The ambulance will hear a reflected sound with a frequency of 855Hz, which is calculated using the Doppler Effect.
Explanation:The subject of this question is the Doppler Effect, which states that the observed frequency of a wave depends on the relative speed of the source and the observer. In this case, the source of the sound is the ambulance and the observer is also the ambulance (after reflection of the sound from the hospital). We can use the formula for Doppler Effect in this case:
F' = F *(v+v0) / (v-vS)
Where:
F' is the apparent frequency (the frequency heard by the observer), F is the source frequency (the frequency of the emitted sound), v is the speed of sound on this day, v0 is the speed of the observer (which is zero since the sound after reflection is observed at the same spot), vS is the speed of the source (which is the speed of the ambulance converted to m/s).
Putting the given values into this formula, we can calculate:
F' = 798Hz * (343m/s + 0) / (343m/s - 108km/h converted to m/s) = 855Hz
Therefore, the frequency of the reflected sound heard by the ambulance is 855Hz. The correct option is b. 855Hz.
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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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You have a foam rod rubbed with felt and a small aluminum foil ball attached to a thread. Describe what happens when you slowly approach the ball with the rod and then touch the ball. Explain why this happens.
Answer:
as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.
Explanation:
When you rub a foam rod that is an insulator with felt it acquires some loads, when separated these two the barrel is loaded
When approaching the loaded rubber rod, suppose positively, to the aluminum ball that is a metal, the positive charge attracts the free electors of the aluminum until a balance of the electric force and the tension of the wire in which it is hung is established The aluminum foil therefore the aluminum foil approaches the rubber ball, due to the electrostatic attraction, the closer the aluminum wave deflects.
When the ball touches the sheet, the free charges of the metal pass to the insulator neutralizing the existing charges, whereby the electric field and the electric force decrease. Consequently the aluminum sheet descending and moving away from the rubber ball.
In summary, as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.
Due do the friction charged develop in the ball, the foil and ball attract.
When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.
What is charge by friction?
When the insulating material are rubbed against each other, the electric charge generate between them.
The effect of friction charge between rod and ball leads to,
In these two materials, the martial which gains the electron besoms negatively charged and the martial which losses the electron becomes positively charged.Now the foam rob rubbed with felt it losses the electron and becomes positively charged. Now when the aluminum foil ball is approached with the charged foam rob the attracted to the foil, as the positive charge attracts the aluminum foil.This happens due to negative charged electron of the aluminum foil which attracts towards the ball.As the rod touches the ball the charges passes thought the insulator and the charged over the ball neutralized due to this contact. Now the attraction between them is disappear and they ball start to move.
Hence due to the friction charged develop in the ball, the foil and ball attract.
When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.
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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 spaceship is on a straight-line path between Earth and its moon. At what distance from Earth is the net gravitational force on the spaceship zero?
To solve this problem we will start from the constants given on the two celestial bodies, that is, the earth and the moon. From there we will seek the balance of forces, where the gravitational force is zero. That distance will remain incognito and will be the one we will seek to find, this is
Given that,
[tex]M_E =5.97*10^24 kg[/tex]
[tex]m_m = 7.36*10^22 kg[/tex]
[tex]r_E = 6370 km[/tex]
[tex]r_m = 1737 km[/tex]
The distance between Earth and Moon
[tex]x = 384403*10^3 m[/tex]
As the net gravitational force is zero we have that
[tex]\frac{GMm'}{d^2} = \frac{Gmm'}{(x -d)^2}[/tex]
[tex]\frac{M}{d^2} = \frac{m}{(x -d)^2}[/tex]
Replacing with our values
[tex]\frac{(5.97*10^{24} )}{d^2} = \frac{(7.36*10^{22})}{(x -d)^2}[/tex]
[tex]\frac{(x- d)^2}{d^2}= 0.0123[/tex]
[tex]x^2 + d^2 - 2xd = 0.0123 d^2[/tex]
Replacing x as the distance betwen moon and earth,
[tex](384403 x 10^3)^2+d^2-2(384403 x 10^3)d=0.0123 d^2[/tex]
Solving the polynomial we have that
[tex]d =4.3387*10^8 \text{ or } 3.4506*10^{8}[/tex]
So the minimum value is [tex]3.4506*10^{8}m[/tex]
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.
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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You have a box of filled with gas and by some process you double the number of atoms in the box, but now each one only has (on average) half as much energy as before the process started.
What has happened to the thermal energy and temperature of the gas, respectively?
-Thermal energy increased, temperature decreased.
-Thermal energy decreased, temperature unchanged.
-Thermal energy unchanged, temperature unchanged.
-Thermal energy unchanged, temperature decreased.
Answer:
Option-(D):Thermal energy unchanged, temperature decreased.
Explanation:
Thermal energy and Temperature,T:
Thermal energy of the system is remained unchanged while the temperature,T of the objects present inside the system is more associated with the objects fundamental properties which each matter shows or exhibits in nature.
While the thermal energy is the flow or exchange of thermal or heat energy between the different objects in interaction to each other.
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?
To solve this problem we will apply the considerations of equilibrium on a body, we will decompose the tensions into their respective components and apply the sum of forces and solving equations for the given system.
