Answer:
a) The angular speed of the wheel after 1.30 seconds is [tex]2.47\,\frac{rad}{s}[/tex], b) The tangential speed of the spot after 1.30 seconds is [tex]0.543\,\frac{m}{s}[/tex], c) The magnitude of the total acceleration of the spot after 1.30 seconds is [tex]1.406\,\frac{m}{s^{2}}[/tex], d) The angular position of the spot is 2.130 radians (122.011°).
Explanation:
a) Given that tire accelerates at constant rate, final angular speed can be predicted by using the following formula:
[tex]\omega = \omega_{o} + \alpha \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Final angular speed, measured in radians per second.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\Delta t[/tex] - Time, measured in seconds.
Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex] (starts at rest), [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex] and [tex]\Delta t = 1.30\,s[/tex], the final angular speed is:
[tex]\omega = 0\,\frac{rad}{s} + \left(1.90\,\frac{rad}{s^{2}} \right) \cdot (1.30\,s)[/tex]
[tex]\omega = 2.47\,\frac{rad}{s}[/tex]
The angular speed of the wheel after 1.30 seconds is [tex]2.47\,\frac{rad}{s}[/tex].
b) The tangential speed of the spot is the product of the distance between the center of the wheel and spot. That is:
[tex]v = r \cdot \omega[/tex]
Where r is the distance between the center of the wheel and spot. The tangential speed of the spot after 1.30 seconds is:
[tex]v = (0.220\,m)\cdot \left(2.47\,\frac{rad}{s} \right)[/tex]
[tex]v = 0.543\,\frac{m}{s}[/tex]
The tangential speed of the spot after 1.30 seconds is [tex]0.543\,\frac{m}{s}[/tex].
c) The magnitude of the total acceleration of the spot is the magnitude of the vectorial sum of radial and tangential accelerations (both components are perpendicular to each other), which is determined by the Pythagorean theorem, that is:
[tex]a = \sqrt{a_{r}^{2} + a_{t}^{2}}[/tex]
Where [tex]a_{r}[/tex] and [tex]a_{t}[/tex] are the radial and tangential accelerations.
[tex]a = r\cdot \sqrt{\omega^{4} + \alpha^{2}}[/tex]
If [tex]r = 0.220\,m[/tex], [tex]\omega = 2.47\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex], then, the resultant acceleration is:
[tex]a = (0.220\,m)\cdot \sqrt{\left(2.47\,\frac{rad}{s} \right)^{4}+\left(1.90\,\frac{rad}{s^{2}} \right)^{2}}[/tex]
[tex]a \approx 1.406\,\frac{m}{s^{2}}[/tex]
The magnitude of the total acceleration of the spot after 1.30 seconds is [tex]1.406\,\frac{m}{s^{2}}[/tex].
d) Let be 30° (0.524 radians) the initial angular position of the spot with respect to center. The final angular position is determined by the following equation of motion:
[tex]\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta - \theta_{o})[/tex]
Final angular position is therefore cleared:
[tex]\theta - \theta_{o} = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
[tex]\theta = \theta_{o} + \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex]
Given that [tex]\theta_{o} = 0.524\,rad[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 2.47\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.90\,\frac{rad}{s^{2}}[/tex], the angular position of the spot after 1.30 seconds is:
[tex]\theta = 0.524\,rad +\frac{\left(2.47\,\frac{rad}{s} \right)^{2} - \left(0\,\frac{rad}{s}\right)^{2}}{2\cdot \left(1.90\,\frac{rad}{s^{2}} \right)}[/tex]
[tex]\theta = 2.130\,rad[/tex]
[tex]\theta = 122.011^{\circ}[/tex]
The angular position of the spot is 2.130 radians (122.011°).
In the rotational motion of an object, the angular acceleration is always towards the center, and the further discussion is as follows:
Rotational motion:The tangential acceleration of the object keeps changing its direction as the object rotates, always directed toward the tangent of the circle passing through the position of the object.
