Answer:
[tex]1.9\times 10^5\ J[/tex]
[tex]1.8\times 10^{2}\ kcal[/tex]
[tex]1.3\times 10^{2}\ kcal[/tex]
[tex]1.2\times 10^{3}\ W[/tex]
Explanation:
m = Mass of person = 60 kg
g = Acceleration due to gravity = 9.81 m/s²
t = Time taken = 10 min and 49 s
Total height
[tex]h=3.7\times 86\\\Rightarrow h=318.2\ m[/tex]
Potential energy is given by
[tex]U=mgh\\\Rightarrow U=60\times 9.81\times 318.2\\\Rightarrow U=187292.52\ J[/tex]
The gravitational potential energy is [tex]1.9\times 10^5\ J[/tex]
The energy in the climb
[tex]\dfrac{187292.52}{0.25}=749170.08\ J[/tex]
Converting to kcal or Cal
[tex]\dfrac{749170.08}{4184}=179.05594\ kcal[/tex]
The amount of energy used to climb [tex]1.8\times 10^{2}\ kcal[/tex]
Amount gone to heat
[tex]179.05594\times 0.75=134.291955\ kcal[/tex]
The amount burned [tex]1.3\times 10^{2}\ kcal[/tex]
Power is given by
[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{749170.08}{10\times 60+49}\\\Rightarrow P=1154.34526\ W[/tex]
The power is [tex]1.2\times 10^{3}\ W[/tex]
The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? a. During each of the three different stages of its motion, the car is moving with constant acceleration. b. During each of the three different stages of its motion, the car is moving with constant velocity. c.The highway is straight (i.e., there are no curves). d. The highway is level (i.e., there are no hills or valleys).Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD.
The assumptions we can reasonably make in this scenario include the highway being straight and level, corresponding to options C and D. Assuming constant acceleration or velocity for each stage of the car's motion, options A and B, is not necessarily accurate.
Explanation:In analyzing the motion of a car in different stages, we can make reasonable assumptions to simplify the problem. For each stage, assuming the car is moving with constant acceleration (option a) or velocity (option b) is not necessarily accurate because acceleration and velocity may change due to various factors like interaction with driver, road conditions, or appearance of a police car. The assumptions more likely to hold are that the highway is straight (option c), meaning there are no curves, and the highway is level (option d), indicating no hills or valleys. Thus, the correct answers would be C and D.
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The position of a particle is given by the function x=(4t3−6t2+12)m, where t is in s.
A.) at what time does the particle reach its minimum velocity
B.) what is (vx)min
C.) at what time is the acceleration zero
Answer
given,
x = 4 t³ - 6 t² + 12
velocity, [tex]v = \dfrac{dx}{dt}[/tex]
[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(4t^3-6t^2+12)[/tex]
[tex]v =12t^2-12t[/tex]
For minimum velocity calculation we have differentiate it and put it equal to zero.
[tex]\dfrac{dv}{dt} =\dfrac{d}{dt}12t^2-12t[/tex]
[tex]\dfrac{dv}{dt} =24t-12[/tex].........(1)
putting it equal to zero
24 t - 12 =0
t = 0.5 s
At t = 0.5 s velocity will be minimum.
b) minimum velocity
v = 12t² -12 t
v = 12 x 0.5² -12 x 0.5
v = -3 m/s
c) derivative of velocity w.r.t. time is acceleration
from equation 1
a = 24 t - 12
time at which acceleration will be zero
0 = 24 t - 12
t = 0.5 s
At t = 0.5 s acceleration will be zero.
Part A. The particle reaches its minimum velocity at 0.5 seconds.
Part B. The minimum velocity of the particle is -3 m/s.
Part C. The acceleration of the particle will be zero at the time t = 0.5 seconds.
How do you calculate the minimum velocity and acceleration?Given that the position of a particle is given by the function x.
[tex]f(x) = 4t^2 -6t^2 +12[/tex]
The function of the velocity of a particle can be obtained by the time function.
[tex]v= \dfrac {dx}{dt}[/tex]
[tex]v = \dfrac {d}{dt} ( 4t^3-6t^2 +12)[/tex]
[tex]v = 12t^2 -12 t[/tex]
The velocity function of the particle is [tex]v = 12t^2 - 12t[/tex].
Part AThe minimum velocity of the particle is obtained by the differentiation of velocity function with respect to the time and put it equal to zero.
[tex]\dfrac {dv}{dt} = \dfrac {d}{dt} (12t^2 - 12t) = 0[/tex]
[tex]\dfrac {dv}{dt} = 24 t-12 = 0[/tex]
[tex]t = 0.5\;\rm s[/tex]
Hence we can conclude that the particle reaches its minimum velocity at 0.5 seconds.
Part BThe velocity function is [tex]v = 12t^2 - 12t[/tex]. Substituting the value of t = 0.5 s to calculate the minimum velocity.
[tex]v = 12(0.5)^2 - 12(0.5)[/tex]
[tex]v = 3 - 6[/tex]
[tex]v = -3 \;\rm m/s[/tex]
The minimum velocity of the particle is -3 m/s.
Part CThe acceleration is defined as the change in the velocity with respect to time. Hence,
[tex]a = \dfrac {dv}{dt}[/tex]
[tex]a = 24 t-12[/tex]
Substituting the value of a = 0, we get the time.
[tex]0 = 24t - 12[/tex]
[tex]t = 0.5 \;\rm s[/tex]
The acceleration of the particle will be zero at the time t = 0.5 seconds.
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During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass. a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?
Complete Question:
During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass, which is 62 kg.
a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?
