A raft is made of 14 logs lashed together. Each log is 42 cm in diameter and a length of 6.4 m. 42% of the log volume is above the water when no one is on the raft. Determine the following: the specific gravity of the logs.

Final answer:

For the return flight, the bird's average velocity is 6.38 m/s, while for the whole episode, the average velocity is 0 m/s.

Explanation:

a) To find the bird's average velocity for the return flight, we need to calculate the displacement and divide it by the time taken. The displacement is the distance between the release point and the nest, which is 5220 km. The time taken is 13.3 days, which can be converted to seconds by multiplying by 24 (hours in a day) and 60 (minutes in an hour). Therefore, the average velocity is:

Average velocity = displacement / time = 5220 km / (13.3 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute) = 6.38 m/s

b) To find the bird's average velocity for the whole episode, we need to calculate the total displacement and divide it by the total time taken. The total displacement is 0 km, as the bird returns to its nest. The total time taken is the time taken for the return flight plus the time taken for the outward flight, which is 2 * 13.3 days. Therefore, the average velocity is:

Average velocity = total displacement / total time = 0 km / (2 * 13.3 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute) = 0 m/s

Answer:

The initial charges on the spheres are  [tex]6.796\ 10^{-6}\ c[/tex] and [tex]-3.898\ 10^{-6}\ c[/tex]

Explanation:

Electrostatic Force

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

[tex]\displaystyle F=\frac{K\ q_1\ q_2}{d^2}[/tex]

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

[tex]\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537[/tex]

Rearranging

[tex]\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}[/tex]

Solving for q1.q2

[tex]\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1][/tex]

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

[tex]\displaystyle q=\frac{q_1+q_2}{2}[/tex]

Using this second condition:

[tex]\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756[/tex]

[tex]\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C[/tex]

Solving for q2

[tex]\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1[/tex]

Replacing in [1]

[tex]\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}[/tex]

Rearranging, we have a second-degree equation for q1.  

[tex]\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0[/tex]

Solving, we have two possible solutions

[tex]\displaystyle q_1=6,796.10^{-6}\ c[/tex]

[tex]\displaystyle q_1=-3.898.10^{-6}\ c[/tex]

Which yields to two solutions for q2

[tex]\displaystyle q_2=-3.898.10^{-6}\ c[/tex]

[tex]\displaystyle q_2=6.796.10^{-6}\ c[/tex]

Regardless of their order, the initial charges on the spheres are [tex]6.796\ 10^{-6}\ c[/tex] and [tex]-3.898\ 10^{-6}\ c[/tex]

To convert the equatorial diameter of the Moon from kilometers to miles, you can use the given conversion factor:

Equatorial diameter in miles = Equatorial diameter in kilometers × Conversion factor

Given:

Equatorial diameter of the Moon = 3476 kilometers

Conversion factor = 0.6214 miles/kilometer

Now, plug in the values:

Equatorial diameter in miles = 3476 km × 0.6214 miles/km

[tex]\[ \text{Equatorial diameter in miles} \approx 2160.9264 \, \text{miles} \][/tex]

So, the Moon's equatorial diameter is approximately 2160.93 miles.

Final answer:

To convert the Moon's diameter from 3476 kilometers to miles, multiply by the conversion factor of 0.6214 miles per kilometer, resulting in approximately 2160.6344 miles.

Explanation:

The question asks for the conversion of the Moon's diameter from kilometers to miles. The given diameter of the Moon is 3476 kilometers. To convert kilometers to miles, we use the provided conversion factor where 1 kilometer equals 0.6214 miles.

Here's how you can calculate the Moon's diameter in miles:

1. Multiply the Moon's diameter in kilometers by the conversion factor:
3476 kilometers × 0.6214 miles/kilometer

2. Calculating this gives:

2160.6344 miles
So, the Moon's diameter in miles is approximately 2160.6344.

Answer:

A net force of 11 kg is needed to produce an acceleration of 7.0 m/s²

Explanation:

Newton's Second Law: It states that the the rate of change of momentum of a body, is directly proportional to the applied force, and takes place along the direction of the the force.

From Newton's second law of motion, we can deduced that,

F = ma ......................... Equation 1.

Where F = Net force acting on the package, m = mass of the package, a = acceleration of the page.

From the question, when

F = 77 N, m = 11 kg.

a = F/m

a = 77/11

a = 7 m/s².

