What is the mass of a body? How is it measured?
​

Answer:

The penny will hit the ground at 6.39 seconds

Explanation:

Free Fall

The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:

\(\displaystyle y=\frac{gt^2}{2}\)

Where \(g=9.8\ m/s^2\), and t is the time.

Solving for t:

\(\displaystyle t=\sqrt{\frac{2y}{g}}\)

Using the value y=200:

\(\displaystyle t=\sqrt{\frac{2*200}{9.8}}\)

t=6.39 sec

The penny will hit the ground at 6.39 seconds

The energy stored in the circuit at any time t is given by \(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)The units are in seconds.

The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.

Initially, the capacitor is uncharged, so the second term is zero.

Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.

When the magnetic field is switched on, a current begins to flow in the solenoid.

This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.

Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.

The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.

Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.

This is a first-order linear differential equation, which can be solved to give \(Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))\), where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.

The current in the solenoid is given by I(t) = \(dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).\)

The energy stored in the circuit at any time t is given by\(U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).\)

The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, \((1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.\)

This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.

The solution is t = 1.74 ms.

The units are in seconds.

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Answer:

32J

Explanation:

Heat H = MC∆T = 4 × 1.0 × 8 = 32J

Answer:

I = 0.65 kgm²

Explanation:

Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.

T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.

Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation

So, T = 2.186 s

We now find I by making it subject of the formula in the equation for T.

So,

T = 2π√(I/mgh)

dividing both sides by 2π, we have

T/2π = √(I/mgh)

squaring both sides, we have

(T/2π)² = [√(I/mgh)]²

T²/4π² = I/mgh

multiplying both sides by mgh, we have

T²mgh/4π² = I

I = T²mgh/4π²

substituting the values of the variables into the equation, we have

I = T²mgh/4π²

I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²

I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²

I =  29.539 kgm²/4π²

I = 0.748 kgm²

Now I = I' + mh²  (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.

So, I' = I - mh²

Substituting the values of the variables into the equation, we have

I' = I - mh²

I' = 0.748 kgm² - 4.07 kg × (0.155 m)²

I' = 0.748 kgm² - 4.07 kg × 0.02403 m²

I' = 0.748 kgm² - 0.098 kgm²

I = 0.65 kgm²

The description that tells two processes that scientists think move Earth's lithospheric plates is convection in the mantle and pressure of magma on the edge of the plate.

The Earth's lithosphere is the rocky outer part of Earth which is made up of the brittle crust and the top part of the upper mantle.

The Earth's lithosphere deflects the convections and as the convections churn clockwise of anticlockwise, they drag the  lithosphere with it via friction an this is what is stipulated to cause tectonic plate movements.  

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Answer: convection in the mantle and pressure of magma on the edge of the plate

Explanation: I took the unit test

A bubbler needs 7"/12" of space to create bubbles in water that is 4 feet, 7 inches deep.

Bubbler Systems gauge water level by measuring the force required to force an air bubble through a plastic tube and into the body of water. This pressure, often known as the "line pressure," must adjust to the water's elevation.

2.31 psig/ft

1.98 psig would balance the head's feet. 4'7" 4.583/2.31 = 1.98. .583 is 7"/12"

A method known as bubbler hydroponics introduces extra oxygen to the water and nutrient solution in which your plants are grown. Imagine it being comparable to the air stone seen in fish tanks. we have a bubbler in hydroponics (air stone) set on the bottom of growing container.

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In a DC generator, the generated electromotive force (emf) is directly proportional to the rotational speed of the generator's armature and the strength of the magnetic field within the generator.

This relationship is described by the equation for the generated emf in a DC generator:

Emf = Φ * N * A * Z / 60

Where:

Emf is the generated electromotive force (in volts),

Φ is the magnetic flux density (in Weber/meter^2\(meter^2\) or Tesla),

N is the number of turns in the armature winding,

A is the effective area of the armature coil (in square meters),

Z is the total number of armature conductors, and

60 is a constant representing the conversion from seconds to minutes.

From this equation, we can see that the generated emf is directly proportional to the magnetic flux density (Φ) and the product of the number of turns (N), effective area (A), and the total number of armature conductors (Z). This means that increasing any of these factors will result in a higher generated emf.