Answer:
a) t= 46.5 msec. b) 0.35 m
Explanation:
Applying Newton´s 2nd Law to the mass supported by the knee (20% of the total mass), we can get the maximum acceleration allowable in order to avoid an injury, as follows:
a = F/m = 4000 N / 0.2*62 kg = 322.6 m/s²
Applying the definition of acceleration, and taking into account that the knee finally come to an stop, we have:
a = vf - v₀ / Δt = -15 m/s / Δt
Solving for Δt :
Δt = -15 m/s / -322.6 m/s² = 0.0465 sec = 46.5 msec.
b) Assuming the acceleration remains constant during this time interval, we can find the distance needed to come to an ⇒stop, applying any of the kinematic equations, as this one:
vf² - v₀² = 2*a*Δx
⇒Δx = (vf²-v₀²) / 2*a
⇒Δx = -(15 m/s)² / 2*(-322.6 m/s²) = 0.35 m
This question involves the concepts of the equations of motion and Newton's Second Law of Motion.
a) The minimum stopping time interval needed to avoid knee injury is "0.05 s".
b) The minimum stopping distance is "0.37 m".
a)
First, we will use Newton's Second Law of Motion to find out the acceleration:
[tex]F=ma\\\\a=\frac{F}{m}[/tex]
where,
a = acceleration = ?
m = mass supported by knee = 20% of total mass = (0.2)(65 kg) = 13 kg
Assuming the average mass of a person to be 65 kg.
Force = - 4000 N (reaction force)
Therefore,
[tex]a=\frac{-4000\ N}{13\ kg}\\\\a=-307.7\ m/s^2[/tex] (negative sign shows decelration)
Now, for the minimum time interval, we will use the first equation of motion:
[tex]v_f=v_i+at\\\\t=\frac{v_f-v_i}{a}[/tex]
where,
t = time interval = ?
vf = final speed = 0 m/s
vi = initial speed = 15 m/s
Therefore,
[tex]t=\frac{0\ m/s-15\ m/s}{-307.7\ m/s^2}[/tex]
t = 0.05 s = 50 ms
b)
Now, we will use the second equation of motion to find out the stopping distance:
[tex]s=v_it+\frac{1}{2}at^2\\s=(15\ m/s)(0.05\ s)+\frac{1}{2}(-307.7\ m/s^2)(0.05\ s)^2\\[/tex]
s = 0.75 m - 0.38 m
s = 0.37 m
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The attached picture shows the equations of motion.
How much work must I do to assemble a charge distribution consisting of three point charges of -1.00 nC, 2.00nC, and 3.00 nC, located at the vertices of an equilateral triangle? The edges of the triangle are each 20.0 cm in length. Initially the three point charges are infinitely far apart.
Answer:
[tex]W_{total}=4.5*10^{-8}J[/tex]
Explanation:
Remember that electric potential can be written as:
[tex]V=\frac{kQ}{r}[/tex],
Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:
[tex]V=\frac{W}{q}[/tex],
where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.
From all this we can get an expression for the work:
[tex]W=\frac{kQq}{r}[/tex]
We are going to let
[tex]q_{1}=-1.00nC\\q_{2}=2.00nC\\q_{3}=3.00nC[/tex]
To take the first charge [tex]q_{1}[/tex] from infinitely far apart to one of the vertices of the triangle, since there is no electric field and charges, we make no work.
Next, we will move [tex]q_{2}[/tex] . Now, [tex]q_{1}[/tex] is a vertice of the triangle, and we want [tex]q_{2}[/tex] to be 20.cm apart from [tex]q_{1 }[/tex] so the work we need to do is
[tex]W_{12}=\frac{kq_{1}q_{2}}{(0.20)}\\\\W_{12}=\frac{(9*10^{9})(-1*10^{-9})(2*10^{-9})}{0.20}\\\\W_{12}=-9*10^{-8}J[/tex]
Now, we move the last point charge. Here, we need to take in account the electric potential due to [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. So
[tex]W=W_{13}+W_{23}\\\\\\W=\frac{kq_{1}q_{3}}{0.20}+\frac{kq_{2}q_{3}}{0.20}\\\\W=q_{3}k(\frac{q_{1}}{0.20}+\frac{q_{2}}{0.20})\\\\W=(3*10^{-9})(9*10^{9})(\frac{-1*10^{-9}}{0.20}+\frac{2*10^{-9}}{0.20})\\\\[/tex]
[tex]W=1.35*10^{-7}J[/tex]
Now, the only thing left to do is to find the total work, this can be easily done by adding [tex]W_{12}[/tex] and [tex]W[/tex]:
[tex]W_{total}=W_{12}+W\\W_{total}=1.35*10^{-7}-9*10^{-8}\\W_{total}=4.5*10^{-8}J[/tex]
A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?
Answer:
π*R²*E
Explanation:
According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.
As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.
The flux across the open surface can be expressed as follows:
[tex]\int\ {E} \, dA = E*A = E*\pi *R^{2}[/tex]
As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².
⇒Flux = E*π*R²
Final answer:
The electric flux through a hemisphere surface is calculated using Gauss's Law, where the electric field lines are parallel to the area element vector on the surface. The flux through a closed spherical surface is independent of its radius. Choosing a Gaussian surface with appropriate symmetry aids in evaluating the flux integral.
Explanation:
The magnitude of the electric flux through the hemisphere surface can be calculated using Gauss's Law. For a hemispherical surface of radius R in a uniform electric field,
The electric field lines are parallel to the area element vector on the surface. Electric flux is the product of the electric field magnitude and the area.The flux through a closed spherical surface is independent of the radius of the surface. Every field line piercing one radius also pierces another, resulting in constant flux.Choosing a suitable Gaussian surface simplifies the flux integral evaluation. A spherical surface is ideal due to uniform field angle and magnitude.(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface is 2.9 m/s2?
