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Anton [14]
3 years ago
8

In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o

f 4 m/s. When the ball impacts the target, it sticks to it and they drift off together.
1) How much energy is in the translational energy of the block+ball system after the collision?
2) How much of the initial translational kinetic energy of the ball was converted to internal energy in this collision? KEinitial-KEfinal= ___ J
3) The initial temperatures of the ball and block are 300K. By how much does the entropy of the entire system increase in the collision? Assume all the energy stays in the system, which reaches thermal equilibrium. You can also assume that the temperature is constant in the calculation.
Sfinal-Sinitial = ____ J/K
4) How much did the dimensionless entropy, S/k, change?
Physics
1 answer:
ioda3 years ago
6 0

Answer:

Explanation:

We shall apply law of conservation of  momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1  )  

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

2 )

Initial kinetic energy of the system

= 1/2 x 0.1 x 4²

= 0.8 J

Final  kinetic energy of the system = 7.3 x 10⁻²

Loss of energy = .8 - .073

= .727 J

This energy was converted into internal energy of the system .

3 )

increase in entropy = dQ / T

Here dQ = .727 J

T  = 300 ( Constant )

dQ / T = 2.42 X 10⁻³ J/K

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A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal sp
gulaghasi [49]

Answer:

C

Explanation:

horizintal speed stays same

only vertical speed changes

so H speed will stay 30 m/s

6 0
2 years ago
You can tell from reading this passage that -
Romashka-Z-Leto [24]

Answer: B or "diamonds are rare,valuable gems".

Explanation:

since the link does not show very much, I don't know if A is correct or not. D is incorrect because that  is an opinion. C is incorrect.

4 0
3 years ago
(b) Two containers made of insulating material contain the same volume of water at room
masha68 [24]

Answer:

the volume of liquid decreased due to evaporation from the exposed free surface of water so molecules got evaporated .

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5 0
2 years ago
A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical
Alex_Xolod [135]

Answer:

109.32 N/m

Explanation:

Given that

Mass of the hung object, m = 8 kg

Period of oscillation of object, T = 1.7 s

Force constant, k = ?

Recall that the period of oscillation of a Simple Harmonic Motion is given as

T = 2π √(m/k), where

T = period of oscillation

m = mass of object and

k = force constant if the spring

Since we are looking for the force constant, if we make "k" the subject of the formula, we have

k = 4π²m / T², now we go ahead to substitute our given values from the question

k = (4 * π² * 8) / 1.7²

k = 315.91 / 2.89

k = 109.32 N/m

Therefore, the force constant of the spring is 109.32 N/m

8 0
3 years ago
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