Radius of the spot, r = 0.220
minitial angle from the horizontal, θ = 30°
angular acceleration, α = 1.9 rad/s²
(a) from the first equation of motion we get:
ω = ω₀ + αt where
ω is the final angular speed
ω₀ is the initial angular speedand
t is the time = 1.3sω = 1.9×1.3 rad/sω = 2.47 rad/s
(b) tangential speed (v) is given by:
v = r×ωv = 0.220×2.47 m/sv = 0.5434 m/s
(c) The instantaneous tangential acceleration is given by:
[tex]a_t[/tex] = rω²so the resultant acceleration will be:
[tex]a=\sqrt{a_t^2+\alpha^2}\\\\a =\sqrt{r^2\omega^4+\alpha^2}\\\\a= \sqrt{(0.220)^2(2.47)^4+(1.9)^2}\\\\a = 1.4 \ \frac{m}{s^2}[/tex]
(d)
The angular displacement is given by:
θ = θ₀t + ¹/₂αt²θ₀ = 30° = 0.524
rad θ = 0.524×1.3 + ¹/₂×1.9×1.3²θ = 2.286 radθ = 131°
Following are the solution for points:
For a)
The angular speed is 2.47 rad/s
For b)
The tangential speed is 0.5434 m/s
For c)
Total acceleration is 1.4 m/s²
For d)
The final angular position is 131°
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A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,010 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 166 years
Explanation:
In order to calculate the amount of years that electrons take to cross the complete transmission line. You first calculate the drift speed of the electrons by using the following formula:
[tex]v_d=\frac{I}{nqA}[/tex] (1)
I: current on the wire = 1,010A
n: free charge density = 8.50*10^28 electrons/m^3
A: cross-sectional area of the transmission line = π*r^2
r: radius of the cross-sectional area = 2.00cm = 0.02m
You replace the values of the parameters in the equation (1):
[tex]v_d=\frac{1,010A}{(8.50*10^{28}electron/m^3)(1.6*10^{-19}C)(\pi (0.02m)^2)}\\\\v_d=5.9*10^{-5}\frac{m}{s}[/tex]
Next, you use the following formula:
[tex]t=\frac{x}{v_d}[/tex] (2)
x: length of the line transmission = 310km = 310,000m
You replace the values of vd and x in the equation (2):
[tex]t=\frac{310,000m}{5.9*10^{-5}m/s}=5.24*10^9s[/tex]
Finally, you convert the obtained t to seconds
[tex]t=5.24*10^9s*\frac{1\ year}{3.156*10^7s}=166.03\ years[/tex]
The electrons take approximately 166 years to travel trough the complete transmission line
An ideal transformer has LaTeX: N_1 = 1000 N 1 = 1000 (number of windings on the primary side), and LaTeX: N_2 = 8000 N 2 = 8000 (number of windings on the secondary side). If the rms voltage on the primary side is LaTeX: V_{rms}=100V V r m s = 100 V , what is the rms voltage on the secondary side? Give your answer in terms of Volts (rms), without entering the units.
Answer:
V₂ = 800 Volts (rms)
Explanation:
The turns ratio of an ideal transformer is given by the following formula:
N₁/N₂ = V₁/V₂
where,
N₁ = No. of turns in primary coil of the transformer = 1000 N
N₂ = No. of turns in secondary coil of the transformer = 8000 N
V₁ = rms Voltage on primary side of the transformer = 100 V
V₂ = rms Voltage on secondary side of the transformer = ?
Therefore, using these values in equation, we get:
1000 N/8000 N = 100 V/V₂
V₂ = (100)(8) Volts
V₂ = 800 Volts (rms)
Calculate the electric potential due to a dipole whose dipole moment is 5.2×10−30 C⋅m at a point r = 2.8×10−9 m away. Suppose that r≫ℓ, where ℓ is the distance between the charges in the dipole.