Answer:
1777.92 m/s
Explanation:
R = Radius of asteroid = 545 km
M = Mass of planet
g = Acceleration due to gravity = 2.9 m/s²
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
Acceleration due to gravity is given by
[tex]g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}[/tex]
The expression of escape velocity is given by
[tex]v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s[/tex]
The escape speed is 1777.92 m/s
Answer:
1.78 km/s
Explanation:
radius, R = 545 km = 545000 m
acceleration due to gravity, g = 2.9 m/s²
The formula for the escape velocity is given by
[tex]v=\sqrt{2gR}[/tex]
[tex]v=\sqrt{2\times 545000\times 2.9}[/tex]
v = 1777.92 m/s
v = 1.78 km/s
Thus, the escape velocity on the surface of asteroid is 1.78 km/s.
A simple pendulum in a science museum entry hall is 3.58 m long, has a 1.23 kg bob at its lower end, and swings with an amplitude of 10.9 ∘. How much time does the pendulum take to swing from its extreme right side to its extreme left side?
Answer:
1.897825 seconds
Explanation:
L = Length of pendulum = 3.58 m
g = Acceleration due to gravity = 9.81 m/s²
The amplitude of a pendulum is given by
[tex]T=2\pi \sqrt{\dfrac{L}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{3.58}{9.81}}\\\Rightarrow T=3.79565\ s[/tex]
It can be seen that the time period does not depend on amplitude
Time period includes the time to go from extreme left to right and back to extreme left.
So, time taken to go from extreme right to left is [tex]\dfrac{T}{2}=\dfrac{3.79565}{2}=1.897825\ s[/tex]
The period of a simple pendulum is determined by its length and the acceleration due to gravity. It is independent of other factors such as mass and amplitude.
Explanation:The period of a simple pendulum is determined by its length and the acceleration due to gravity. The period is independent of other factors such as mass and amplitude. For a small amplitude of less than 15 degrees, the period of a simple pendulum can be calculated using the formula:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
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Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far does it go in this time?
Answer:
(A) [tex]a=2.0.37m/sec^2[/tex]
(B) s = 146.664 m
Explanation:
We have given car starts from the rest so initial velocity u = 0 m /sec
Final velocity v = 88 km/hr
We know that 1 km = 1000 m
And 1 hour = 3600 sec
So [tex]88km/hr=88\times \frac{1000}{3600}=24.444m/sec[/tex]
Time is given t = 12 sec
(A) From first equation of motion v = u+at
So [tex]24.444=0+a\times 12[/tex]
[tex]a=2.0.37m/sec^2[/tex]
So acceleration of the car will be [tex]a=2.0.37m/sec^2[/tex]
(b) From third equation of motion [tex]v^2=u^2+2as[/tex]
So [tex]24.444^2=0^2+2\times 2.037\times s[/tex]
s = 146.664 m
Distance traveled by the car in this interval will be 146.664 m
The amplitude of an electric field is 0.4 v/m 23 km from a radio transmitter. What is the total power emitted by the transmitter, if one assumes the radiation is emitted isotropically. The impedence of free space is 377 ohms. Answer in units of W.
The total power emitted by a radio transmitter can be calculated by first determining the intensity from the electric field and then multiplying by the surface area of the sphere covered by the isotropic radiation.
Explanation:The total power emitted by a radio transmitter can be calculated by using the relation between the electric field and the intensity of the radiation. The intensity (I) can be obtained by squaring the electric field strength (E) and dividing it by two times the impedance of free space (Z0). The formula is I = E²/2Z0. After finding the intensity, we can multiply it by the surface area of the sphere which is 4πr² (as it's isotropically radiated, it forms a sphere). The formula becomes Power (P) = I * 4πr². Substituting the given values, we calculate the total power.
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In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.
A. QC > 0; QH > 0
B. QC > 0; QH < 0
C. QC = 0; QH > 0
D. QC < 0; QH > 0
E. QC < 0; QH < 0
Answer:
B. QC > 0; QH < 0
Explanation:
Given that there are two reservoir of energy.
Sign convention for heat and work :
1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
From hot reservoir heat is going out that is why it is taken as negative
[tex]Q_H<0[/tex]
From cold reservoir heat is coming inside the reservoir that is why it is taken as positive
[tex]Q_C>0[/tex]
That is why the answer will be
[tex]Q_H<0[/tex] ,[tex]Q_C>0[/tex]
In the context of energy transfers with hot and cold reservoirs, the sign convention is
Option A (QC > 0; QH > 0)Given that there are two reservoir of energy. Sign convention for heat and work :
1) If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2) If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
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An electron and a proton are each placed at rest in a uniform electric field of magnitude 560 N/C. Calculate the speed of each particle 46.0 ns after being released. electron 4.5e^-6 Incorrect.
Answer:
The speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s.
Explanation:
Given that,
Electric field, E = 560 N/C
To find,
The speed of each particle (electrons and proton) 46.0 ns after being released.
Solution,
For electron,
The electric force is given by :
[tex]F=qE[/tex]
[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]
Let v is the speed of electron. It can be calculated using first equation of motion as :
[tex]v=u+at[/tex]
u = 0 (at rest)
[tex]v=\dfrac{F}{m}t[/tex]
[tex]v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}[/tex]
[tex]v=4.52\times 10^6\ m/s[/tex]
For proton,
The electric force is given by :
[tex]F=qE[/tex]
[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]
Let v is the speed of electron. It can be calculated using first equation of motion as :
[tex]v=u+at[/tex]
u = 0 (at rest)
[tex]v=\dfrac{F}{m}t[/tex]
[tex]v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}[/tex]
[tex]v=2468.02\ m/s[/tex]
So, the speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.