From the above, a net force of 11 kg is needed to produce an acceleration of 7.0 m/s²

Final answer:

To determine the net force needed to accelerate an 11-kg package at 7.0 m/s², apply Newton's second law using the formula Fnet = ma, which yields a net force of 77 N.

Explanation:

The student is asking how to use Newton's second law to calculate the net force required to produce a certain acceleration for an object of known mass. This is a physics problem that involves using the formula Fnet = ma (where Fnet is the net force, m is the mass, and a is the acceleration).

In this case, the mass m of the package is 11 kg, and the desired acceleration a is 7.0 m/s². To find the net force Fnet, we rearrange the formula as it is already in the form solving for the net force and simply substitute the known values:

Fnet = (11 kg) × (7.0 m/s²)

Now, we multiply:

Fnet = 77 N

Therefore, a net force of 77 N is indeed needed to accelerate an 11-kg package at 7.0 m/s².

Answer:

10.84406 Nm/rad

0.068625 kgm²

2.00066 rad/s

0.49983 s

Explanation:

F = Force = 4.29 N

R = Radius = [tex]\dfrac{30}{2}=15\ cm[/tex]

[tex]\theta[/tex] = Angle = [tex]3.4\ ^{\circ}[/tex]

m = Mass of disk = 6.1 kg

Torsional constant is given by

[tex]J=\dfrac{\tau}{\theta}\\\Rightarrow J=\dfrac{FR}{\theta}\\\Rightarrow J=\dfrac{4.29\times 0.15}{3.4\times \dfrac{\pi}{180}}\\\Rightarrow J=10.84406\ Nm/rad[/tex]

The torsion constant is 10.84406 Nm/rad

Moment of inertia is given by

[tex]I=\dfrac{1}{2}mr^2\\\Rightarrow I=\dfrac{1}{2}6.1\times 0.15^2\\\Rightarrow I=0.068625\ kgm^2[/tex]

The moment of inertia is 0.068625 kgm²

Frequency is given by

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{J}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{10.84406}{0.068625}}\\\Rightarrow f=2.00066\ rad/s[/tex]

The frequency is 2.00066 rad/s

Time period is given by

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{2.00066}\\\Rightarrow T=0.49983\ s[/tex]

The time period is 0.49983 s

Answer:

d. 500 m/s

Explanation:

Momentum: This is the product of the mass of a body to its velocity, The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv....................... equation 1

Where M = momentum, m = mass, v = velocity

deduced from the question,

Momentum of the car = momentum of the barrier.

MV = mv ............................. Equation 1

Where M = mass of the car, V = velocity of the car, m = mass of the barrier, v = velocity of the barrier.

making v the subject of the equation,

v = MV/m........................ Equation 2

Given: M = 1000 kg, V = 10 m/s, m = 20 kg.

Substitute into equation 2

v = 1000(10)/20

v = 500 m/s.

Hence the speed of the barrier = 500 m/s

The right option is d. 500 m/s

To warm the cold air inhaled during a breath to body temperature requires about 35.05J of energy or heat. This results in a total loss of around 44.16 KJ of heat due to respiration per hour.

This is a question about the heat exchange between the human body and the surrounding air when breathing, in cold weather conditions. We can solve it using the concepts of heat transfer and specific heat capacity.

A. To find the amount of heat necessary to warm the cold air to body temperature, we first need to determine the change in temperature. This would be (37-(-19)) = 56°C. The mass of the air breathed in with each breath can be calculated as 0.470L * 1.3g/L = 0.611g. Converting this to kg, we get 0.000611 Kg. The quantity of heat, Q, can be found using the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, we get Q = 0.000611 Kg * 1020 J/Kg°C * 56°C = 35.05J.

B. The total amount of heat lost per hour due to respiration can be calculated by first finding the heat lost per minute: Q per minute = Q per breath * respiration rate = 35.05J/breath * 21 breaths/minute = 736.05 J/minute. Converting this to an hourly rate gives us 736.05 J/minute * 60 minutes/hour = 44,163J/hour or about 44.16 KJ/hour. So, about 44.16 KJ of heat is lost via respiration each hour.

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The amount of heat needed to warm to internal body temperature the air exchanged with each breath on a cold day is 74.3 J per breath. The total heat loss per hour for a respiration rate of 21.0 breaths per minute is approximately 93.6 kJ/hour.