According to the graph, the sum of the horizontal components would be
[tex]\sum T_x = 0[/tex]
[tex]TSin35\° -TSin35\° = 0[/tex]
[tex]0 = 0[/tex]
The sum of the vertical components is
[tex]\sum T_y = 0[/tex]
[tex]2TCos35\° - mg = 0[/tex]
[tex]2TCos35\° = 600N[/tex]
[tex]T = \frac{600N}{2Cos35\°}[/tex]
[tex]T = 366.2N[/tex]
Therefore the net force on the new vine is 366.2N
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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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]
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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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]
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 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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(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 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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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.
On Planet X, a ball on a massless, rigid rod oscillates as a simple pendulum with a period of 2.0s. If the pendulum is taken to the moon of Planet X, where the free-fall acceleration g is half as big, the period will be
Answer:
Time period at the moon will be 2.828 sec
Explanation:
Let initially the length of the pendulum is l and acceleration due to gravity is g
And time period is given T = 2 sec
So time period of the pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]-------------eqn 1
Now on the planet acceleration due to gravity [tex]g_{new}=\frac{g}{2}[/tex]
So time period of the pendulum at this planet [tex]T_{new}=2\pi \sqrt{\frac{l}{g_{new}}}[/tex]--------eqn 2
Now dividing eqn 1 by eqn 2
[tex]\frac{T}{T_{new}}=\sqrt{\frac{\frac{g}{2}}{g}}[/tex]
[tex]T_{new}=\sqrt{2}T=1.414\times 2=2.828sec[/tex]
So time period at the moon will be 2.828 sec
If the acceleration due to gravity decreases by a half, the period of a simple pendulum increases by √2, or approximately 1.41. Therefore, the period of the pendulum on the moon of Planet X, where gravity is half, would be 2.8 seconds.
Explanation:The period of a simple pendulum in physics is influenced by its length and the acceleration due to gravity (g). It is calculated by the formula: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravity. To find the effect of a change in g on the period of the pendulum, we can look at the relationship between them in the formula. Because the formula for the period of a pendulum involves the square root of the gravity, if gravity is reduced by a half, the period will increase by a factor of √2, which is approximately 1.41.
Thus, if the pendulum's period on Planet X was 2.0 s, then on its moon, where gravity is halved, the period will be approximately 2.0 s * 1.41 = 2.8 s.
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One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.370 kg . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 2.05 m and 1.06 m, and the period of the oscillation is 2.15 s, find the force constant of the spring, the maximum mass of the glider and the magnitude of the maximum acceleration of the glider.
The force constant of the spring is 1.932 N/m. The maximum mass of the glider can be calculated using Newton's second law. The magnitude of the maximum acceleration of the glider can be calculated using the equation for acceleration in SHM.
Explanation:In simple harmonic motion (SHM), the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium position. Let's calculate the force constant of the spring by using Hooke's Law. We know that the force exerted by the spring is given by F = kx, where F is the force, k is the force constant, and x is the displacement. At the maximum displacement, the force exerted is F = k * 2.05 m. At the minimum displacement, the force exerted is F = k * 1.06 m. Equating these two forces, we get k * 2.05 = k * 1.06. Solving this equation, we find the force constant of the spring to be k = 2.05 / 1.06 = 1.932 N/m.
The maximum mass of the glider can be found by calculating the maximum force that can be exerted by the spring. The maximum force is given by F = k * x, where F is the force, k is the force constant, and x is the maximum displacement. Plugging in the values, we get F = 1.932 N/m * 2.05 m = 3.961 N. Using Newton's second law, F = m * a, where F is the force, m is the mass, and a is the acceleration, we can calculate the maximum mass. Rearranging the equation gives m = F / a. Since the acceleration in SHM is given by a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period, we can plug in the given values to find the maximum mass.
The magnitude of the maximum acceleration of the glider can be found by using the equation a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period. Plugging in the given values, we can calculate the magnitude of the maximum acceleration.
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.
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 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]
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.
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
Assuming the same percent uncertainty and the same measurement device, what is the uncertainty in a blood pressure measurement of 71 mm Hg ?
Final answer:
The percent uncertainty of a blood pressure measurement of 120 mm Hg with an uncertainty of ±2 mm Hg is 1.67%. Using this percentage, the uncertainty in a blood pressure measurement of 71 mm Hg would be ±1.2 mm Hg, after rounding to one decimal place.
Explanation:
To calculate the uncertainty in a blood pressure measurement of 71 mm Hg, given the percent uncertainty is the same as that for a measurement of 120 mm Hg with an uncertainty of ± 2 mm Hg, we first find the percent uncertainty of the original measurement.
The percent uncertainty is calculated by dividing the uncertainty by the actual measurement and then multiplying by 100 to get a percentage:
Percent Uncertainty = (Uncertainty / Measurement) × 100
For the measurement of 120 mm Hg, the percent uncertainty is:
(2 mm Hg / 120 mm Hg) × 100 = 1.67%
Assuming the same percent uncertainty for a blood pressure measurement of 71 mm Hg, you would use the same percentage to find the uncertainty:
Uncertainty = Percent Uncertainty × Measurement / 100
Uncertainty in 71 mm Hg = 1.67% × 71 mm Hg / 100 = 1.187 mm Hg
Since we typically round to the nearest whole unit or significant digit the uncertainty would be ± 1.2 mm Hg (rounded to one decimal place).
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.
(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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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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