A fullback with a mass of 100kg and a velocity of 3.5 m/s duewest collides head-on with a defensive back with a mass of 80 kgand a velocity of 6 m/s due east.
a.)What is the initial momentum of each player?b.)What is the total momentum of the system before thecollision?c.)If they stick together and external forces can be ignored,what direction will they be traveling immediately after theycollide?
Answer
given,
mass of full back, M = 100 Kg
velocity of full back in east = 3.5 m/s
mass of the defensive back,m = 80 Kg
velocity of defensive back, due east = 6 m/s
a) initial momentum of each player
for full back
P₁ = M v = 100 x 3.5 = 350 kg.m/s
for defensive back
P₂= m v = 80 x 6 = 480 kg.m/s
b) total momentum before collision
P = P₁ + P₂
taking west direction as positive
P = 350 + (- 480 )
P = -130 kg.m/s
c) speed of the them when they stick together.
using conservation of momentum
initial momentum = final momentum
-130 = (M + m ) V
180 V = -130
V = -0.722 m/s
velocity after the collision is equal to 0.722 m/s in direction of east.
a) P₁=350 kg.m/s and P₂= 480 kg.m/s
b) Total momentum before collision is -130 kg.m/s.
c) 0.722 m/s in direction of east
Given,
Mass of full back, M = 100 Kg
Velocity of full back in east = 3.5 m/s
Mass of the defensive back, m = 80 Kg
Velocity of defensive back, due east = 6 m/s
Momentum
It is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
To find:
a) initial momentum of each player
For full back
P₁ = M*v = 100 * 3.5 = 350 kg.m/s
For defensive back
P₂= m*v = 80*6 = 480 kg.m/s
b) total momentum before collision
P = P₁ + P₂
taking west direction as positive
P = 350 + (- 480 )
P = -130 kg.m/s
c) speed of the them when they stick together.
Using conservation of momentum
initial momentum = final momentum
-130 = (M + m ) V
180 V = -130
V = -0.722 m/s
Velocity after the collision is equal to 0.722 m/s in direction of east.
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12 of 15 Constants The predominant frequency of a certain fire truck's siren is 1500 Hz when at rest.
Answer:
The detect frequency is 1622.72 Hz.
Explanation:
Given that,
Frequency = 1500 Hz
Suppose you move with a speed of 27.0 m/s toward the fire engine. what frequency do you detect ?
We need to calculate the frequency
Using formula of frequency
[tex]f=(\dfrac{v+v_{0}}{v})f_{0}[/tex]
Where, v = speed of sound
v₀ = speed of source
f₀ = frequency of siren
Put the value into the formula
[tex]f=\dfrac{330+27.0}{330}\times1500[/tex]
[tex]f=1622.72\ Hz[/tex]
Hence, The detect frequency is 1622.72 Hz.
A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° with the vertical. In such a condition of static equilibrium, what is the net force on the new vine? ANS: 366 N
Answer:
[tex]T=366.23\ N[/tex]
Explanation:
Given:
mass of monkey, [tex]w=600\ N[/tex]angle of vine from the vertical, [tex]\theta=35^{\circ}[/tex]Now follow the schematic to understand the symmetry and solution via Lami's theorem.
The weight of the monkey will be balanced equally by the tension in both the vines:
Using Lami's Theorem:
[tex]\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}[/tex]
[tex]\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }[/tex]
[tex]T=366.23\ N[/tex]
In a hydrogen atom (i.e., one stationary proton and one orbiting electron), the angular velocity of the electron is 10 x 10^6 radians per second. What is the radius of the electron orbit? Assume uniform circular motion.
Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,
[tex]F_c = k \frac{e^2}{R^2}[/tex]
Here,
k = Coulomb's Constant
R = Distance
e = Electron charge
And by the centripetal force,
[tex]F_c = m_e R\omega^2[/tex]
[tex]m_e[/tex] = Mass of electron
R = Radius
[tex]\omega[/tex] = Angular velocity
Equation both expression,
[tex]k \frac{e^2}{R^2} = m_e R\omega^2[/tex]
Replacing,
[tex]R^3 = \frac{(9*10^9)(1.6*10^{-19})^2}{(9.11*10^{-31})(10^{6})^2}[/tex]
[tex]R^3 = 2.52908*10^{-10} m[/tex]
[tex]R = 6.32394*10^{-4}m[/tex]
Therefore the radius of the electron orbit is [tex]6.32394*10^{-4}m[/tex]
Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out over their facing surfaces. How many excess electrons are on the negative surface if the electric field between the plates has a magnitude of 18,000 N/C? (k = 1/4 pi epsilon_0 = 9.0 times 10^9 N m^2/C^2, e = 1.6 times 10^-19 C)
Answer:
The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]
Explanation:
Given that,
Distance =1.5 cm
Side = 22 cm
Electric field = 18000 N/C
We need to calculate the capacitance in the metal plates
Using formula of capacitance
[tex]C=\dfrac{\epsilon_{0}A}{d}[/tex]
Put the value into the formula
[tex]C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}[/tex]
[tex]C=0.285\times10^{-10}\ F[/tex]
We need to calculate the potential
Using formula of potential
[tex]V=Ed[/tex]
Put the value into the formula
[tex]V=18000\times1.5\times10^{-2}\ V[/tex]
[tex]V=270\ V[/tex]
We need to calculate the charge
Using formula of charge
[tex]Q=CV[/tex]
Put the value into the formula
[tex]Q=0.285\times10^{-10}\times270[/tex]
[tex]Q=76.95\times10^{-10}\ C[/tex]
Here, the charge on both the positive and negative plates
[tex]Q=+76.95\times10^{-10}\ C[/tex]
[tex]Q=-76.95\times10^{-10}\ C[/tex]
We need to calculate the number of excess electrons are on the negative surface
Using formula of number of electrons
[tex]n=\dfrac{q}{e}[/tex]
Put the value into the formula
[tex]n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}[/tex]
[tex]n=4.80\times10^{10}\ electrons[/tex]
Hence, The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]
A helicopter is flying horizontally with a speed of 339 m/s over a hill that slopes upward with a 2% grade. What is the component of the helicopter's velocity perpendicular to the sloping surface of the hill?