To calculate the amount of heat needed to warm the air inhaled during breathing to the internal body temperature, we can use the formula:

Q = mcΔT

Where:

The mass of 1.0 L of air at 1.3 g/L is 0.0013 kg. The change in temperature needed to warm air from -19°C to 37°C is 56°C or 56 K since the size of 1 degree on both scales is the same. Thus:

Q = 0.0013 kg * 1020 J/kg·K * 56 K

Q = 74.3 J per breath.

To find the heat lost per hour at a respiration rate of 21.0 breaths per minute, we do:

Heat loss per hour = Q * number of breaths per hour = 74.3 J * (21 breaths/min * 60 min/hour)

Heat loss per hour = 74.3 J * 1260 breaths/hour = 93618 J/hour, or approximately 93.6 kJ/hour.

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The definition of volumetric charge density is given as the ratio between the load per unit volume, while the linear load is the same ratio of the load but per unit length. Applying these concepts then we have that the volumetric density of charge is,

Here,

q = Charge

V = Volume

Replacing we have,

[tex]\gamma = \frac{0.51}{\pi r^2 l}[/tex]

[tex]\gamma = \frac{0.51}{\pi (4*10^{-3})^2(0.33)}[/tex]

[tex]\gamma =30475.84C/m^3[/tex]

And the linear charge density is

[tex]\rho = \frac{q}{l}[/tex]

[tex]\rho = \frac{0.51}{0.33}[/tex]

[tex]\rho = 1.54C/m[/tex]

To determine the volume and linear charge densities, divide the total charge by the rod's volume and length, respectively, after calculating the volume using the formula for a cylinder.

The volume charge density (ρ) and the linear charge density (λ) are physical quantities that represent the distribution of electric charge in a material. To find the volume charge density, we divide the total charge (Q) by the volume (V) of the rod. The formula is ρ = Q/V. The linear charge density is found by dividing the charge by the length (L) of the rod, using the formula λ = Q/L.

Given that the charge Q is 0.51 C, distributed uniformly along a cylindrical rod with a length of 33 cm (or 0.33 m) and a radius of 4 mm (or 0.004 m), we first need to calculate the volume of the cylinder using the formula V = πr²h, where r is the radius and h is the height (length) of the cylinder. The volume V is then π ×(0.004 m)² ×0.33 m.

After computing the volume, we calculate ρ as ρ = 0.51 C / V. To find λ, we simply divide the charge by the length, λ = 0.51 C / 0.33 m. These computations will yield the volume and linear charge densities for the rod.

Answer:

[tex]L_f=26.8108 ft[/tex]

Part B:

For Final Reduction

[tex]v_f=48.5436ft/min[/tex]

Explanation:

Part A:

At each step 0.8 (100-20)% of thickness is left

Final Thickness t_f:

[tex]t_f=(0.80)(0.80)(0.80)*3\\t_f=1.536 in[/tex]

Width increases by 0.03 in each step so (100+3)%=1.03

Final Width w_f:

[tex]w_f=(1.03)(1.03)(1.03)*12\\w_f=13.1127 in[/tex]

Conservation of volume:

[tex]t_ow_oL_o=t_fw_fL_f\\L_f=\frac{t_ow_oL_o}{t_fw_f} \\L_f=\frac{3*12*(15*12)}{1.536*13.1127}\\L_f=321.730 in\\L_f=26.8108 ft[/tex]

Part B:

[tex]t_ow_ov_o=t_fw_fv_f[/tex]

At First reduction exit Velocity:

[tex]v_f=\frac{t_ow_oL_o}{t_fw_f} \\v_f=\frac{3*12*(40)}{0.8*3*1.03*12}\\v_f=48.5439ft/min[/tex]

At 2nd Reduction:

[tex]v_f=\frac{0.8*3*1.03*12*40}{0.8^2*3*1.03^2*12}\\v_f=48.5436ft/min[/tex]

For Final Reduction:

[tex]v_f=\frac{0.8^2*3*1.03^2*12*40}{0.8^3*3*1.03^3*12}\\v_f=48.5436ft/min[/tex]

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g=[tex]=\frac{450}{1000}=0.45 kg[/tex]

Where 1 kg=1000 g

Frequency of oscillation=[tex]\nu=1.2Hz[/tex]

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=[tex]E=\frac{1}{2}m\omega^2A^2[/tex]

Where [tex]\omega=2\pi\nu[/tex]=Angular frequency

A=Amplitude

[tex]\pi=\frac{22}{7}[/tex]

Using the formula

[tex]0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2[/tex]

[tex]A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398[/tex]

[tex]A=\sqrt{0.0398}=0.199[/tex]

Hence, the amplitude of oscillation=A=0.199

The amplitude of oscillation will be "0.199".

