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AlekseyPX
3 months ago
11

A 5.0-g marble is released from rest in the deep end of a swimming pool. An underwater video reveals that its terminal speed in

the water is 0.30 m/s. (a) What is the acceleration of the marble at the instant it is released? (b) What is the acceleration of the marble when it has reached its terminal speed? (c) How long does it take the marble to reach half its terminal speed?
Physics
1 answer:
serg [3.5K]3 months ago
3 0
a) The marble's acceleration when released is a = -g = 9.8 m/s²; b) When it reaches terminal velocity, the acceleration is a = 0 m/s²; c) The time taken to reach half of its terminal speed is t1 = 0.0213 s. When the marble is dropped, its velocity is zero, meaning there is no resistive force acting on it; the only force present is gravity, resulting in an acceleration equivalent to gravity's rate: a = -g = 9.8 m/s² At terminal velocity, all forces balance out, which leads to zero net acceleration: a = 0 m/s² Assuming a specific resistive force for a liquid, this force is generally proportional to the object's velocity. The relevant equation for this scenario is: v = mg/b (1 - e^(-bt/m)) Long term, the exponential approaches zero, so terminal velocity is expressed as: v_terminal = mg/b and, subsequently, b = mg/v_terminal Substituting gives: b = 5 x 10^-3 kg × 9.8 m/s² / 0.3 m/s = 0.163 Using this data, we can calculate the time to reach v = 0.5 × v_terminal: 0.5 × v_terminal = mg/b (1 - e^(-bt/m)) Thus, 0.5 = 1 - e^(-0.163t1/(5 x 10^-3)) e^(-32.6t1) = 0.5 Taking the natural logarithm gives: -32.6t1 = ln(0.5) t1 = -1/(32.6)(-0.693) = 0.0213 s.
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A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself.
ValentinkaMS [3465]

What is the coefficient of kinetic friction between the block and the surface? Provide your answer using two significant figures.

Response:

0.39

To elaborate:

The provided data includes

m1=1.7 Kg

m2=0.011 Kg

v2=670 m/s

d=2.4 m

The relationship m2v2=(m1+m2)v applies here hence v=\frac {m2v2}{m1+m2} also a=\frac {v^{2}}{2d}

Friction's deceleration can be calculated using

F=\mu_k N=\mu_k W=\mu_k (m1+m2)g

F=(m1+m2)a=\mu_k (m1+m2)gthus a=\mu_k g

a=\frac {v^{2}}{2d}=\mu_k g

\mu_k=\frac {1}{2dg}(\frac {m2v}{m1+m2})^{2}=\frac {1}{2*2.4*9.81}\times (\frac {0.011*670}{1.7+0.011})^{2}

\mu_k=0.39

7 0
2 months ago
All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is
Sav [3153]
With all forces acting on the bullets balanced, the resultant force on a bullet becomes zero, leading to zero acceleration and establishing a state identified as constant velocity. The bullet maintains a steady velocity. Since acceleration signifies the change in velocity over time, having no acceleration indicates a constant velocity.
8 0
3 months ago
Read 2 more answers
Suppose you sketch a model of an atom using the ones here as a guide. How would you build a model that is ionized? How would you
Maru [3345]

Answer:

Explanation:

An atom comprises three distinct particles: electrons, protons, and neutrons.

These particles vary in mass and electric charge, contributing to the atom's characteristics.

Electrons bear a negative charge, protons are positively charged, and neutrons possess no charge. An atom is electrically neutral when it has an equal number of electrons and protons, but this can change if particles are removed.

1: Ionized atom model - an ionized atom carries a net charge, which can be positive or negative.

To illustrate an ionized atom, one would need to reduce the number of either electrons or protons.

2: Radioactive atom model: A radioactive atom is characterized as unstable and retains excess energy in its nucleus, often due to added neutrons or protons.

3 0
3 months ago
An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
Sav [3153]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

\texttt{ }

Additional explanation

The Ideal Gas Law that should be remembered is:

\large {\boxed {PV = nRT} }

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

\texttt{ }

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

Next, we will determine the work performed on the gas:

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

Step B:

By utilizing the methodology mentioned earlier:

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

Next, we will ascertain the work completed on the gas:

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

Ultimately, we can calculate the total work done and heat supplied as follows:

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

Learn more

  • Minimum Coefficient of Static Friction:
  • The Pressure In A Sealed Plastic Container:
  • Effect of Earth’s Gravity on Objects:

\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0
3 months ago
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