We will use the trigonometric ratios to know the helicopter's deceleration speed. Later applying the concept of speed as a vector component of the value found we will find the vertical speed.
The 2% grade indicates that for every 100 meters traveled in the x direction, there is an ascent / descent of 2 meters. Therefore we will have the relationship
[tex]\rightarrow \frac{y}{x} = \frac{2}{100}[/tex]
Now you would have the value o[tex]v_y = 6.78m/s[/tex]f the angle tangent would be
[tex]tan \theta = \frac{y}{x}[/tex]
[tex]tan \theta = \frac{2}{100}{[/tex]
[tex]tan\theta = 0.02[/tex]
From this relationship we could conclude that the vertical speed would be
[tex]v_y = v*tan\theta[/tex]
[tex]v_y = 339*0.02[/tex]
[tex]v_y = 6.78m/s[/tex]
Therefore the component of the helicopter's velocity perpendicular to the sloping surface of the hill is 6.78m/s
The component of the helicopter's velocity perpendicular to the sloping surface of the hill is approximately 6.78 m/s.
Explanation:In this scenario, we can analyze the helicopter's velocity vector into two components: the parallel component along the sloping surface of the hill and the perpendicular component. The component of the velocity perpendicular to the sloping surface is equal to the total velocity multiplied by the grade percentage as a decimal.
Given that the helicopter's speed is 339 m/s and the hill has a 2% grade, we can calculate the perpendicular component as follows:
Perpendicular component = Total velocity × Grade percentage
Perpendicular component = 339 m/s × 0.02
Perpendicular component ≈ 6.78 m/s
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(I) The springs of a 1700-kg car compress 5.0 mm when its 66-kg driver gets into the driver's seat If the car goes over abump, what will be the frequency of oscillations? Ignore damping.
Answer:
[tex]f=7 \ Hz[/tex]
Explanation:
Simple Harmonic Motion
The spring-mass system is a typical case of a simple harmonic motion, since the distance traveled by the mass describes an oscillatory behaviour. The natural angular frequency of a spring-mass system is computed by
[tex]{\displaystyle \omega ={\sqrt {\frac {k}{m}}}}[/tex]
And the frequency is
[tex]{\displaystyle f=\frac {w}{2\pi}[/tex]
Thus
[tex]{\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {k}{m}}}}[/tex]
The total mass of the car and the driver is
[tex]m=1700+66=1766\ kg[/tex]
They both weigh
[tex]W=m.g=1766\ kg*9.8\ m/s^2[/tex]
[tex]W=17306.8\ N[/tex]
We need to know the constant of the spring. It can be found by using the formula of the Hook's law:
[tex]F=k.x[/tex]
We know the spring stretches 5 mm (0.005 m) when holding the total weight of the car and the driver. Solving for k
[tex]\displaystyle k=\frac{F}{x}[/tex]
[tex]\displaystyle k=\frac{17306.8}{0.005}[/tex]
[tex]k=3,461,360\ N/m[/tex]
Thus, the frequency of oscillations is
[tex]{\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {3,461,360}{1,766}}}}[/tex]
[tex]\boxed{f=7 \ Hz}[/tex]
The frequency of oscillation of a car going over a bump can be calculated using principles of simple harmonic motion. By calculating the spring constant from the car's weight and the amount the springs are compressed, we can find the angular frequency and then the normal frequency of the oscillation.
Explanation:The question is about understanding the oscillation concepts of a car once it goes over a bump. It can be treated as a simple harmonic oscillator. To calculate the frequency, we need to first find the spring constant (k) based on Hooke's Law which states that the force F on the spring exerted by the car's weight equals k multiplied by the distance the spring is compressed (x), or F = kx.
In this case, the force F is the weight of the car plus the driver: (1700kg + 66kg) * 9.81m/s² gravitational acceleration. The spring compression x is 5.0mm, or 0.005m. Solving F = kx for k gives us the spring constant.
The angular frequency w of oscillation is equal to √(k/m), where m is the total mass: 1700 kg (car mass) + 66 kg (driver mass), and k is the spring constant we found earlier. The frequency (f) can then be found by dividing w by 2π.
This frequency measures the number of oscillations the car makes per second after going over a bump.
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Why isn't Coulomb's law valid for dielectric objects, even if they are spherically symmetrical?
Answer:
Explanation:
The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.
In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert force on moving charges, therefore, no longer can consider only the electrostatic force.
An electron falls through a distance d in a uniform electric field of magnitude E. Thereafter, the direction of the field is reversed (keeping its magnitude the same) and now a proton falls through the same distance. Compare, using quantitative reasoning, the time of fall in each case. Contrast this situation with that of objects falling freely under gravity.
Answer:
Please see below as the answer is self-explanatory
Explanation:
Assuming that the electric field is pointing upward this will produce a downward force on the electron. Neglecting the effect of gravity, according to Newton's 2nd Law, the force on the electron due to the field, produces an acceleration, that can be found solving the following equation:
F = me*a = qe*E ⇒ a = qe*E / me
If the electric field is uniform, the acceleration that produces is constant, so, we can use the kinematic equation that relates displacement and acceleration with time:
x = v₀*t + 1/2*a*t² = v₀*t +1/2*(qe*E/me)*t²
Now, for a proton falling, if the direction of the field is reversed (pointing downward) it will accelerate the proton downward.