The magnetic flux density (Φ) can be increased by using stronger permanent magnets or increasing the strength of the field windings in the generator.

The number of turns (N) and the effective area (A) are design parameters and can be optimized for a specific generator. Increasing the number of turns or the effective area will result in a higher generated emf.

Similarly, the total number of armature conductors (Z) can be increased to enhance the generated emf.

By controlling and optimizing these factors, the generated emf in a DC generator can be increased, resulting in higher electrical output. However, it is important to note that there are practical limits to these factors based on the design and construction of the generator.

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Given that two large parallel plates are separated by 0.20 m and that the potential difference is 12V.

(a) Describe the motion of an electron released from rest at a distance "d" from the negative plate.

(b) What distance from the positive plate will the electron have a speed of 1 x 10^6 m/s?

For part (a):

The magnitude of an electric field can be given as  \(||\vec E||=\frac{\Delta V}{d}\), where "ΔV" is the potential difference and "d" is the distance between the plates.

So, \(||\vec E||=\frac{12 \ V}{0.20 \ m} \Longrightarrow \boxed{||\vec E||=60 \ \frac{N}{C} }\)

An electric field is created between the plates pointing from positive towards negative. We know that negative charges accelerate opposite the direction of electrical fields. So the electron placed "d" meters away from the negative plate will accelerate towards the positive plate at a constant rate.

For part (b):

We know that...

- the charge of an electron is \(\bold{-1.602 \times10^{-19} \ C}\).

- the mass of an electron is \(\bold{9.11 \times10^{-31} \ kg}\).

- \(\vec F_e=q\vec E\)

- \(\vec F =m\vec a\)

\(\Longrightarrow \vec F_e=(-1.602 \times10^{-19} \ C)(60 \ \frac{N}{C} }) \Longrightarrow \boxed{\vec F_e= -9.612 \times10^{-18} \ N}\)

\(\Longrightarrow \vec F =m\vec a \Longrightarrow \vec a=\frac{\vec F}{m} \Longrightarrow \vec a=\frac{-9.612 \times10^{-18}}{9.11 \times10^{-31} \ kg} \Longrightarrow \boxed{\vec a=-1.06 \times10^{13} \ m/s^2}\)

Kinematic Equation: \(\vec v_f^2=\vec v_0^2+2\vec a \Delta \vec x\)

\(\Longrightarrow 1 \times10^{12} \ m^2/s^2=-2.11 \times10^{13} \ m/s^2 \Delta \vec x \Longrightarrow \Delta \vec x= \frac{1 \times10^{12} \ m^2/s^2}{-2.12 1\times10^{13} \ m/s^2}\)

\(\Longrightarrow \boxed{\Delta \vec x= -0.047 \ m}\)

The distance from the positive plate we'll call, "D."

\(D=0.20+\Delta \vec x\)

\(\Longrightarrow D=0.20+\Delta \vec x \Longrightarrow D=0.20 \ m+(-0.047 \ m) \Longrightarrow \boxed{D=0.153 \ m} \therefore Sol.\)

The value of  the missing resistance, R₃ = 10.35 ohms.

The value of the missing voltages, V₁ = 6 V, V ₃ = 24 V.

The value of the missing currents, I₁ = 3 A, I₃ = 2.32 A.

The values of the missing component of the circuit is calculated by applying the following formula.

The total resistance of the circuit;

For R₂, R₃, 1/R = 1/R₂ + 1/R₃

1/R = 1/12 +  1/R₃

1/R = (R₃ + 1)/(12R₃)

R = 12R₃ / (R₃ + 1)

For, R₁, R₂ and R₃, total resistance;

R = 12R₃ / (R₃ + 1) + R₁

R = [12R₃ / (R₃ + 1)] + 2

R = (12R₃ + 2(R₃ + 1) ) / (R₃ + 1)

R = (12R₃ + 2R₃ + 2 ) / (R₃ + 1)

R = (14R₃ + 2 ) / (R₃ + 1)

The total current in circuit is calculated as;

I = V/R

I = 30 / R

I = ( 30 ) / (14R₃ + 2 ) / (R₃ + 1)

I = (30R₃ + 30) / (14R₃ + 2) ------- (1)

The voltage in parallel circuit is the same

V₂ = V₃ = 24 V

V₃ = IR₃

24 = IR₃

I = 24/R₃  --------- (2)

Solve (1) and (2) together as follows;

24/R₃ = (30R₃ + 30) / (14R₃ + 2)

30R₃² - 306R - 48 = 0

Solve the quadratic equation, using formula method.

