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A force of 200N acts on a body that moves along a horizontal plane in the same direction of movement. The body moves 30m. What is the work done by that force?

Answers

Answer 1

Answer:

Work = Force times Distance

Work = 200 x 30

Work = 6000

The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.

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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:

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

What is 902 in proper scientific notation?

Answers

Hope it help you 9.02x10^2

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 162 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.

Answers

Answer:

0.00124 V

Explanation:

Parameters given:

Initial circumference = 162 cm

Rate of decrease of circumference = 14 cm/s

Magnetic field, B = 0.5 T

Time, t = 8 secs

The magnitude of the EMF induced in the loop is given as:

V = (-NBA) / t

Where N = number of turns = 1

B = magnetic field

A = area of loop

t = time taken

First, we need to find the area of the loop.

To do this, we will find the radius after the loop circumference has decreased for 8 secs.

The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:

14 * 8 = 112 cm

The new circumference is:

162 - 112 = 50 cm = 0.5 m

To get radius:

C = 2 * pi * r

r = C / (2 * pi)

r = 0.5 / (2 * 3.142)

r = 0.0796 m

The area is:

A = pi * r²

A = 3.142 * 0.0796²

A = 0.0199 m²

Therefore, the EMF induced is:

V = (-1 * 0.5 * 0.0199) / 8

V = -0.00124V

This is the EMF induced in the coil.

The magnitude is |-0.00124| V = 0.00124 V.

If a barometer reads 772 mm hg, what is the atmospheric pressure expressed in pounds per square inch?

Answers

15.23.....................

I think it would be 15.23 not so sure but
hope this helps! (:

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

A frictionless pendulum is made with a bob of mass 12.6 kg. The bob is held at height = 0.650 meter above the bottom of its trajectory, and then pushedforward with an initial speed of 4.22 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?

Answers

The answer to your question is 55

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