Given:

Mass of spring,

or,

           = 0.45 kg

Frequency,

Total energy,

As we know the formula,

→ [tex]E = \frac{1}{2} m \omega^2 A^2[/tex]

By putting the values, we get

→ [tex]0.51= \frac{1}{2}\times 045 (2\times \frac{22}{7}\times 1.2 )^2 A^2[/tex]

→   [tex]A^2 = \frac{2\times 0.51}{0.45(2\times \frac{22}{7}\times 1.2 )^2}[/tex]

          [tex]= 0.0398[/tex]

→     [tex]A = \sqrt{0.0398}[/tex]

          [tex]= 0.199[/tex]

Thus the above response is right.

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The can of soft drink in a refrigerator is modelled as a closed system because while heat energy is exchanged with its surroundings, no matter is exchanged. An open system, in contrast, allows both energy and matter to be exchanged.

In the scenario where a can of soft drink is placed into a refrigerator to cool, you would model the can of soft drink as a closed system. A closed system is one in which energy can be exchanged with its surroundings, but matter cannot. In this case, the can of soft drink is transferring heat energy to its surroundings, i.e., the refrigerator, until equilibrium is achieved, but the soda itself remains within the can - no matter is exchanged.

An open system, on the other hand, allows both energy and matter to be exchanged with its environment. The dissolving of CO2 in soft drinks, as mentioned in the reference, is an example of an open system, where matter (CO2) is able to escape from the soft drink when the can or bottle is opened.

It's important to note that while practical real-world systems are ultimately open due to unavoidable interactions with the surroundings, we often model systems as closed (or even isolated) in order to simplify the analysis.

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The correct answer is that the can of soft drink would be modeled as a closed system.

In thermodynamics, a closed system is defined as a system that can exchange energy (such as heat) with its surroundings, but not matter. This means that while the can of soft drink can absorb or release heat to the refrigerator (its surroundings), no matter (like the soft drink itself) is exchanged between the can and the refrigerator.

Therefore, when the can of soft drink is placed in the refrigerator to cool, it is appropriately modeled as a closed system because only energy in the form of heat is transferred between the can and the refrigerator, and there is no transfer of matter.

Answer:

False

Explanation:

There is less pressure higher in the atmosphere, which means that air will expand, and thus cool

Answer:

The answer is zero. The total change in momentum is equal to the sum of the areas within the time interview 0-4s. The area under the graph is Force x time (F being the breadth and time the length)

The sun of the areas is zero.

Explanation:

See the attachment below for the full solution.

Thank you for reading this post. I hope it is helpful to you.

Answer:

a) 88.1 rad/s

b) 0.286

Explanation:

given information:

diameter, d = 8 cm = 0.08 m

sphere's mass, m = 400 g = 0.4 kg

the distance from rest to the tip, h = 2.1 m

incline angle, θ = 25°

a) What is the sphere's angular velocity at the bottom of the incline?

mg(h sinθ) = 1/2 Iω² + 1/2mv²

I of solid sphere = 2/5 mr², thus

mg(h sinθ) = 1/2 (2/5 mr²) ω² + 1/2 mv², now we can remove the mass

g h sin θ = 1/5 r² ω² + 1/2 v²

ω = v/r, v = ωr

so,

g h sin θ = 1/5 r² ω² + 1/2 (ωr)²

g h sin θ = (7/10) r² ω²

ω² = 10 g h sin θ/7 r²

ω = √10 g h sin θ/7 r²

   = √10 (9.8) (2.1) sin 25° / 7 (0.04)²

   = 88.1 rad/s

b) What fraction of its kinetic energy(KE) is rotational?

