Unless otherwise specified in the question, responses to numerical question must be within 5% of the correct answer. Try out different responses to the following question, which has an answer of 20, where a correct response is between 19 and 21.

What is the average velocity (in kilometers per hour) of a cart that travels 100 km in 5 hours?

2.An exception to the 5% tolerance rule arises for simple calculations that can be done in your head that come out as integers or or simple decimals. So if the calculation involves dividing 5 by 2 with the result of 2.5, an answer of 2.6 may not be graded as correct.

Give the value of 6.5 times 2.

Try it several times giving the answer (13) in different forms (say, 13 or 1.3e1) and with different accuracies (say, 13.001 or 13.2). Since this is a simple calculation, two of these four will be graded as incorrect.

Answers

Answer 1

Part of the problem involves not only answering the question with the correct degree of accuracy, but also making approximations to the correct answer. For example, for the first case we know that the speed is equivalent to the distance traveled in a given time. Therefore it would be defined as

]1)  [tex]V = \frac{x}{t}[/tex]

x = Displacement

t = Time

The displacement value is 100km and the time value is 5 hours. Therefore the speed value would be

[tex]V = \frac{100km}{5h} = 20km/h[/tex]

We know that the answer margin is within 5% of the value, then 5% of 20 would be 1.   That is, we have a margin of error of '1km / h' to answer the question. Any value that falls within that range can be added or subtracted from the response and the response will be valid. Values included within this value would be

[tex]V_1 = 20km/h +1km/h = 21km/h[/tex]

[tex]V_2 = 20km/h - 1km/h = 19km/h[/tex]

[tex]V_3 = 20km/h + 0.5km/h = 20.5km/h[/tex]

[tex]V_4 = 20km/h -0.5km/h = 19.5km/h[/tex]

[tex]V_5 = 20km/h +0.01km/h = 20.01km/h[/tex]

2) For the second case the margin of tolerance for the response is 5%, so if we multiply the given value we would have a response of.

[tex]x = 6.5*2 = 13[/tex]

5% of 13 is 0.65. Therefore, any value that falls within that range will be a correct answer. The value could then be

[tex]x_1 = 13+0.65 = 13.65[/tex]

[tex]x_2 = 13-0.65 =12.35[/tex]

Incorrect values will be

[tex]x_3 = 13+1 = 14[/tex]

[tex]x_4 = 13-1=12[/tex]


Related Questions

g what are the bands of electromagneitc radiation and how are they related to energy frequency and wavelength

Answers

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.

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?

Answers

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]

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?

Answers

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

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.

Answers

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]

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.

Answers

Final answer:

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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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?

Answers

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.

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?

Answers

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.

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.

Answers

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]

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

Answers

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 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

Answers

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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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

Answers

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 A

The 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 B

The 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 C

The 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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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?

Answers

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.

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?

Answers

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.

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?

Answers

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 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?

Answers

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.

Velocity

Velocity 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^2

Velocity 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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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.

Answers

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.

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?

Answers

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

Why isn't Coulomb's law valid for dielectric objects, even if they are spherically symmetrical?

Answers

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.

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)

Answers

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?

Answers

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

Final answer:

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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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?

Answers

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

Final answer:

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?

Answers

Final answer:

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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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?

Answers

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.

Wavelength

Wavelength 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).

Frequency

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Propagation speed

Finally, 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 waves

In this case, you know:

v= 5 [tex]\frac{m}{s^{2} }[/tex]f= ?λ= 40 m

Replacing 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 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?

Answers

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]

Final answer:

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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(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?

Answers

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?

Answers

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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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.

Answers

Final answer:

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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12 of 15 Constants The predominant frequency of a certain fire truck's siren is 1500 Hz when at rest.

Answers

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.

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.

Answers

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.

(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.

Answers

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]

Final answer:

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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