Answer:
V = 8.01*10^-12 V
Explanation:
In order to calculate the electric potential produced by the dipole you use the following formula:
[tex]V=k\frac{p}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
p: dipole moment = 5.2×10−30 C⋅m
r: distance to the dipole = 2.8*10^-9m
You replace the values of the parameters:
[tex]V=(8.98*10^9Nm^2/C^2)\frac{5.2*10^{-30}Cm}{(2.8*10^{-9}m)}\\\\V=8.01*10^{-12}V[/tex]
The electric potential of the dipole is 8.01*10^-12V
Rock can melt at a depth of about
below Earth's surface.
O A. 50 m
B. 500 km
C. 100 km
O D. 2000 km
Rock can melt at a depth of about below Earth's surface 100 km
What role does surface physics play in the world?On the other hand, interface physics offers a wide variety of spectroscopic & microscopic techniques to characterize that deposition & structure creation process upon that sub-nanometer size and, therefore, to pave the way for effective manufacturing techniques.
How does Surface Chemistry work?It is the study of both the chemical processes that take place at the meeting point of two surfaces, such as solid-liquid, solid-gas, sturdy, liquid-gas, etc. Surface engineering refers to a few surface chemistry applications.
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A boxcar at a rail yard is set into motion at the top of a hump. The car rolls down quietly and without friction onto a straight, level track where it couples with a flatcar of smaller mass, originally at rest, so that the two cars then roll together without friction. Consider the two cars as a system from the moment of release of the boxcar until both are rolling together.
(a) is the mechanical energy of the system conserved?
(b) is the momentum of this system conserved?
how much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate
Answer:
2479 NewtonSolution,
Mass=100 kg
Acceleration due to gravity(g)=24.79 m/s^2
Now,.
[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]
hope this helps ..
Good luck on your assignment..
g A The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.) Express the frictional force in terms of m, v0, and x1. View Available Hint(s) Ff = nothing Part B After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1.
Answer:
fr = m v₀² / 2 (x₁-x₀)
Explanation:
a) For this exercise we use Newton's second law
X axis
- fr = ma
Y Axis
N-W = 0
N=W
let's look for acceleration with expressions of kinematics
v² = v₀² - 2 a Δx
at the point where stop v = 0
a = v₀² / 2 Δx
let's replace
-fr = m (- v₀² / 2 (x₁-x₀))
fr = m v₀² / 2 (x₁-x₀)
b)they ask for the same
in this case part of rest
v₁² = 0 + 2 a Δx
a = v₁² / 2ΔX
we write Newton's second law
F - fr = m a
fr = F - ma
fr = F - m v₁² / 2Δx
which of the following terms refers to the amount of thermal energy need to change 1 kg of a substance from a liquid to a gas at its boiling point
Answer:
See the answer below.
Explanation:
"Latent Heat", also called the "Heat of Vaporization", is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure.
Best Regards!
Immediately outside a conducting sphere(i.e. on the surface) of unknown charge Q and radius R the electric potential is 190 V, and 10.0 cm further from the sphere, the potential is 140 V. What is the magnitude of the charge Q on the sphere
Answer:
Q = 5.9 nC (Approx)
Explanation:
Given:
Further distance = 10 cm
Electric potential(V) = 190 v
Potential difference(V1) = 140 v
Find:
Magnitude of the charge Q
Computation:
V = KQ / r
190 = KQ / r.............Eq1
V1 = KQ / (r+10)
140 = KQ / (r+10) ............Eq2
From Eq2 and Eq1
r = 28 cm = 0.28 m
So,
190 = KQ / r
190 = (9×10⁹)(Q) / 0.28
53.2 = (9×10⁹)(Q)
5.9111 = (10⁹)(Q)
Q = 5.9 nC (Approx)
The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that the acceleration an object experiences is
Answer:
According to Newtons 2nd law of motion ;
The acceleration an object experiences is as a result of the net force which is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Explanation:
This law is simply saying ;
Force = Mass ×Acceleration
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A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Answer:
a
[tex]KE = 7.17 *10^{7} \ J[/tex]
b
[tex]t = 6411.09 \ s[/tex]
Explanation:
From the question we are told that
The radius of the flywheel is [tex]r = 1.50 \ m[/tex]
The mass of the flywheel is [tex]m = 430 \ kg[/tex]
The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]
The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]
Generally the moment of inertia of the flywheel is mathematically represented as
[tex]I = \frac{1}{2} mr^2[/tex]
substituting values
[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]
[tex]I = 483.75 \ kgm^2[/tex]
The kinetic energy that is been stored is
[tex]KE = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]
[tex]KE = 7.17 *10^{7} \ J[/tex]
Generally power is mathematically represented as
[tex]P = \frac{KE}{t}[/tex]
=> [tex]t = \frac{KE}{P}[/tex]
substituting the value
[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]
[tex]t = 6411.09 \ s[/tex]
The water level in identical bowls, A and B, is exactly the same. A contains only water; B contains floating ice as well as water. When we weigh the bowls, we find that Group of answer choices
Answer:
We know that the density of the ice is smaller than the density of the water (and this is why the ice floats in water).