fraction of its kinetic energy = rotational KE / total KE

total KE = total potential energy

             = m g h sin θ

             = 0.4 x 9.8 x 2.1 sin 25°

             = 3.48 J

rotational KE = 1/2 Iω²

                      = 1/5 mr²ω²

                      = 1/5 0.4 (0.04)²(88.1)²

                      = 0.99

fraction of its KE = 0.99/3.48

                            = 0.286

A) The sphere's angular velocity at the bottom of the incline is; ω = 88.1 rad/s

B) Fraction of its kinetic energy that is rotational is; 0.286

We are given;

Diameter; d = 8 cm = 0.08 m

Mass of sphere; m = 400 g = 0.4 kg

Distance from rest to the tip; h = 2.1 m

Angle of inclination; θ = 25°

a) To get the sphere's angular velocity at the bottom of the incline, we will use the expression;

mg(h*sinθ) = ¹/₂Iω² + ¹/₂mv²

where;

I of solid sphere = ²/₅mr²

Thus;

mg(h*sinθ) = ¹/₂(²/₅mr²)ω² + ¹/₂mv²

The mass m will cancel out to give;

gh*sin θ = ¹/₅r²ω² + ¹/₂v²

where v = ωr

Thus;

gh*sin θ = ¹/₅r²ω² + ¹/₂r²ω²

gh*sin θ = ⁷/₁₀r²ω²

ω = √[(¹⁰/₇)*g*h*(sin θ)/r²]

ω = √[(¹⁰/₇)*9.8*2.1*(sin 25)/(0.04)²]

ω = 88.1 rad/s

b) Fraction of its kinetic energy that is rotational = rotational KE/total KE

But, total KE = total potential energy

Thus;

KE_tot = mgh*sin θ

KE_tot = 0.4 * 9.8 * 2.1 sin 25°

KE_tot = 3.48 J

KE_rot = ¹/₂Iω²

I of solid sphere = ²/₅mr². Thus;

KE_rot = ¹/₅mr²ω²

KE_rot = ¹/₅ * 0.4 * 0.04² * 88.1²

KE_rot = 0.99 J

Fraction of its kinetic energy that is rotational = 0.99/3.48 = 0.286

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The extension in the rubber band for the restoring force of 3 newtons is 1.5 cm.

The force according to Hooke's law is given as:

[tex]F=kx[/tex]

Here F is the restoring force, k is the proportionality constant and x is the stretch or compression.

Given:

Restoring force, [tex]F= 2\ N[/tex]

Stretch in the band, [tex]x=1\ cm[/tex]

The value of the k is computed as:

[tex]F=kx\\2=k \times 1cm\\k= 2\ N/cm[/tex]

The stretch in the band for a restoring force of 3 N is computed as:

[tex]F=kx\\x=\frac{F}{k}\\x= \fract{3}{2}\\x = 1.5 cm[/tex]

Therefore, the extension in the rubber band for the restoring force of 3 newtons is 1.5 cm.

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According to Hooke's Law, a force of 3 newtons will stretch the rubber band 1.5 centimeters.

To answer this question, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, we are given that when a rubber band is stretched 1 cm beyond its natural length, it exerts a restoring force of 2 newtons. This means that the force constant of the rubber band is 2 newtons per centimeter (N/cm).

To find out how far a force of 3 newtons will stretch the rubber band, we can use the formula for Hooke's Law: F = kx, where F is the force, k is the force constant, and x is the displacement. Rearranging the formula, we have x = F/k. Plugging in the values, we get:

x = 3 N / (2 N/cm) = 1.5 cm

Therefore, a force of 3 newtons will stretch the rubber band 1.5 centimeters.

Answer:

0.26315 s

Explanation:

The frequency of the ball tied to a string system is 3.8 rev/s.

That means in one second the ball will complete 3.8 revolutions.

The time period will be the reciprocal of this frequency

[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{3.8}\\\Rightarrow T=0.26315\ s[/tex]

The time period is 0.26315 s

It can be also solved in the following way

[tex]1\ s=3.8\ rev\\\Rightarrow 1\ rev=\dfrac{1}{3.8}\ s\\\Rightarrow 1\ rev=0.26315\ s[/tex]

The time period is 0.26315 s

Answer:

At twice the distance, the electric potential is V/2.

Explanation:

The electric potential at a certain distance d from a point charge q can be represented by V:

V = kq/d .....1

Where q = charge ,d = distance between them, k = coulomb's constant.

When the distance is doubled d1 = 2d

Let V1 represent the electric potential after the distance have been doubled.