R₃ = 10.35 ohms

I₃ = V₃/R₃

I₃ = 24 V / 10.35

I₃ = 2.32 A

If the voltage drop at R₂ and R₃ = 24 V, the voltage drop at R₁ = 30V - 24 V = 6 V

The current in R₁ = V₁/R₁ = 6 V / 2 V = 3 A

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Answer:

Photovoltaic detection is a technique that converts light into electrical energy. It is a process that involves the use of a photovoltaic cell, which is made up of semiconductor materials, to generate an electric current when exposed to light.

The photovoltaic cell absorbs the photons of light, which then knock electrons out of their orbits, creating a flow of electricity. The amount of electricity produced is proportional to the intensity of the light. The photovoltaic cell is commonly used in solar panels to generate electricity from sunlight. The efficiency of the photovoltaic cell is dependent on several factors, including the type of semiconductor material used, the purity of the material, and the thickness of the cell.

The photovoltaic cell has many applications, including in solar power generation, telecommunications, and remote sensing. The technique of photovoltaic detection is an important area of research, as it has the potential to provide a clean and renewable source of energy that can help mitigate climate change.

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Answer:

\(\huge\boxed{Refracted}\)

Explanation:

When a ray of light passes from glass to water, it

1) is Slightly refracted (bending of light)

2) moves away from the normal.

Whenever a light ray travels from a denser medium to a rarer medium, it bends away from the normal.

Answer:

refraction

Explanation:

Answer:

2 amperes

Explanation:

V = IR

V/R = I

120 v / 60 ohms = 2 amps

Please see attachment

Approximately 0.123 kg of liquid water at 26 degrees Celsius would be needed to melt 41 grams of ice.

To calculate the minimum amount of liquid water required to melt 41 grams of ice at 0°C, we need to consider the energy required for the phase change from solid to liquid, which is known as the specific heat of fusion of ice.

The energy required to melt 1 kg of ice is 3.33×105 J/kg.

Therefore, the energy required to melt 41 grams of ice is (3.33×105 J/kg) × (41/1000) kg = 13653 J.

To calculate the amount of liquid water required, we use the specific heat capacity of water, which is 4180 J/kg/°C.

Assuming the initial temperature of water is 26°C, the amount of water needed can be calculated as (13653 J) ÷ (4180 J/kg/°C) ÷ (26°C) = 0.123 kg or approximately 123 ml of water.

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The increase in temperature of the hammer head is determined as 0.034 ⁰C.

The kinetic energy of the hammer is calculated as follows;

K.E = ¹/₂mv²

K.E =0.5 x 1.45 x 6.43²

K.E = 29.98 J

Q = mcΔθ

Δθ = Q/mc

where;

Δθ = (0.0072)/(1.45 x 0.145)

Δθ = 0.034 ⁰C

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Find Kinetic energy

\(\\ \rm\Rrightarrow KE=\dfrac{1{2}mv^2\)

\(\\ \rm\Rrightarrow KE=\dfrac{1}{2}(1.45)(6.43)^2\)

\(\\ \rm\Rrightarrow KE=29.98J=0.0072kcal\)

Now

\(\\ \rm\Rrightarrow Q=mc\Delta T\)

\(\\ \rm\Rrightarrow 0.0072=1.45(0.145)T\)

\(\\ \rm\Rrightarrow \Delta T=0.03°C\)

A force of 1.5 x 10⁴ N must be applied to the small piston to support the weight of the car.

The hydraulic lift works on the principle of Pascal's law, which states that pressure applied to an enclosed fluid is transmitted uniformly throughout the fluid and acts with equal force on all surfaces of the container. Therefore, the pressure applied to the small piston will be transmitted to the large piston and support the weight of the car.