Using the same reasoning as above, we get the value of the acceleration as follows:
F = mp*a = qp*E ⇒ a = qp*E / mp
The equation for displacement is just the following:
x = v₀*t + 1/2*a*t² = v₀*t +1/2*(qp*E/mp)*t²
We know that qe = qp = 1.6*10⁻¹⁹ coul, but mp = 1,836 me, so, for the same displacement, the time must be much less for the electron, that has an acceleration 1,836 times higher.
When both objects fall freely the same distance under the sole influence of gravity, if the initial velocity is the same, the time must be the same also, as the fall time doesn't depend on the mass of the object.
Answer:
Explanation:
We know that the electric force can be expressed as: F=qE. According to Newton's Second Law of Motion, force can also be expressed as: F=ma. Therefore: a=F/m. We can substitute the electric force expression for "F" in this equation. We get: a=qE/m. We can see from this equation that acceleration is inversely proportional to mass and directly proportional to the electric field and charge. Since the electric field is being reversed and since the charges on the proton and electron differ only by the + or - sign respectively, the numerator of this fraction will remain constant in this scenario. The only variables that are effectively changing are the mass and the resultant acceleration. From the inverse relationship of acceleration and mass, we can say that the proton - having a significantly larger mass than the electron - should experience a smaller acceleration, and should thus take longer to fall distance "d." The electron, on the other hand, should experience a greater acceleration due to its significantly smaller mass, and should fall through distance "d" in a shorter amount of time.
Under the influence of gravity (on the surface of the Earth, for example), objects released from the same height should fall freely with the same acceleration at any given time, regardless of mass. It makes sense, however, that subatomic particles interacting with the electric field are hardly affected by gravity, given how weak gravitational forces are on the microscopic scale.
A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows through a capacitor that has an initial charge of 2μC (micro Coulombs). After two microseconds, how much is the magnitude of the net electric charge (in μC) of the capacitor?
Answer:
[tex]8\mu C[/tex]
Explanation:
t = Time taken = [tex]2\mu s[/tex]
i = Current = 3 A
q(0) = Initial charge = [tex]2\mu C[/tex]
Charge is given by
[tex]q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C[/tex]
The magnitude of the net electric charge of the capacitor is [tex]8\mu C[/tex]
An 8900-pF capacitor holds plus and minus charges of 1.85×10−7 C . Part A What is the voltage across the capacitor?
The voltage across the capacitor is 20.79 V.
Explanation:Capacitance is a measure of how much charge can be stored in a capacitor per unit potential difference. The voltage across the capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. In this case, the charge is 1.85×10-7 C and the capacitance is 8900 pF. Converting the capacitance to farads, we get C = 8900 × 10-12 F. Plugging in these values, we have V = 1.85×10-7 C / (8900 × 10-12 F), which simplifies to V = 20.79 V.
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Two carts are colliding on an airtrack (neglect friction). The first cart has a mass of m1=40 g and an initial velocity of v1=2 m/s. The second cart has a mass of m2=47 g and an initial velocity of v2=-5 m/s. Two experiments are conducted. In the first experiment, the first cart has a final velocity of v1'=-1.11 m/s.
What is the velocity of the second cart?
For the second experiment, the bumpers of the carts are modified, but the carts are started with the same initial velocities as before. Now the first cart has a final velocity of v1'=-3.8942 m/s, and the second cart a final velocity of v2'=0.0163 m/s.
How much (if any) energy was lost in the collision?
In good approximation, what kind of collision was the second experiment?
Answer:
v₂ = -2.35m/s
This is an Inelastic collision
Explanation:
Law of conservation of momentum
This states that for a collision occurring between two object in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
Given from the question
m₁=40 g
initial velocity of m₁: v₁=2 m/s
m₂=47 g
initial velocity of m₂: v₂ = -5 m/s
final velocity of m₁: v'₁ = -1.11 m/s.
final velocity of m₂ = ?
second experiment final velocity of m₁ : v₁' = -3.8942 m/s
second experiment final velocity of m₂ : v₂' = 0.0163 m/s
Considering the first experiment
Apply the knowledge of conservation of momentum
[tex] m₁v₁ + m₂v₂ =m₁v₁ + m₂v₂[/tex]
[tex](0.040)(2) + (0.047)(-5) = (0.040)(-1.11) + (0.040)v₂[/tex]
[tex]v₂ = -2.35m/s[/tex]
Considering the second experiment
initial kinetic energy [tex]KEₓ = (1/2)m₁v₁² + (1/2)m₂v₂²[/tex]
[tex]KEₓ = (1/2)(0.040)(2)² + (1/2)(0.047)(-5)²[/tex]
[tex]KEₓ = 0.6675 J[/tex]
Final kinetic energy[tex] KEₙ = (1/2)m₁v₁² + (1/2)m₂v₂² [/tex]
[tex]KEₙ = (1/2) (0.040) (-3.8942)² + (1/2) (0.047) (0.0163)² [/tex]
[tex]KEₙ = 0.3033 J [/tex]
Loss in kinetic energy [tex]ΔKE = KEₓ - KEₙ [/tex]
ΔKE = 0.6675 - 0.3033
ΔKE = 0.3642 J
This collision is a perfectly inelastic collision because the maximum kinetic energy is lost this means that the kinetic energy before the collision is not the same as the kinetic energy after the collision.Though momentum is conserved kinetic energy is not conserved.
While floating in space a 100.-kg robot throws a 0.800-kg wrench at 12.0 m/s toward its human partner working on the spaceship. How fast will the robot recoil?