V1 = kq/2d = (kq/d)/2

V1 = V/2

Therefore, at twice the distance the electric potential is halved V1 = V/2

The electric potential (V) decreases as you increase your distance from the point charge. Therefore, at twice the distance from a charge, the electric potential will be halved, becoming V/2.

The electric potential (V) at a certain distance from a point charge, according to Coulomb's Law, is directly related to the charge and inversely related to the distance from it. Therefore, if we double the distance, the electric potential will be half as much. So, the electric potential at twice the distance from the point charge is V/2.

Think of it like this, if you're twice as far from the point charge, the effect of the charge (which in this case is represented by the electric potential) is lessened.

Therefore, the potential decreases as you increase your distance from the charge. Hence, at twice the distance from a charge, the electric potential will be halved, i.e., V/2.

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Answer: Pπr2

Explanation: Force is any interaction that affects an object,that if not opposed or treated will cause some changes in the object. Pressure is directly proportional to force,which means an increase in force will create a corresponding increase in the pressure of a system.

The force on the cloth will be The PRESSURE EXERTED ON THE CLOT (P) *THE PI( which is the RATIO of the circumference of the cylinder i.e3.14159) *the square of the ratio. Which is Pπr2.

The force exerted on the artery clot can be determined by the formula F = P x A, where F is the force, P is the pressure and A is the area. In the case of a cylindrical clot, A = πr². Therefore, the force F = P x πr².

The force exerted on the clot by the pressure in the artery can be determined by using the formula for the force exerted by a pressure on a surface, which is Force (F)= pressure (P) x Area (A). The pressure is given as P. The area of the surface the pressure is exerted upon can be found by considering the cross-sectional area of the cylinder (clot), that can be expressed as the product of pi and the square of the radius (r), or πr².

Therefore, the force on the clot can be calculated by substituting the cross-sectional area (A = πr²) into the equation F = P x A, giving us F = P x πr². In this way, we can understand how the pressure in the artery results in a force on the clot.

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Answer:

h= 32059.37 m

Explanation:

Assuming the missing in formation as

A certain lead-acid storage battery has a mass of 33 kg . Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.

Now, Applying energy conservation ( Electrical to potential)

Electrical Energy E= I×V×t

I = correct , V= voltage , t= time of flow of current

E = 6×24×20×60×60.

E= 10368 KJ

Now this energy is used to lift the battery with 100% efficiency

Hence,

electrical energy E= potential energy P

P= mgh

m=mass of the battery , g= the acceleration due to gravity is 9.8 m/s^2

h= height

mgh = 10368 kJ

33×9.8×h= 10368×1000

h = 10368×1000/(33×9.8)

h= 32059.37 m

Answer:

The right leg (oil on top) is higher

Explanation:

Given:

The mass density of oil is lesser than the mass density of water.

This is justified by the equation:

[tex]P=\rho.g.h[/tex]

where:

[tex]\rho =[/tex] density of the liquid

[tex]g=[/tex] acceleration due to gravity

[tex]h=[/tex] height of the liquid column

Answer:

Remains the same

Explanation:

The speed of waves of higher and lower frequency both will be same.

the speed of sound in a medium is constant  and independent of it's frequency. Moreover, when the frequency changes wavelength changes accordingly, such that their product remains constant.

we know that

υ×λ = constant = velocity

υ= frequency

λ= wavelength.

Final answer:

The work done when a 185g tomato is lifted 15.0m is 27.261 J. If 82.1% of this work is converted to kinetic energy, the velocity of the tomato when it hits the ground is approximately 15.54 m/s.

Explanation:

To determine the work done when lifting a 185g tomato 15.0m, we use the formula for gravitational potential energy (GPE), which is equivalent to the work done against gravity.

Work (W) = mass (m) × gravitational acceleration (g) × height (h)
m = 185 g = 0.185 kg (since 1g = 0.001 kg)
g = 9.8 m/s²
h = 15.0 m

W = 0.185 kg × 9.8 m/s² × 15.0 m
W = 27.261 J

To determine the velocity (v) when the tomato hits the ground, assuming 82.1% of the work done is converted to kinetic energy (KE), we calculate:

KE = 0.821 × W
KE = 0.821 × 27.261 J
KE = 22.379 J

The formula for KE is:

KE = 1/2 m v²
Solving for v gives:

v = √(2 × KE / m)
v = √(2 × 22.379 J / 0.185 kg)
v ≈ 15.54 m/s

Thus, the velocity of the tomato when it hits the ground, assuming 82.1% conversion of energy, is approximately 15.54 m/s.