The pressure applied to the fluid is given by the formula:

pressure = force / area

Since the area of both pistons is the same, the pressure applied to the fluid will be the same on both pistons. Therefore, we can write:

pressure = force on small piston / area of small piston = force on large piston / area of large piston

We know the area of both pistons, so we can rearrange this equation to solve for the force on the small piston:

force on small piston = pressure x area of small piston

The force on the large piston is equal to the weight of the car, which is 1.5 x 10⁴ N. The pressure applied to the fluid is the same on both pistons, so we just need to calculate the pressure using the formula:

pressure = force on large piston / area of large piston

pressure = (1.5 x 10⁴ N) / (0.50 m²) = 3.0 x 10⁴Pa

Now we can calculate the force on the small piston using the formula we derived earlier:

force on small piston = pressure x area of small piston

force on small piston = (3.0 x 10⁴Pa) x (0.50 m²) = 1.5 x 10⁴ N

Therefore, a force of 1.5 x 10⁴ N must be applied to the small piston to support the weight of the car.

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The tea absorbs 2613420 J of heat energy when it is placed in sunlight until it reaches an equilibrium temperature of 32.4°C.

To calculate the heat energy absorbed by the tea, we can use the formula:

Q = mcΔT

where Q is the heat energy absorbed by the tea, m is the mass of the tea, c is the specific heat capacity of water, and ΔT is the temperature change of the tea.

Using the given values, we get:

m = 500 g

c = 4186 J/kg·°C

ΔT = 32.4°C - 20°C = 12.4°C

Substituting these values into the formula, we get:

Q = (500 g)(4186 J/kg·°C)(12.4°C) = 2613420 J

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--The complete Question is, A jar of tea with a mass of 500 g is initially at a temperature of 20°C. If the jar is placed in sunlight until it reaches an equilibrium temperature of 32.4°C, how much heat energy is absorbed by the tea? Assume the specific heat capacity of the tea to be that of pure liquid water, which is 4186 J/kg·°C.--

Answer:

The acceleration of the car is 10.16m/s²

Explanation:

Given parameters:

  Initial velocity = 8.77m/s

   Final velocity = 47.8m/s

   Time duration  = 3.84s

Unknown:

Acceleration of the car = ?

Solution:

To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;

     Acceleration  = \(\frac{V - U}{T}\)

V is the final velocity

U is the initial velocity

T is the time taken

  Input the parameters and solve for acceleration;

      Acceleration  = \(\frac{47.8 - 8.77}{3.84}\)   = 10.16m/s²

The acceleration of the car is 10.16m/s²

The formula for calculating the speed of a wave is expressed as

Speed = frequency x wavelength

From the information given,

frequency = 0.25

wavelength = 6

By substituting these values into the formula,

Speed = 0.25 x 6

Speed = 1.5 m/s

Answer:

A) C_{eq} = 15 10⁻⁶  F,  B)   U₃ = 3 J,  U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

         \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}\)            (you has an mistake in the formula)

         \(\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}\)

         \(\frac{1}{C_{eq1}}\) = 0.1   10⁶

         \(C_{eq1}\) = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

          \(\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}\)

          \(\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6\)

          \(\frac{1}{C_{eq2} }\) = 0.2 10⁶

          \(C_{eq2}\) = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

          C_{eq} = C_{eq1} + C_{eq2}

          C_{eq} = (10 + 5) 10⁻⁶

          C_{eq} = 15 10⁻⁶  F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

         Q = C_{eq} V

         Q = 15 10⁻⁶ 6

         Q = 90 10⁻⁶ C

the charge on each branch is

         Q₁ = Ceq1 V

         Q₁ = 10 10⁻⁶ 6

          Q₁ = 60 10⁻⁶ C

         Q₂ = C_{eq2} V

         Q₂ = 5 10⁻⁶ 6

         Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant    Q = Q₁ = Q₃

          U₃ = \(\frac{Q^2}{2 C_3}\)

          U₃ =\(\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}\)

          U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

          U₄ = \(\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}\)

          U₄ = 0.5 J

Answer:

