. B. moral development was made up of constant, gradual growth.C. the stages of development were distinct; they did not build on one another.D. children did not necessarily progress through the stages in order.
Answer:
A. moral development progressed through distinct, qualitatively different stages.
Explanation:
Kolhberg’s theory of moral development states that we progress through three levels of moral thinking that build on our cognitive development.
Is this statement true or false? The next generation of nuclear power plants being built in California and South Africa are even safer and more efficient than previous generations of power plants.
Answer:
True
Explanation:
Modern safer and cheaper nuclear reactors can not only meet the range of our long term energy demands, they can also fight global warming.
Modern techniques provide ways to reduce radioactive waste amount. "A closed fuel cycle may be switched on for new kinds of nuclear plants. Alternatively, the waste is chemically dissuaded to transform the reusable element into fuel. This implies that nuclear waste would not be buried.
Answer:
Truee
Explanation:
Wayne exerts a force of 63 N to pull a 308 N sled along a snowy path using a rope that makes a 33° angle with the ground. The sled moves 11.3 m in 3.1 s. What is Wayne’s power? Answer in units of W.
Answer:
P = 192.6 Watt
Explanation:
given,
Force of Pull, F = 63 N
Weight of the sled, W = 308 N
Angle made with ground, θ = 33°
Movement of sled,s= 11.3 m
time, t = 3.1 s
Power of Wayne = ?
Work done = F .s cos θ
W = 63 x 11.3 x cos 33°
W = 597 J
We know.
[tex]Power = \dfrac{Work\ done}{time}[/tex]
[tex]P= \dfrac{597}{3.1}[/tex]
P = 192.6 Watt
The Wayne's Power is equal to P = 192.6 Watt
Calculate the discharge (in cubic meters/second) if a large stream's average velocity is 3 meters/second, its stage is 14 meters, and its width as measured bank-to-bank is 27 meters.
Answer:
Explanation:
Given
average velocity of stream is [tex]v_{avg}=3\ m/s[/tex]
Stage of stream i.e. depth [tex]d=14\ m[/tex]
width of stream [tex]w=27\ m[/tex]
We know that discharge of a stream is given by
[tex]Q=A\cdot v[/tex]
where A=area of cross-section
v=average velocity
[tex]A=14\times 27\ m^2[/tex]
Therefore discharge is given by
[tex]Q=14\times 27\times 3[/tex]
[tex]Q=1134\ m^3/s[/tex]
Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Isaac, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed?
Answer: The question is incomplete or missing details. here is the remaining part of the question ;
1. impossible to determine
2. half of Isaac’s
3. the same as Isaac’s
4. twice Isaac’s
The angular speed of feng will be the same as that of Isaac. Hence the answer is option 3
Explanation:
Since we have been told that both feng and isaac are riding on a merry go round i.e in a circular motion, irrespective of how fast one ride above the other, the angular speed will be constant since they are riding on a merry go round, as such both feng and isaac will maintain equal angular speed, hence the angular speed of feng will be the same as that of Isaac.
When the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed will be same as Isaac's.
Since, both Feng and Isaac are riding on the same marry-go-round in a circular motion. The speed of the marry-go-round is constant.
Angular speed depends on the frequency of the rotation and frequency of the rotation is irrespective of the position in a circle.
Therefore, when the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed will be same as Isaac's.
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How much charge will have accumulated on the plates of a charging capacitor after a length of time equal to one time constant?
Answer: 63% of the final charge
Explanation:
When a capacitor is connected to a battery/power supply, the capacitor charges following the exponential law:
[tex]Q(t)=Q_0 (1-e^{-\frac{t}{\tau}})[/tex]
where
[tex]Q_0[/tex] is the final charge of the capacitor, which is
[tex]Q_0 =CV_0[/tex]
where C is the capacitance and [tex]V_0[/tex] the potential difference of the battery
t is the time
[tex]\tau[/tex] is the time constant of the circuit
Re-writing the equation,
[tex]Q(t)=CV_0 (1-e^{-\frac{t}{\tau}})[/tex]
After a time equal to one time constant,
[tex]t=\tau[/tex]
Therefore the charge on the capacitor will be
[tex]Q(\tau)=CV_0 (1-e^{-\frac{\tau}{\tau}})=CV_0(1-e^{-1})=0.63CV_0[/tex]
Which means 63% of the final charge.
Final answer:
After one time constant, the charge on a capacitor in an RC circuit will be approximately 63% of its maximum value. The time constant τ is calculated as the product of resistance R and capacitance C. The formula for the charge at any time is Q(t) = Q_{max}(1 - e^{-t/RC}).