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy  = 1 / 7

λy / λx = 7 / 1

The ratio of wavelengths is λ(y)/λ(x) = 1/49.

The energy of a wave is directly proportional to the square of the amplitude of the wave.

   E ∝ [tex]A^{2}[/tex] , where E is the energy of the wave and A is the apmlitude

We also know that enegy

E = hc/λ

E ∝ 1/λ

According to the question:

Let wave X has energy [tex]E_x[/tex] and amplitude [tex]A_x[/tex]

and wave Y has energy [tex]E_y[/tex] and amplitude [tex]A_y[/tex]

[tex]\frac{E_x}{E_y}=\frac{A_x^2}{A_y^2} \\\\\frac{E_x}{E_y}= \frac{(A_y/7)^2}{A_y}\\\\\frac{E_x}{E_y}=\frac{1}{49}[/tex] since it is given that [tex]A_x=\frac{1}{7}A_y[/tex]

{1/λ(x)} / {1/λ(y)} = 1/49

λ(y)/λ(x) = 1/49 is the ratio of the wavelengths.

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Answer:

[tex]K=\frac{GmM}{2R}[/tex]

Explanation:

The kinetic energy is defined as:

[tex]K=\frac{mv^2}{2}(1)[/tex]

Here, m is the object's mass and v its speed. In this case the speed of the satellite is the orbital speed, which is given by:

[tex]v_{orb}=\sqrt\frac{GM}{R}(2)[/tex]

Here, G is the gravitational constant, M is the mass of the object that the satellite is orbiting and R is the radius of its circular orbit. Replacing (2) in (1):

[tex]K=\frac{mv_{orb}^2}{2}\\K=\frac{m(\sqrt\frac{GM}{R})^2}{2}\\K=\frac{GmM}{2R}[/tex]

The kinetic energy of a satellite in a circular orbit can be found using the equation K=GMm/2r, involving the gravitational constant, mass of the satellite, mass of the body being orbited, and the radius of the orbit. This kinetic energy is half the potential energy and equal to the total energy of the satellite.

The kinetic energy K of a satellite with mass m in a circular orbit with radius R can be defined using the equation of the kinetic energy K = 1/2 * mv², where m is the mass of the satellite and v is its speed. However, in the context of gravitational forces, we must consider that the gravitational force provides the centripetal force necessary for the satellite to maintain its orbit with speed v. Therefore, we have GMm/r² = mv²/r where G is the gravitational constant and M is the mass of the body the satellite is orbiting.

When we solve for the speed v, we find that v=sqrt(GM/R), which leads us to an equation for the kinetic energy of K = GMm/2r. These equations demonstrate that the kinetic energy of the satellite is dependent not only on its own mass and speed, but also on the mass of the body it is orbiting and the radius of its orbit.

The concept that the kinetic energy of a satellite in circular orbit is half the magnitude of the potential energy, and the same as the magnitude of the total energy, is an important aspect of understanding these calculations. We also note that the gravitational constant G is by far the least well determined of all fundamental constants in physics.

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The car's speed at the top of the loop is approximately 0.41 m/s.

At the top of the loop-the-loop, the normal force equals the magnitude of the gravitational force, which means the net force acting on the roller coaster car is zero.

This condition occurs when the car is just about to lose contact with the track due to insufficient normal force.

Using the centripetal force formula, [tex]\( F_{\text{net}} = \frac{mv^2}{r} \)[/tex], where \( m \) is the mass of the car, v is its speed, and \( r \) is the radius of the loop.

At the top of the loop, the net force is zero, so we equate the gravitational force and the centripetal force:

[tex]\[ mg = \frac{mv^2}{r} \][/tex]

Solving for v, we get:

[tex]\[ v = \sqrt{gr} \][/tex]

Substituting the given radius [tex]\( r = \frac{33 \, \text{mm}}{2} = 0.0165 \, \text{m} \)[/tex] and the acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex], we find:

[tex]\[ v = \sqrt{(9.8 \, \text{m/s}^2)(0.0165 \, \text{m})} \approx 0.41 \, \text{m/s} \][/tex]

Therefore, the car's speed at the top of the loop is approximately 0.41 m/s.