Dw > Di
Da is the density of the water and Di is the density of the ice
Since in Bowl A we have a volume V, only of water, then the mass of the bowl A is:
Dw*V.
Now, in the bowl B we have a combination of water and ice, suppose that Vw is the volume of water and Vi is the volume of ice, and we know that:
Vw + Vi = V.
Then the mass in this second bowl is:
Dw*Vw + Di*Vi = Dw*(V - Vi) + Di*Vi = Dw*V + (Di - Dw)*Vi
and we know that Dw > Di, then the left term is a negative term, then the mass of bowl B is smaller than the mass of bowl A.
The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at an acceleration of 2.5 × 104 m. s2 If the acceleration is constant, what impulse is delivered to a pollen grain with a mass of 1.0 × 10−7g?
Answer:
I = 7.5*10^-10 kg m/s
Explanation:
In order to calculate the impulse you first take into account the following formula:
[tex]I=m\Delta v=m(v-v_o)[/tex] (1)
m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg
v: final speed of the pollen grain = ?
vo: initial speed of the pollen grain = 0 m/s
Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:
[tex]v=v_o+a t[/tex] (2)
a: acceleration = 2.5*10^4 m/s^2
t: time = 0.30ms = 0.30*10^-3 s
[tex]v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}[/tex]
Next, you replcae this value of v in the equation (1) and calculate the impulse:
[tex]I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}[/tex]
The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s
A surface is bombarded by particles, each of mass small 'm', which have velocity
normal to the surface. On average, n particles strike unit area of the surface each second
and rebound elastically. What is the pressure on the surface?
A. nmv
B. 2nmv
C. nmv²
D. 1/2nmv²
Answer:
B. 2nmv
Explanation:
Pressure is force over area.
P = F / A
Force is mass times acceleration.
F = ma
Acceleration is change in velocity over change in time.
a = Δv / Δt
Therefore:
F = m Δv / Δt
P = m Δv / (A Δt)
The total mass is nm.
The change in velocity is Δv = v − (-v) = 2v.
A = 1 and Δt = 1.
Plugging in:
P = (nm) (2v) / (1 × 1)
P = 2nmv
WHO WANTS BRAINLIEST THEN ANSWER THIS QUESTION
look at my previous last question they relate
so
the car slows down to 50 mph
stae the new speed of the car relative to the lorry
if the velocity of the car reduces from 70km/h to 50km/h then the speed of the car will be equal to the speed of the lorry...
thus the relative velocity will be 0
The figure shows an arrangement of four charged particles, with θ = 20.0° and d1 = 3.00 cm, which is the distance from the origin to a charge q1. Charge q1 is unknown, but q2= +7.00×10‒19 C and q3 = q4 = ‒2.00×10‒19 C. If there is no nett electrostatic force on q1 due to the other charges (the nett electrostatic force on q1 is zero), calculate the distance from the origin to q2, given by d2, in cm. Assume that all forces apart from the electrostatic forces in the system are negligible
Answer:
[tex]d_2=3.16cm[/tex]
Explanation:
So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.