Explanation:
When a capacitor charges in an RC circuit, it goes through a process described by an exponential function. After a time period equal to one time constant, represented by the symbol τ and given by the formula τ = RC (where R is the resistance in ohms and C is the capacitance in farads), the charge on a capacitor will reach approximately 63% of its maximum charge. This characteristic time, the time constant, determines how quickly the capacitor charges up to its maximum value, which is limited by the supply voltage.
The charge Q on the capacitor can be determined by the equation Q(t) = Q_{max}(1 - e^{-t/RC}) where Q(t) is the charge at time t, Q_{max} is the maximum charge, R is the resistance, C is the capacitance, and e is the base of the natural logarithm. When t equals the time constant τ (t = RC), the charge on the capacitor is Q(τ) = Q_{max}(1 - e^{-1}) which is approximately 63% of Q_{max}.
An unknown weak acid with a concentration of 0.090 M has a pH of 1.80. What is the Ka of the weak acid?
Answer : The value of [tex]K_a[/tex] of the weak acid is, [tex]3.36\times 10^{-3}[/tex]
Explanation : Given,
Initial concentration = 0.090 M
pH = 1.80
First we have to calculate the hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]1.80=-\log [H^+][/tex]
[tex][H^+]=0.0158M[/tex]
Now we have to calculate the [tex]K_a[/tex] of the weak acid.
The dissociation reaction of weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Initial conc. 0.090 0 0
At eqm. (0.090-x) x x
x = 0.0158 M
The expression for dissociation constant is:
[tex]K_a=\frac{(x)\times (x)}{(0.090-x)}[/tex]
Now put all the given values in this expression, we get:
[tex]K_a=\frac{(0.0158)\times (0.0158)}{(0.090-0.0158)}[/tex]
[tex]K_a=3.36\times 10^{-3}[/tex]
Thus, the value of [tex]K_a[/tex] of the weak acid is, [tex]3.36\times 10^{-3}[/tex]
The value of Ka of the weak acid with a concentration of 0.090 M and has a pH of 1.80 is 3.36 × 10-³.
How to calculate Ka of an acid?To calculate the Ka of an acid, we have to calculate the hydrogen ion concentration of the acid using the following expression:
pH = -log {H+}
1.80 = -log {H+}
{H+} = 0.0158M
The dissociation equation is given as follows:
HA ⇌ H+ + A-
Ka = 0.0158²/(0.090 - 0.0158)
Ka = 2.49 × 10-⁴/7.42 × 10-²
Ka = 0.336 × 10-²
Ka = 3.36 × 10-³
Therefore, the value of Ka of the weak acid with a concentration of 0.090 M and has a pH of 1.80 is 3.36 × 10-³.
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A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at a speed of 6 m/s at the top. The force exerted by the track on the cart at the top of the loop is ________.
Final answer:
The force exerted by the track on the cart at the top of the loop is 220 N, calculated by the difference between the centripetal force required for circular motion and the gravitational force acting on the cart.
Explanation:
The student is asking for the force exerted by the track on a 100 kg cart at the top of a vertical loop, where the cart has a speed of 6 m/s and the loop's radius is 3 m. At the top of the loop, the cart's weight and the normal force from the track provide the centripetal force necessary to keep the cart moving in a circle. To calculate the total force exerted by the track on the cart, we need to account for both the cart's weight (gravitational force) and the centripetal force required to keep it in circular motion.
The gravitational force (Fg) can be calculated using the equation Fg = m × g, where m is the mass of the cart and g is the acceleration due to gravity. The centripetal force (Fc) required for circular motion at the top of the loop can be calculated with the formula Fc = m × v² / r, where v is the speed of the cart and r is the radius of the loop.
The force exerted by the track (Ft) is then the difference between the centripetal force and the gravitational force because, at the top of the loop, the track has to push down on the cart to provide the inward centripetal force while also supporting the cart's weight.
Let's calculate each force:
Gravitational Force (Fg): Fg = 100 kg × 9.8 m/s² = 980 N
Centripetal Force (Fc): Fc = (100 kg) × (6 m/s)² / (3 m) = 1200 N
Now, we calculate the total force exerted by the track at the top of the loop:
Total Force by Track (Ft):
Ft = Fc - Fg
Ft = 1200 N - 980 N = 220 N
Therefore, the force exerted by the track on the cart at the top of the loop is 220 N.
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
Several students in the group made comments to the Coach about the back pack. Which student has the correct analysis of the forces and motion of the back pack?
