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:

  1. \rm \left ((dx)/(dt)\right )_(t=t_o)=0.
  2. \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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Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.7Ω and 15.9 W, 30.4Ω and 9.12 W, and 16.3Ω and 12.3 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

Answers

Answer:

a) greatest voltage = 29.25 V

b) power = 16 W

Explanation:

The total resistance R of the three resistors in series is:

R = (6.7 + 30.4 + 16.3) \Omega = 53.4 \Omega  

a) The greatest current I is the one that will burn the resistor with lower power rating, which is 9.12 W:

P_(max) = I_(max)^2 R = I_(max)^2 30.4\Omega = 9.12W\nI_(max) = 0.54 A

The voltage is:

V_(max)=IR = 0.54*53.4V= 29.25 V

b) When the current is 0.54 A, the power is:

P = RI^2=53.4*0.3 W = 16W

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Answers

The work (\(W_f\)) done by the friction force between the ramp and the skateboarder is given by \(-\mu_k \cdot m_s \cdot g \cdot h_y\).

The workdone by the friction force (\(W_f\)) can be calculated using the formula for work, which is the product of the force applied (\(F_f\)) and the displacement (d) over which the force is applied:

\[W_f = F_f \cdot d\]

In this scenario, the frictionforce works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of skateboarder (\(m_s\)) = 54 kg

Height of the ramp (\(h_y\)) = 3.3 m

Final velocity (\(v_f\)) = 6.2 m/s

Coefficient of kineticfriction (\(\mu_k\)) between skateboarder and ramp

Acceleration due to gravity (g) = \(9.81 \, \text{m/s}^2\)

The normal force (\(F_{\text{normal}}\)) is equal to the weight of the skateboarder:

\[F_{\text{normal}} = m_s \cdot g\]

The displacement (d) is the vertical distance (\(h_y\)) that the skateboarder descends down the ramp.

Now we can write the expression for the work done by the friction force (\(W_f\)):

\[W_f = -\mu_k \cdot F_{\text{normal}} \cdot d\]

Substitute the expression for the normal force:

\[W_f = -\mu_k \cdot (m_s \cdot g) \cdot h_y\]

Thus, this expression represents the work done by the friction force between the ramp and the skateboarder in terms of the given variables.

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Your question seems incomplete, the probable complete question is:

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  

Final answer:

The momentum of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

Explanation:

To find the momentum of the box, we can use the formula:

Momentum = mass x velocity

a. The momentum of the box with respect to the floor is: 5 kg x 5 m/s = 25 kg·m/s

b. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

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How long would it take a 500. W electric motor to do 15010 J of work?

Answers

time = energy / power = 15010 / 500 = .... seconds

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answers

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=(1)/(2) kx^(2) +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

(1)/(2) kx^(2) +PE1=PE2

PE2-PE1=(1)/(2) kx^(2)

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

mgh=(1)/(2) kx^(2)

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

2100*9.8*h=(1)/(2)*2200*0.11^(2)

20580*h=13.31

h=(13.31)/(20580)

⇒h=0.0006467m=6.5e-4

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J

So, the change in its gravitational potential energy is 1.76 J.

You're carrying a 3.6-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A Part complete How much force must you exert to keep the pole motionless in a horizontal position? Express your answer in newtons. F = 114 N Previous Answers

Answers

Final answer:

This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.

Explanation:

This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.

To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.

Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:

0.35*F2 = 3.25*F1   (1)

Because the net force should also be zero in equilibrium, we have:

F1 + F2 = 205.8    (2)

Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.

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Final answer:

To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a force of approximately 114N is required

Explanation:

The subject question is a classic example of Torque problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The weight of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.

This torque is calculated as Torque = r * F = 1.8m (distance from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.

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