Recall: earth applies _____________ on Earth or equivalently

Can two objects exert a force on each other without touching? example

lara [203]

Answer:

where is the example...?

7 0

2 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

II) A 0.40-kg ball, attached to the end of a horizontal ord, is rotated in a circle of radius 1.3 m on a friction- less horizont

Lostsunrise [7]

Hi there!

In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.

The net force is equivalent to the centripetal force, so:

∑F = T

mv²/r = T

Solve for v:

v = √rT/m

v = 13.96 m/s

3 0

2 years ago

A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Lunna [17]

Answer:

option A

Explanation:

given,

height of the drop of stone = 9.44 m

speed of the stone = ?

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2} m v^2$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 9.44}$

$v = \sqrt{185.024}$

v = 13.60 m/s

Hence, the correct answer is option A

3 0

3 years ago

Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$ F be the temperature and T that of the body

Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = k(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = kdt

In [T-70]= kt + $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for k using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

K = $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230 $e^{-0.19018}$

4 0

2 years ago