In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:
[tex]\sum{F_{x}}=0[/tex]
so:
[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]
Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:
[tex]-F_{12}+2F_{13x}=0[/tex]
we can now solve this for [tex]F_{12}[/tex] so we get:
[tex]F_{12}=2F_{13x}[/tex]
Now we can substitute with the electrostatic force formula, so we get:
[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]
We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]
so the simplified equation is:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]
From the given diagram we know that:
[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]
so when solving for [tex]r_{13}[/tex] we get:
[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]
and if we square both sides of the equation, we get:
[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]
and we can substitute this into our equation:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]
so we can now solve this for [tex]r_{12}[/tex] so we get:
[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
which can be rewritten as:
[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
and now we can substitute values.
[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]
which solves to:
[tex]r_{12}=6.16cm[/tex]
now, we must find [tex]d_{2}[/tex] by using the following equation:
[tex]r_{12}=d_{1}+d_{2}[/tex]
when solving for [tex]d_{2}[/tex] we get:
[tex]d_{2}=r_{12}-d_{1}[/tex]
when substituting we get:
[tex]d_{2}=6.16cm-3cm[/tex]
so:
[tex]d_{2}=3.16cm[/tex]
A 0.40-kg particle moves under the influence of a single conservative force. At point A, where the particle has a speed of 10 m/s, the potential energy associated with the conservative force is 40 J. As the particle moves from A to B, the force does 25 J of work on the particle. What is the value of the potential energy at point B
Answer:
The value of the potential energy of the particle at point B is 85 joules.
Explanation:
According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:
[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]
Where:
[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.
[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.
[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.
[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.
The initial kinetic energy of the particle is:
[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity, measured in meters per second.
If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:
[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]
[tex]K_{A} = 20\,J[/tex]
Finally, the value of the potential energy at point B is:
[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]
[tex]U_{B} = 85\,J[/tex]
The value of the potential energy of the particle at point B is 85 joules.
The potential energy of the particle at point B is 85 J.
Given to us:
Mass of the particle, [tex]m=0.40\ kg[/tex]
velocity of the particle, [tex]v= 10\ m/s[/tex]
potential energy of the particle, [tex]PE= 40\ J[/tex]
Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]
Calculating the kinetic energy of the particle,
[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]
According to the Principle of Energy Conservation,
The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,
Also,
Total Energy at point A ,
[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]
Total Energy at point B,
[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]
As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,
[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]
Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.
Hence, the potential energy of the particle at point B is 85 J.
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1. Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. Radius of tire is 50 cm. What angle did the tire move through in those 5 secs
Answer:
[tex]\theta=65.18rad[/tex]
Explanation:
The angle in rotational motion is given by:
[tex]\theta=\frac{w_o+w_f}{2}t[/tex]
Recall that the angular speed is larger than regular frequency (in rpm) by a factor of [tex]2\pi[/tex], so:
[tex]\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}[/tex]
The wheel spins from rest, that means that its initial angular speed is zero([tex]\omega_o[/tex]). Finally, we have to convert the given time to minutes and replace in the first equation:
[tex]t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad[/tex]
A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux through it is approximately Group of answer choices
Answer:
126 mWb
Explanation:
Given that:
length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.
We assume that the magnetic field in the solenoid is constant.
The magnetic flux is given as:
[tex]\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb[/tex]
A 200 turn coil is in a uniform magnetic field that is decreasing at the rate 0.20 T/s. The coil is perpendicular to the field and its dimensions are 0.20 m by 0.40 m. What is the magnitude of the induced emf in the coil
Answer:
emf = 3.2V
Explanation:
In order to calculate the magnitude of the induced emf in the coil you use the following formula:
[tex]emf=-N\frac{d\Phi_B}{dt}[/tex] (1)
N: turns of the coil = 200
ФB: magnetic flux = A*B
A: area of the coil = (0.20m)(0.40m) = 0.08m²
B: magnitude of the magnetic field
You take into account that the area of the coil is constant, while magnetic field changes on time. Then, the equation (1) becomes:
[tex]emf=-NA\frac{dB}{dt}[/tex] (2)
dB/dt = rate of change of the magnetic field = -0.20T/s (it is decreasing)
You replace the values of all parameters in the equation (2):
[tex]emf=-200(0.08m^2)(-0.20T/s)=3.2V[/tex]
The induced emf in the coil is 3.2V
Jack and Jill went up the hill to fetch a pail of water. Jack, who’s mass is 75 kg, 1.5 times heavier than Jill’s mass, fell down and broke his crown after climbing a 15 m high hill. Jillcame tumbling after covering the same distance as Jack in 1/3rd of the time.Required:a. Who did the most work climbing up the hill? b. Who applied the most power?
