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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer 1

Answer:

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=(2c)/(3b).$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left ((dx)/(dt)\right )_(t=t_o)=0.$
$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Applying both these conditions,

$\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).$

For $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = (2c)/(3b)$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.$

Here,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Thus, the particle reach its maximum x value at time $\rm t_o = (2c)/(3b).$

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A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and, at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna

Answers

Answer:

-0.00152 V

Explanation:

Parameters given:

Diameter of the loop = 11 cm = 0.11m

Rate of change of magnetic field, dB/dt = 0.16 T/s

Radius of the loop = 0.055m

The area of the loop will be:

A = pi * r²

A = 3.142 * 0.055²

A = 0.0095 m²

The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:

EMF = - N * A * dB/dt

In this case, there's only one loop, so N = 1.

Therefore:

EMF = -1 * 0.0095 * 0.16

EMF = -0.00152 V

The negative sign indicates that the current flowing through the loop acts opposite to the change in the magnetic field.

Answer:

The induced emf is 0.00152 V

Explanation:

Given data:

d = 11 cm = 0.11 m

$r=(d)/(2) =(0.11)/(2) =0.055m$

The area is:

$A=\pi r^(2) =\pi *(0.055^(2) )=0.0095m^(2)$

The induced emf is:

$E=-A(dB)/(dt) =-(0.0095)*(0.16)=-0.00152V$

The negative indicates the direction of E.

3. Which object has more inertia?A. A tractor trailer rig moving at 2 m/s
B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s

Answers

Answer:

A. A tractor trailer rig moving at 2 m/s

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a tractor trailer rig, the greater will be its tendency to continuously remain at rest.

Hence, the object that has more inertia is a tractor trailer rig moving at 2 m/s because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia.

An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due east and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?

Answers

The average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Total displacement

Since the unladen swallow that weighs 0.03 kg flies straight northeast (that is at a bearing of 45°) a distance of 125 km in 4.0 hours.

Its position vector after 4.0 hours is d = (125kmcos45)i + (125kmsin45)j = (125000 × 1/√2)i + (125000 × 1/√2)j

= (62500√2)i + (62500√2)j.

If the initial position of the swallow is d' = 0i + 0j, then its total displacement after 4 hours is, D = d - d'

= (62500√2)i + (625000√2)j - (0i + 0j)

= (62500√2)i + (62500√2)j m

Average velocity

The unladen swallow's average velocity, v = D/t where

D = total displacement = (62500√2)i + (62500√2)j m and
t = time = 4.0 hours = 4 × 60 min/hr × 60 s/min = 14400 s

So, v = [(62500√2)i + (62500√2)j m]/14400 s = (88388.35)i/14400 + (88388.35)j /1440

= 6.14i + 6.14j m/s

Average momentum

The average momentum of the unladen swallow is p = mv where

m = mass of unladen swallow = 0.03 kg and
v = average velocity = 6.14i + 6.14j m/s

So, p = mv

p = 0.03 kg × (6.14i + 6.14j m/s)

p = (0.1842i + 0.1842j) kgm/s

So, the average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Learn more about average momentum here:

brainly.com/question/25941500

Answer:

The average momentum of the bird is 0.26 kgm/s

Explanation:

The formula to be used here is that of momentum which is

momentum (in kgm/s) = mass (in kg) × velocity (in m/s)

The velocity of the bird is

velocity (in m/s) = distance (in meter) ÷ time (in seconds)

distance in meters = 125km × 1000 = 125,000 m

time in seconds = 4 hrs × 60 × 60 = 14,400 secs

velocity = 125000/14400

velocity = 8.68 m/s

momentum (p) = 0.03 × 8.68

p = 0.26 kgm/s

The average momentum of the bird is 0.26 kgm/s

A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?

Answers

Answer:

Energy, $E=4.002* 10^(-19)\ J$

Explanation:

It is given that,

Wavelength of the photon, $\lambda=500\ nm=5* 10^(-7)\ m$

We need to find the photon representing the particle interpretation of this light. it is given by :

$E=(hc)/(\lambda)$

$E=(6.67* 10^(-34)* 3* 10^8)/(5* 10^(-7))$

$E=4.002* 10^(-19)\ J$

So, the energy of the photon is $4.002* 10^(-19)\ J$ . Hence, this is the required solution.

An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?

Answers

Answer:

Δ KE = - 8.75 x 10⁻⁴ J

Explanation:

given,

mass of applesauce = 7 g = 0.007 Kg

initial velocity, u = 0.5 m/s

final velocity, v = 0 m/s

Decrease in kinetic energy = ?

initial kinetic energy

$KE_1=(1)/(2)mu^2$

$KE_1=(1)/(2)* 0.007 * 0.5^2$

KE₁ = 8.75 x 10⁻⁴ J

final kinetic energy

$KE_2=(1)/(2)mv^2$

$KE_2=(1)/(2)* 0.007 * 0^2$

KE₂ =0 J

Decrease in kinetic energy

Δ KE = KE₂ - KE₁

Δ KE = 0 - 8.75 x 10⁻⁴

Δ KE = - 8.75 x 10⁻⁴ J

decrease in kinetic energy of the applesauce is equal to 8.75 x 10⁻⁴ J

Final answer:

The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.

Explanation:

This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.

When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).

Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.

Learn more about Kinetic Energy here:

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?