Answer:

423m/s

Explanation:

Suppose after the impact, the bullet-block system swings upward a vertical distance of 0.4 m. That's means their kinetic energy is converted to potential energy:

[tex]E_p = E_k[/tex]

[tex]mgh = mv^2/2[/tex]

where m is the total mass and h is the vertical distance traveled, v is the velocity right after the impact at, which we can solve by divide both sides my m

Let g = 9.81 m/s2

[tex]gh = v^2/2[/tex]

[tex]v^2 = 2gh = 2 * 9.81* 0.4 = 7.848[/tex]

[tex]v = \sqrt{7.848} = 2.8m/s[/tex]

According the law of momentum conservation, momentum before and after the impact must be the same

[tex]m_uv_u + m_ov_o = (m_u + m_o)v[/tex]

where [tex]m_u = 0.01, v_u[/tex] are the mass and velocity of the bullet before the impact, respectively.[tex]m_ov_o[/tex] are the mass and velocity of the block before the impact, respectively, which is 0 because the block was stationary before the impact

[tex]0.01v_u + 0 = (0.01 + 1.5)*2.8[/tex]

[tex]0.01v_u = 4.23[/tex]

[tex]v_u = 4.23 / 0.01 = 423 m/s[/tex]

The initial velocity of the bullet just before it struck the block was 422 m/s.

To determine the velocity of a bullet just before it strikes a block of wood and causes a ballistic pendulum motion, we follow these steps:

[tex]p_{initial}[/tex] = [tex]P_{final}[/tex]
(0.01 kg)u = (1.51 kg)(2.8 m/s)
Simplifying this:
u = 1.51 kg * 2.8 m/s / 0.01 kg
u = 422 m/s

Thus, the initial velocity of the bullet just before it struck the block was 422 m/s.

Answer:

0.4rad/s²

Explanation:

Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)

w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m

a = ∆w/t

a = (w2 - w1)/t

a = (2 - 4)/5 = -2/5 =

a = - 0.4rad/s²

The -ve sign indicates a deceleration in the motion

Good luck

The void ratio of the soil is 1.2045, the specific gravity of solids is 2.6329, and the saturated unit weight is 82.7586 lb/ft3.

The void ratio of the soil can be determined using the formula:

e = (1 + w) / (1 - w)

where e is the void ratio and w is the moisture content. Plugging in the values, we get:

e = (1 + 0.17) / (1 - 0.17) = 1.2045

To calculate the specific gravity of solids, we can use the formula:

Gs = (1 + e) * (1 - S) / (1 - e * S)

where Gs is the specific gravity of solids and S is the degree of saturation. Substituting the given values:

Gs = (1 + 1.2045) * (1 - 0.6) / (1 - 1.2045 * 0.6) = 2.6329

The saturated unit weight can be found using the equation:

Gamma_sat = Gamma_dry / (1 + w)

where Gamma_sat is the saturated unit weight and Gamma_dry is the dry unit weight. Given that the unit weight of the soil is 96 lb/ft3, we have:

Gamma_sat = 96 / (1 + 0.17) = 82.7586 lb/ft3

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Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

The correct conclusion would be 0.248 meters for the length of a pendulum with a period of 1 second. Therefore, the SI unit of length, the meter, would have had to be 0.752 m shorter had Huygens' suggestion to define it as the length of such a pendulum been followed.

The question proposes a discussion between Tamsen and Vera about how much shorter the SI unit of length, the meter, would have had to be if Christian Huygens' suggestion to define an international unit of length as the length of a simple pendulum with a period of 1 s, had been followed. Huygens' suggestion corresponds to the second pendulum or gravitational pendulum, which was an arrangement constituted by a simple pendulum adjusted so that its time of oscillation is exactly 2 seconds, meaning 'going and coming back'.

Mathematically, the formula to calculate the length of a pendulum with a given periodic time is given by L = g×(T/2×pi)² where g is the acceleration due to gravity and T is the period of the pendulum. For T = 1 second and g approximated to 9.81 m/s², the length derived would be approximately 0.248 meters. Therefore, among the given options, Vera and Tamsen's correct conclusion would be c) 0.248 m.

This means the SI unit of length, the meter, would have had to be 1 meter - 0.248 m = 0.752 m shorter had Huygens' suggestion been followed.

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