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3 years ago

Why is it possible to remove a paper without moving the object on it.(apply newton's law)

Physics

1 answer:

klasskru [66]3 years ago

7 0

This is possible due to inertia of motion. which is nothing but newton's first law.

according to this law , an object tries to retain its state of motion or rest unless acted upon by an external force.

consider an object placed on a paper, initially both the object and paper are at rest. to pull the paper , we apply force on the paper and paper gains velocity. but the object keeps its motion of rest and hence the paper can be removed without moving the object.

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a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th

ludmilkaskok [199]

Answer:

Position at t= 4 seconds is 144 m

Explanation:

It is given that acceleration, a = 18 t, where t is the time.

We know that Velocity, $v = \int { a} \, dt$

Substituting value of a,

Velocity, $v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c$

We know that at t = 0, v = -12 m/s

So, $9*0^2+c=-12\\ \\ c=-12m/s$

So velocity, $v = (9t^2-12)m/s$

We also know that displacement, $x = \int { v} \, dt$

Substituting value of v,

Displacement, $x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c$

We know that at t = 0, particle is at origin, x =0.

So, $0=3*0^3-12*0+c\\ \\ c=0$

Displacement, $x = 3t^3-12t$

At t = 4 seconds

$x = 3*4^3-12*4=192--48=144m$

Position at t= 4 seconds is 144 m

4 0

3 years ago

A car traveling at 20 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is it’s acceleration?

Yuki888 [10]

I think the answer is -2.9m/s2.

6 0

3 years ago

A straight 1.20 m long conductor has a 2.00 A current travelling toward the East. Earth's magnetic field in this location is 4.9

KATRIN_1 [288]

Answer:

Force's magnitude $=1.176\,10^{-4}\,N$

Direction: down (towards the center of the Earth)

Explanation:

Recall that the magnetic force on a conductor of length L carrying a current I in a magnetic field B is given by the equation: $F=I\,L\,B$ in the case the magnetic field B and the direction of the current are at 90 degrees from each other (which is our case). The direction of the force will be given by the "right hand rule" associated with the vector product that defines this force.

Since the current is moving East, and the magnetic field of the Earth goes from North to South, the resultant Force vector will be pointing towards the Earth (and perpendicular to the plane defined by the current's direction and the magnetic field B)

The magnitude of the force, is given by the formula above, and given that all quantities to be considered are is SI units, it will result in Newtons (N):

$F=I\,L\,B\\F=2\,*\,1.2\,*\,4.9\,10^{-5}\,N\\F=1.176\,10^{-4}\,N$

8 0

4 years ago

A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.

Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0

3 years ago

A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C

GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0

3 years ago