The student who has the correct analysis of the forces and motion of the backpack understands Newton's laws of motion and can apply them to the situation.
Explanation:The student who has the correct analysis of the forces and motion of the backpack would be the one who understands the principles of Newton's laws of motion and can apply them to the situation.
Newton's first law of motion states that an object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction, unless acted upon by an external force. This law applies to the backpack on the group's graph, as it initially moves forward and then stops.
Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The student who correctly recognizes the forces and acceleration involved in the motion of the backpack would have the correct analysis.
In which situations can you conclude that the object is undergoing a net interaction with one or more other objects? (Select all that apply.
1)A car travels at constant speed around a circular race track.
2)A book slides across the table and comes to a stop.
3)A hydrogen atom remains at rest in outer space.
4)A proton in a particle accelerator moves faster and faster.
5)A spacecraft travels at a constant speed toward a distant star.
Answer:
4, A proton in a particle accelerator moves faster and faster.
Explanation:
A particle accelerator according to Wikipedia.com is a machine that uses electromagnetic fields to propel charged particles to very high speeds and energies, and to contain them in well-defined beams. Large accelerators are used for basic research in particle physics.
As the proton moves around, it collides with other proton and the wall of the accelerator thereby undergoing net interaction.
Tech A says clutch slippage can best be diagnosed by loading the engine while the clutch is released with the transmission in a high gear, such as fourth. Tech B says a slipping clutch can cause friction surfaces warpage or hot spots. Who is correct?a. Technician Ab. Technician Bc. Both Technician A and Technician Bd. Neither Technician A nor Technician B
Answer: c. Both Technician A and Technician B.
Explanation:
In motor enginnering, in case of a clutch slippage. It can best be diagnosed by loading the engine while the clutch is released with the transmission in a high gear, such as fourth gear whilst releasing the clutch pedal slowly.
It's also true that a slipping clutch can cause friction surfaces warpage or hot spots.
This therefore makes both Technicians A and B right in their arguments.
Both Tech A, who suggests diagnosing clutch slippage by loading the engine in high gear, and Tech B, who states that a slipping clutch can lead to friction surface damage, are correct.
The subject of this question pertains to automotive technology, specifically the diagnosis of clutch issues in vehicles. When evaluating the statements given by the two technicians:
Tech A is correct in suggesting that clutch slippage can be diagnosed by loading the engine in a high gear while the clutch is released, as this can put enough load on the engine to reveal if the clutch is slipping.
Tech B is correct in saying that a slipping clutch can lead to warpage or the development of hot spots on the friction surfaces.
Therefore, the correct answer is c. Both Technician A and Technician B are correct in their assessments of clutch slippage diagnosis and potential consequences.
What is it's speed at this time? A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7s the rocket is at a height of 86m
Answer: 31.85 m/s
Explanation:
Since we are talking about constant acceleration, we can use the following equation to find the speed [tex]s[/tex] of the rocket:
[tex]s=\frac{d}{t}[/tex]
Where:
[tex]d=86 m[/tex] is the distance the rocket has traveled
[tex]t=2.7 s[/tex] is the time in which the rocket has traveled the distance [tex]d[/tex]
Solving the equation with the given data:
[tex]s=\frac{86 m}{2.7 s}[/tex]
[tex]s=31.85 m/s[/tex] This is the rocket's speed
A startled armadillo leaps upward, rising 0.587 m in the first 0.193 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.587 m? (c) How much higher does it go? Use g=9.81 m/s2.
Answer:
a) Initial speed as it leaves the ground is 3.99 m/sb) Speed at the height of 0.587 m is 2.10 m/sc) Height reached is 0.81 mExplanation:
a) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = ?
Acceleration, a = -9.81 m/s²
Time, t = 0.193 s
Displacement, s = 0.587 m
Substituting
s = ut + 0.5 at²
0.587 = u x 0.193 + 0.5 x -9.81x 0.193²
u = 3.99 m/s
Initial speed as it leaves the ground is 3.99 m/s
b) We have equation of motion v = u + at
Initial velocity, u = 3.99 m/s
Final velocity, v = ?
Time, t = 0.193 s
Acceleration, a = -9.81 m/s²
Substituting
v = u + at
v = 3.99 + -9.81 x 0.193
v = 2.10 m/s
Speed at the height of 0.587 m is 2.10 m/s
c) We have equation of motion v² = u² + 2as
Initial velocity, u = 3.99 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = 0 m/s
Substituting
v² = u² + 2as
0² = 3.99² + 2 x -9.81 x s
s = 0.81 m
Height reached is 0.81 m