W' = 1.66 x 10¹⁴ N

Explanation:

First, we will calculate the mass:

\(W = mg\)

where,

W = weight on earth = 690 N

m = mass = ?

g = acceleration due to gravity on earth = 9.8 m/s²

Therefore,

\(m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg\)

Now, we will calculate the value of g on the neutron star:

\(g' = \frac{GM}{R^2}\)

where,

g' = acceleration due to gravity on the surface of the neutron star = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of the Neutron Star = 1.99 x 10³⁰ kg

R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m

Therefore,

\(g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2\)

Therefore, the weight on the surface of the neutron star will be:

\(W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)\)

W' = 1.66 x 10¹⁴ N

Answer:

s = 13.6 m

Explanation:

Ann is driving at a constant speed. Hence, the formula of uniform motion shall be used here. The formula for the distance traveled during uniform speed motion is given as follows:

s = vt

where,

s = distance covered by the car before it begins to slow down= ?

v = uniform speed of car = (63 km/h)(1000 m/ 1 km)(1 h/3600 s) = 17.5 m/s

t = time to react before applying brakes = 0.777 s

Therefore, using the values in the equation, we get:

s = (17.5 m/s)(0.777 s)

s = 13.6 m

The magnitude of the initial acceleration of the object is 4.2 m/s².

The tension in the string once the object starts moving is 13.65 N.

The magnitude of the initial acceleration of the object is calculated by applying Newton's second law of motion as follows;

F(net) = ma

m₂g - μm₁g cosθ = a(m₁ + m₂)

where;

(4.75 x 9.8) - (0.535 x 3.25 x 9.8 x cos40) = a(3.25 + 4.75)

33.5 = 8a

a = 33.5/8

a = 4.2 m/s²

The tension in the string once the object starts moving is calculated as;

T = m₁a

T = 3.25 x 4.2

T = 13.65 N

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Answer:

These all different sources of energy add to the store of electrical power that is then sent out to different locations via high powered lines. It is the energy from the sun that is harnessed using a range of technologies such as solar heating, solar architecture, photovoltaics, and artificial photosynthesis.

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Answer:206.3

Explanation:

Answer:

4.3 m/s^2

Explanation:

V=0

\(V_{f}\) = 26.41 m/s

t= 6.2 s

a= ?

\(V_{f}\) = \(V_{i}\) + at

26.41 m/s = 0+a (6.2s)

a= \(4.3 m/s^{2}\)

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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The engine weighs four times as much as a freight car. Therefore, the final velocity following connection is 4 km/h.

v′=m1v1+m2v2m1+m2 m1 is the weight of item 1, v1 is indeed the velocity of the object of item 1, m2 is indeed the mass of argument 2, and v2 is the starting velocity of instrument 2 wherein v' is the final speed of a two objects after they travel as one mass after the collision.

The velocity of the special properties in a head-on object with a projectile that is significantly more massive than target the projectile's speed before and after the contact will be roughly equal, and the projectile's speed will practically remain unaltered.

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Answer:

Approximately \(3.967\; {\rm kg\cdot m\cdot s^{-1}}\).

Explanation:

Convert velocity to the standard units (meters per second):

\(\begin{aligned}v &= 246.2 \; {\rm km \cdot h^{-1}} \\ &= 246.2 \; {\rm km \cdot h^{-1}}\times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &\approx 68.389\; {\rm m\cdot s^{-1}}\end{aligned}\).

Convert mass to standard units (kilograms):

\(\begin{aligned} m &= 58\; {\rm g} \\ &= 58\; {\rm g} \times\frac{1\; {\rm kg}}{1000\; {\rm g}}\\ &= 0.058\; {\rm kg}\end{aligned}\).

When an object of mass \(m\) travels at a velocity of \(v\), momentum of that object would be \(p = m\, v\). In standard units, the momentum of this tennis ball would be:

\(\begin{aligned}p &= m\, v \\ &\approx (0.058\; {\rm kg})\, (68.389\; {\rm m\cdot s^{-1}}) \\ &\approx 3.967\; {\rm kg \cdot m\cdot s^{-1}}\end{aligned}\).