Answer:
a) Jack does more work uphill
b) Numerically, we can see that Jill applied the most power downhill
Explanation:
Jack's mass = 75 kg
Jill's mass = [tex]1.5x = 75[/tex]
Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg
distance up hill = 15 m
a) work done by Jack uphill = mgh
where g = acceleration due to gravity= 9.81 m/s^2
work = 75 x 9.81 x 15 = 11036.25 J
similarly,
Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J
this shows that Jack does more work climbing up the hill
b) assuming Jack's time downhill to be t,
then Jill's time = [tex]\frac{t}{3}[/tex]
we recall that power is the rate in which work id done, i.e
P = [tex]\frac{work}{time}[/tex]
For Jack, power = [tex]\frac{11036.25}{t}[/tex]
For Jill, power = [tex]\frac{3*7357.5}{t}[/tex] = [tex]\frac{22072.5}{t}[/tex]
Numerically, we can see that Jill applied the most power downhill
The velocity of an object is given by the expression v(t) = 3.00 m/s + ( 4.00 m/s^3)t^2, where t is in seconds. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.000 s
Answer:
Position of object is;
s(t) = 4t³/3 + 3t + 1
Explanation:
We are told that the velocity has an expression;
v(t) = 3.00 m/s + ( 4.00 m/s³)t²
Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;
s(t) = ∫3 + 4t²
s(t) = 3t + 4t³/3 + c
Now, we were told that at x = 1.00 m, time t = 0.000 s
Thus, plugging the values in;
1 = 3(0) + 4(0³/3) + c
c = 1
Thus,the expression for the position of the object is;
s(t) = 4t³/3 + 3t + 1
A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 29.4 s. The distance between two successive crests is 31.4 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Answer:
(a) 5.88 s
(b) 0.17 Hz
(c) 31.4 m
(d) 5.338 m/s
Explanation:
From the question,
(a) Period = time between successive crest = t/n............ Equation 1
where t = time, n = number of wave crest
Given: t = 29.4, n = 5
therefore,
Period (T) = 29.4/5 = 5.88 s.
(b) Frequency = 1/period
Frequency = 1/5.88
Frequency = 0.17 Hz.
(c) wavelength = distance between two successive crest = 31.4 m
(d) using,
v = λf............... Equation 2
Where v = speed, f = frequency, λ = wavelength
given: f = 0.17 Hz, λ = 31.4 m
Substitute into equation 2
v = 0.17(31.4)
v = 5.338 m/s
g A small car travels up the hill with a speed of v = 0.2 s (m/s) where s is the distance measured from point A in meters. Determine the magnitude of the car’s acceleration when it reaches s = 50 (m) where the road’s radius of curvature is r
Answer:
The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is [tex]2\,\frac{m}{s^{2}}[/tex].
Explanation:
The acceleration can be obtained by using the following differential equation:
[tex]a = v \cdot \frac{dv}{ds}[/tex]
Where [tex]a[/tex], [tex]v[/tex] and [tex]s[/tex] are the acceleration, speed and distance masured in meters per square second, meters per second and meters, respectively.
Given that [tex]v = 0.2\cdot s[/tex], its first derivative is:
[tex]\frac{dv}{ds} = 0.2[/tex]
The following expression is obtained by replacing each term:
[tex]a = 0.2\cdot 0.2\cdot s[/tex]
[tex]a = 0.04\cdot s[/tex]
The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is:
[tex]a = 0.04\cdot (50\,m)[/tex]
[tex]a = 2\,\frac{m}{s^{2}}[/tex]
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Answer:
The answer is 3.48 seconds
Explanation:
The kinematic equation
y= y0+V0*t+1/2*a*(t*t)
-50=0+(0)t+1/2(-9.8)*(t*t)
t=3.194 seconds
During ribbons ball,
x=x0+ Vt+1/2*a*(t*t)
x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)
x= 47.9157m
So, distance (D) = 100-47.9157= 52.084m
52.084m=0+15(t)+1/2*(0)(t*t)
t=52.084/15=3.472286= 3.48seconds
Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of oneâs birth. The only known force that a planet could exerts on us is gravitational, so if there is anything to astrology we should expect this force to be significant.
