Energy transferred as heat always moves from an object

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago

Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0

Sati [7]

Answer:

Ex= -17.1 N/C

Ey = +26.9 N/C

Explanation:

We apply formula of electric field:

Ep=k*q/d²

Ep: Electric field at point ( N/C)

q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1

Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)

Calculation of the field at point P due to the load q₂

$d_{2} = \sqrt{1.30^{2}+0.450^{2} }$

d₂=1.375 m

E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂

Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β

, then,E₂ tiene componentes x-y en P.

E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C

E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C

Calculation of the electric field at point P located at the origin(0,0)

Ex=E₂x= -17.1 N/C

Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C

4 0

3 years ago

What is longitudinal wave?

nignag [31]

Answer: The definition is a longitudinal wave is a wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave

Explanation:

7 0

3 years ago

A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

<u>a) calculating work done by the force over the journey of the train</u>

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago

While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary.

alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

For any object placed on a slide, there are 3 external forces acting on it:

Fg = m*g (always downward)
N (normal force, always perpendicular to the surface of the slide. going upward)
Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)

6 0

3 years ago