Answer:
v = - 0.096 m/s
Explanation:
given,
mass of the robot, M = 100 Kg
mass of wrench, m = 0.8 Kg
speed of the wrench,v' = 12 m/s
recoil of the robot,v = ?
initial speed of the robot and the wrench is equal to zero
using conservation of motion
(M + m) V = M v + m v'
(M + m) x 0 = 100 x v + 0.8 x 12
100 v = -9.6
v = - 0.096 m/s
negative sign shows that velocity of robot is in opposite direction of wrench.
hence, the recoil velocity of the robot is equal to 0.096 m/s
A magnetic field of magnitude 1.30x10-3 T is measured a distance of 0.03 m from a long straight wire. What is the current through the wire?
Final answer:
The current through the wire is 1.95 A.
Explanation:
To find the current through the wire, we can use Ampere's law. Ampere's law states that the magnetic field around a long straight wire is directly proportional to the current through the wire and inversely proportional to the distance from the wire.
So, we can use the equation B = μ0 * I / (2π * r), where B is the magnetic field, μ0 is the magnetic constant, I is the current, and r is the distance from the wire.
Plugging in the given values, we have 1.30x10-3 T = (4πx10-7 T*m/A) * I / (2π * 0.03 m). Solving for I, we get I = 1.30x10-3 * (2*0.03)/(4x10-7) = 1.95 A.
A 0.105-g sample of X2 contains 8.92x1020 molecules. The compound MX3 consists of 54.47% X by mass. What are the identities of M and X, and what is the correct name for MX3?
Answer:
The element X is chlorine.
The element M is Ytterium.
Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]
Explanation:
Molecules of [tex]X_2, N = 8.92\times 10^{20} [/tex]
Moles of [tex]X_2, n= ?[/tex]
[tex]N=n\times N_A[/tex]
[tex]n=\frac{N}{N_A}=\frac{8.92\times 10^{20}}{6.022\times 10^{23} mol^{-1}}[/tex]
n = 0.001481 mol
Mass of [tex]X_2, m=0.105 g[/tex]
Molar mass of [tex]X_2=M[/tex]
[tex]m=M\times n[/tex]
[tex]M=\frac{m}{n}=\frac{0.105 g}{0.001481 mol}[/tex]
M = 70.87 g/mol
Atomic mass of X = [tex]\frac{70.87 g/mol}{2}=35.435 g/mol[/tex]
X is a chlorine atom.
The compound [tex]MX_3[/tex] consists of 54.47% X by mass.
M' = Molar mass of compound [tex]MX_3[/tex]
Percentage of X in compound = 54.47%
[tex]54.47\%=\frac{3\times 35.435 g/mol}{M'}\times 100[/tex]
M' = 195.16 g/mol
Atomic mass of M = a
[tex]195.16 g/mol=a+3\times 35.435 g/mol[/tex]
[tex]a = 195.16 g/mol-3\times 35.435 g/mol[/tex]
a = 88.86 g./mol
The element M is Ytterium.
Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]
The element X is Chlorine and the element M is Ytterium.
Ytterium Chloride is the correct name for [tex]MX_3[/tex] that is [tex]YCl_3[/tex]
Number of molecules of [tex]X_2={8.92*10^{20}[/tex]
Number of moles of X:
[tex]n=\frac{8.92*10^{20}}{6.023*10^{23}}[/tex]
n = 0.001481 mol = 0.105g
Molar mass of [tex]X_2[/tex] molecule
[tex]M= 0.105/0.001481\\\\M=70.87g/mol[/tex]
Therefore molar mass of X atom = molar mass of [tex]X_2[/tex]/2
m = 35.435g/mol
X is a chlorine atom.
Given that the compound consists of 54.47% X by mass.
Let M' = Molar mass of compound [tex]MX_3[/tex]
Percentage of X in compound = 54.47%
54.47 = {amount of X}/{mola mass of compound}*100
[tex]54.47 = \frac{3*35.435}{M_{'} }*100\\\\M^{'}= 195.16g/mol[/tex]
let Atomic mass of M = A
A = 195.16 - 3 × 35.435
A = 88.86 g/mol
The element M is Ytterium.
Therefore the compound [tex]MX_3[/tex] is Ytterium Chloride [tex]YCl_3[/tex]
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A 90 kg man lying on a surface of negligible friction shoves a 69 g stone away from himself, giving it a speed of 4.0 m/s.
What speed does the man acquire as a result?
Answer:
v = 3.06 x 10⁻³ m/s
Explanation:
given,
mass of man,M = 90 kg
mass of stone , m= 69 g = 0.069 Kg
speed of the stone, u = 4 m/s
speed of the man, v = ?
using conservation of momentum
initial velocity of the man and stone is equal to zero
(M + m)V = M v + m u
(M + m) x 0 = 90 x v + 0.069 x 4
90 v = 0.276
v = 0.00306 m/s
v = 3.06 x 10⁻³ m/s
speed of the man on the frictionless surface is equal to 3.06 x 10⁻³ m/s
The man's velocity after shoving the stone is -0.003067 m/s, indicating movement in the direction opposite to the stone, as per the conservation of momentum.
Explanation:This problem is a classic example of the conservation of momentum, which is a fundamental concept in physics. According to the law of conservation of momentum, if no external forces are acting on a system, the total momentum of the system remains constant. In this scenario, the man and the stone together constitute an isolated system because the surface is frictionless and there are no external forces acting on the system.
The total momentum before the man shoves the stone is zero since both are initially at rest. When the man shoves the stone, the stone acquires momentum in one direction, and the man must acquire momentum in the opposite direction to conserve the total momentum of the system. The magnitude of the momentum gained by the stone and the man will be equal because their initial total momentum was zero.
The man's velocity (v) can be calculated using the formula:
Momentum of stone = mass of stone (m_stone) * velocity of stone (v_stone)
Momentum of man = mass of man (m_man) * velocity of man (v)
Since total momentum is conserved, Momentum of stone = -Momentum of man.