Required:
a. Calculate the gravitational force, in Newtons, exerted on a 4.1 kg baby by a 120 kg father who is a distance of 0.18 m away at the time of its birth.
b. Calculate the force on the baby, in Newtons, due to Jupiter (the largest planet, which has a mass of 1.90Ã10^27 kg if it is at its closest distance to Earth, 6.29Ã10^11 m away.
c. What is the ratio of the force of the father on the baby to the force of Jupiter on the baby?
Answer:
Explanation:
Gravitational force between two objects having mass m₁ and m₂ at a distance R
F = G m₁ m₂ / R²
Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²
= 1.01 x 10⁻⁶ N
b )
Force between baby and Jupiter
F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²
= 1.31 x 10⁻⁶ N
c )
Ratio = 1.01 / 1.31
= .77
A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with length 2R that has a kinetic friction coefficient of 0.5. From what height h must the mass be released to stay on the track
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
[tex]Em_{f}[/tex] = U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
[tex]Em_{f}[/tex] = K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)
While on Mars, two astronauts repeat the pendulum experiment you conducted earlier this term in your physics lab. They go off script and plot the pendulum length vs. the square of the period. Their best fit is a line with y = 0.091 x and R^2=1.
Recall that the theoretical period of a pendulum is T = 2pi(L/g)^1/2. Apply your understanding of mathematical modeling to determine what the constant 0.091 in the astronauts' equation of the best fit line is equal to in this case and use that to find their experimental value of g on Mars in m/s^2.
Answer:
The constant 0.091 in the astronauts' equation of the best fit line is equal to [tex]\frac{L}{T^2}[/tex]
The value of g on Mars is [tex]g = 3.593 \ m/s^2[/tex]
Explanation:
From the question we are told that
The line of best fit is defined by the equation [tex]y = 0.091 x \ and \ R^2 = 1[/tex]
Now the equation of a straight line is defined as
[tex]y = mx + c[/tex]
Now comparing the given equation to this we have that
[tex]m = slope = 0.091[/tex]
Now from the graph the formula for the slope is
[tex]m = \frac{L}{T^2}[/tex]
=> [tex]0.091 = \frac{L}{T^2}[/tex]
Now from the question we are told that
[tex]T = 2 \pi \sqrt{\frac{L}{g} }[/tex]
=> [tex]\frac{g}{4\pi r^2} = \frac{L}{T^2} = 0.091[/tex]
=> [tex]g = 4\pi^2 * 0.091[/tex]
=> [tex]g = 3.593 \ m/s^2[/tex]
A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upon impact?
Answer: 56.72 ft/s
Explanation:
Ok, initially we only have potential energy, that is equal to:
U =m*g*h
where g is the gravitational acceleration, m the mass and h the height.
h = 50ft and g = 32.17 ft/s^2
when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Then we have:
K = U
m*g*h = (m/2)*v^2
we solve it for v.
v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s
1. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg a. Calculate the speed of the ball right before the bounce. b. How much energy was converted into heat after the ball bounced off the ground
Answer:
(a) 7.67 m/s.
(b) 4.9 J
Explanation:
(a) From the law of conservation of energy,
P.E = K.E
mgh = 1/2(mv²)
therefore,
v = √(2gh)....................... Equation 1
Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.
Given: h = 3 m, g = 9.8 m/s²
Substitute into equation 1
v = √(2×9.8×3)
v = √(58.8)
v = 7.67 m/s.
(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J
Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J
Amount of energy converted to heat = 14.7-9.8 = 4.9 J