On the position vs time graph a curved line represents ___________

A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the

o-na [289]

Answer:

A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.

Gpe @ A=mgh=40*9.81*2=784.8J

B. At 30degree vertical angle the vertical displacement from lowest position is given by

2-2cos(30)=2-1.73=0.27m

Gpe @B= 40*9.81*0.27=106 J

C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.

8 0

3 years ago

A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe

marin [14]

Answer:

a) $T_f=305.7049\ K$

b) $\Delta S=313.51\ J.K^{-1}$

Explanation:

Given:

mass of copper, $m_c=2.64\ kg$

initial temperature of copper, $T_{ic}=400\ K$

specific heat capacity of copper, $c_c=385\ J.kg^{-1}.K^{-1}$

mass of water, $m_w=4\ kg$

initial temperature of water, $T_{iw}=300\ K$

specific heat capacity of water, $c_w=4200\ J.kg^{-1}.K^{-1}$

a)

∵No heat is lost in the environment and the heat is transferred only between the two bodies:

Heat rejected by the copper = heat absorbed by the water

$2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)$

$T_f=305.7049\ K$

b)

Now the amount of heat transfer:

$Q=m_c.c_c.(T_{ic}-T_{f})$

$Q=2.64\times 385\times (400-305.7049)$

$Q=95841.5841\ J$

∴Entropy change

$\Delta S=\frac{dQ}{T}$

$\Delta S=\frac{95841.5841}{305.7049}$

$\Delta S=313.51\ J.K^{-1}$

5 0

4 years ago

A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an

Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

8 0

3 years ago

How long would it take for a sound impulse to travel through an aluminum rod 1,500 meters long?

NNADVOKAT [17]

I believe the answer is 0.29.

5 0

3 years ago

Read 2 more answers

When does air become turbulent around a thrown ball?

Mrrafil [7]

B, good luck, hope this helps :)

5 0

3 years ago

On the position vs time graph a curved line represents ____________.