We can then solve for the man's velocity:
(m_stone * v_stone) = -(m_man * v)
To find v, we rearrange the equation:
v = - (m_stone * v_stone) / m_man
Plugging in the given values:
v = - (0.069 kg * 4.0 m/s) / 90 kg
v = - (0.276 kg·m/s) / 90 kg
v = -0.003067 m/s
The negative sign indicates that the man's velocity is in the opposite direction of the stone's velocity.
g what are the bands of electromagneitc radiation and how are they related to energy frequency and wavelength
Answer:
The details of bands is given in explanation.
Explanation:
The electromagnetic waves are differentiated into different bands based upon their wavelengths and frequencies. The names of different bands are as follows:
1. Radio Waves
2. Micro Waves
3. Infra-red
4. Visible light
5. Ultra Violet
6. X-rays
7. Gamma Rays
The frequency of every region or rays increases from 1 through 7. The energy of rays also increase from 1 through 7. Since, the wave length is inversely related to energy and frequency, thus the wavelength of rays decrease from 1 through 7.
A detailed information of the bands is provided in the picture attached.
How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s?
Answer:
n = 7.5 times/minute
Explanation:
Given that,
Wavelength of the ocean wave, [tex]\lambda=40\ m[/tex]
The speed of the ocean wave, v = 5 m/s
To find,
Number of times a minute does a boat bob up and down on ocean waves.
Solution,
The relation between the speed of wave, wavelength and frequency is given by :
[tex]v=f\times \lambda[/tex]
[tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f=\dfrac{5\ m/s}{40\ m}[/tex]
f = 0.125 Hz
The number of times per minute the bob moves up and down is given by :
[tex]n=f\times t[/tex]
[tex]n=0.125\times 60[/tex]
n = 7.5 times/minute
So, its will move up and down in 7.5 times/minute. Therefore, this is the required solution.
Taking into account the definition of wavelength, frecuency and propagation speed, the number of times per minute the bob moves up and down is 7.5 times per minute.
WavelengthWavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
FrequencyFrequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Propagation speedFinally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
Amount of times in a minute that a boat bob up and down on ocean wavesIn this case, you know:
v= 5 [tex]\frac{m}{s^{2} }[/tex]f= ?λ= 40 mReplacing in the definition of propagation speed:
5 [tex]\frac{m}{s^{2} }[/tex]= f× 40 m
Solving:
f= 5 [tex]\frac{m}{s^{2} }[/tex]÷ 40 m
f= 0.125 Hz= 0.125 [tex]\frac{1}{seconds}[/tex]
Then, a boat bob up and down on ocean waves 0.125 times in a second.
So, the number of times per minute the bob moves up and down is given by:
n= f× time
n= 0.125 Hz× 60 minutes in 1 second
n=7.5 times per minute
Finally, the number of times per minute the bob moves up and down is 7.5 times per minute.
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A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A and B are given below.Randomized Variables A= -3.5 m/s B= 3.9 m/s^2 a. What is the velocity, in meters per second. of the particle at the time t1= 3.0 s? b. What is the velocity, in meters per second: of the particle when it is at the origm (x=0) at time to> 0?
Answer
given,
position of particle
x(t)= A t + B t²
A = -3.5 m/s
B = 3.9 m/s²
t = 3 s
a) x(t)= -3.5 t + 3.9 t²
velocity of the particle is equal to the differentiation of position w.r.t. time.
[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)[/tex]
[tex]v= -3.5 + 7.8 t [/tex]------(1)
velocity of the particle at t = 3 s
v = -3.5 + 7.8 x 3
v = 19.9 m/s
b) velocity of the particle at origin
time at which particle is at origin
x(t)= -3.5 t + 3.9 t²
0 = t (-3.5 + 3.9 t )
t = 0, [tex]t=\dfrac{3.5}{3.9}[/tex]
t = 0 , 0.897 s
speed of the particle at t = 0.897 s
from equation (1)
v = -3.9 + 7.8 t
v = -3.9 + 7.8 x 0.897
v = 3.1 m/s
To solve the problem we should know about velocity.
VelocityVelocity is the rate of change of its position with respect to time.
[tex]V = \dfrac{dy}{dt}[/tex]
Given to us
x(t)= At+Bt^2A= -3.5 m/s B= 3.9 m/s^2Velocity of Particle
x(t)= At+Bt²
[tex]V(t) = \dfrac{dy}{dt} = \dfrac{d(At+Bt^2)}{dt} = A+2Bt[/tex]
A.) the velocity, in meters per second. of the particle at the time t1= 3.0 s,
Velocity of particle(t = 3.0 s)[tex]V(t) = A +2Bt[/tex]
Substituting the values,
[tex]V(t_1=3) = (-3.5) +2(3.9)(3.0)\\\\V(t_1=3) = 19.9\ m/s[/tex]
B.) the velocity, in meters per second: of the particle when it is at the origin (x=0) at t ≥ 0
Displacement, x = 0[tex]x(t)= At+Bt^2\\\\0 = At+Bt^2\\\\[/tex]
Taking t as common,
[tex]0 = t(A+Bt)\\\\[/tex]
[tex]0 = (A+Bt)\\\\[/tex]
Substituting the values and solving or t,
[tex]0 = A+ Bt\\0 = -3.5 + (3.9)t\\3.5=3.9t\\t= \dfrac{3.5}{3.9}\\\\t = 0.8974\ s[/tex]
Velocity of particle(t = 0.8974 s)Substituting the value in the formula of velocity,
[tex]V(t) = A +2Bt[/tex]
Substituting the values,
[tex]V(t_1=0.8974) = (-3.5) +2(3.9)(0.8974)\\\\V(t_1=3) = 3.5 \ m/s[/tex]
Hence, the velocity of the particle is 3.5 m/s.
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