What is a simple pendulum?

Answer:

(A)chimney

Explanation:

bc all the smoke is going into the chimney

Answer:

[tex]I = 94.33 kg m^2[/tex]

Explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

[tex]\tau = mg dsin\theta[/tex]

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

[tex]Inertia = I[/tex]

now we have

[tex]I \alpha = mg d sin\theta[/tex]

now for small angular displacement we will have

[tex]\alpha = \frac{mgd}{I}\theta[/tex]

so angular frequency of SHM is given as

[tex]\omega = \sqrt{\frac{mgd}{I}}[/tex]

now we will have

[tex]\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}[/tex]

[tex]I = 94.33 kg m^2[/tex]

Final answer:

The moment of inertia from the pivot is approximately [tex]72.4 kg\(\cdot\)m2.[/tex]

Explanation:

To generate an accurate answer for the moment of inertia of the stick about the pivot when it is used as a physical pendulum, we can use the formula for the period of a physical pendulum, which is:

[tex]T = 2\(\pi\)\(\sqrt{I/(mgd)}\),[/tex]

where T is the period, I is the moment of inertia, m is the mass of the rod, g is the acceleration due to gravity (approximately 9.8 m/s2), and d is the distance from the pivot to the center of mass. Given that T = 9 seconds, m = 4.8 kg, and d = 1.4 m, we can rearrange the formula to solve for I:

[tex]I = T2mgd/(4\(\pi\)2).[/tex]

Plugging the given values into this equation:

[tex]I = 92 * 4.8 kg * 9.8 m/s^2 * 1.4 m / (4 * \(\pi\)2).[/tex]

Calculating this yields the moment of inertia I:

[tex]I \(\approx\) 72.4 kg\(\cdot\)m2.[/tex]

Explanation:

The boat's velocity relative to the water is 6.10 m/s, and the water's velocity relative to the shore is 1.95 m/s.  We want the boat's velocity relative to the shore to be perpendicular to the shore.

If we say that θ is the angle the boat makes with the shore, then:

6.10 cos θ - 1.95 = 0

6.10 cos θ = 1.95

cos θ = 1.95 / 6.10

θ = 71.4°

The boat should be steered 71.4° relative to the shore, or 18.6° relative to the vertical.

The angle at which the boat must be steered can be determined by using the trigonometric function sine (sin), taking the river current speed as the opposite side and the boat's speed in still water as the hypotenuse in a right triangle. Calculate θ = arcsin(1.95 m/s / 6.10 m/s) to find the angle.

To find the angle the boat must be steered, we need to consider the velocity of both the boat and the river current. Firstly, consider that the boat's velocity (6.10 m/s) and the river current velocity (1.95 m/s) form a right triangle. The speed in still water is the hypotenuse while the river current speed forms the opposite side of the triangle.

To find the angle θ which is the angle between the boat's direction and the river current, we use the trigonometric function sine (sin), defined as the length of the opposite side divided by the length of the hypotenuse. Therefore, sin(θ) = (river current speed) / (boat speed in still water), sin(θ) = 1.95 m/s / 6.10 m/s. To find the angle θ, find the inverse of sin(θ).

Therefore, θ = arcsin(1.95 m/s / 6.10 m/s), which can be calculated using a scientific calculator to get an accurate degree measurement for the angle.

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(a) [tex]4.3\cdot 10^9 J[/tex]

The kinetic energy of an object is given by:

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

where

m is the mass of the object

v is its speed

For the ship in the problem, we have

[tex]m=6.40\cdot 10^7 kg[/tex] is the mass

[tex]v=11.6 m/s[/tex] is the speed

So its kinetic energy is

[tex]K=\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=4.3\cdot 10^9 J[/tex]

(b) [tex]-4.3\cdot 10^9 J[/tex]

According to the work-kinetic energy theorem, the work done on an object is equal to the change in kinetic energy of the object:

[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]

where

W is the work done

m is the mass of the object

v is the final speed of the object

u is the initial speed of the object

Here we want to find the work done to stop the ship, so the final speed of the ship is v=0, while the initial speed is u=11.6 m/s. So the work done will be

[tex]W= 0 -\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=-4.3\cdot 10^9 J[/tex]

(c) [tex]1.3\cdot 10^6 N[/tex]

The work done on an object can be also written as follows

[tex]W=Fd[/tex]

where

F is the magnitude of the force

d is the displacement of the object

Here we know:

[tex]W=-4.3\cdot 10^9 J[/tex] is the work done

d = 3.20 km = 3200 m is the displacement of the ship

So we can solve the formula to find F, the force exerted on the ship to stop it:

[tex]F=\frac{W}{d}=\frac{-4.3\cdot 10^9 J}{3200 m}=-1.3\cdot 10^6 N[/tex]

and the negative sign simply means that the force is opposite to the displacement of the ship (in fact, the force acts against the motion of the ship).

The ship's kinetic energy is 5.77 × 10^9 J. The work required to stop it is -5.77 × 10^9 J. The magnitude of the force required to stop it is 1.80 × 10^3 N.

(a) The kinetic energy of the ship can be calculated using the formula:

Ek = (1/2)mv2

where m is the mass of the ship and v is its velocity. Plugging in the given values, we get a kinetic energy of 5.77 × 109 J.

(b) To stop the ship, work needs to be done to bring it to a complete stop. The work done is equal to the change in kinetic energy of the ship. Initially, the ship had a kinetic energy of 5.77 × 109 J. When it comes to a stop, its kinetic energy becomes zero. Therefore, the work required to stop the ship is -5.77 × 109 J (negative since work is done on the ship).

(c) The magnitude of the constant force required to stop the ship can be calculated using the work-energy theorem:

Work = Force × Distance

Given that the displacement of the ship is 3.20 km, or 3.20 × 106 m, and the work done on the ship is -5.77 × 109 J, we can solve for the force:

Force = Work / Distance = -5.77 × 109 J / (3.20 × 106 m) = -1.80 × 103 N

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Answer:

Speed of car A is half the speed of car B.

Explanation:

Kinetic energy, [tex]KE=\frac{1}{2}mv^2[/tex]

Car A has twice the mass of car B.

         [tex]m_A=2m_B[/tex]

Car A has the half the kinetic energy of car B

         [tex]KE_A=\frac{1}{2}KE_B[/tex]

We have

          [tex]KE_A=\frac{1}{2}KE_B\\\\\frac{1}{2}m_Av_A^2=\frac{1}{2}\times \frac{1}{2}m_Bv_B^2\\\\\frac{1}{2}\times 2m_Bv_A^2=\frac{1}{2}\times \frac{1}{2}m_Bv_B^2\\\\v_A^2=\frac{v_B^2}{4}\\\\v_A=\frac{v_B}{2}[/tex]

Speed of car A is half the speed of car B.

Answer:

it would be like a perfect triangle but the sides instead of being straight lines they would all be curved

Explanation:

A line with a constant bendiness ... if continued far enough ... forms a circle.

Answer:

Final velocity will be 1.7 m/s in the initial direction of running back

Explanation:

Since there is no external force on the system of two line man so here we can say that momentum of the two person will remain constant

So we can say

initial momentum of the two persons = final momentum of them

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2) v[/tex]

now from the above equation we will have

[tex]v = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}[/tex]

now plug in all the values in the above equation

[tex]v = \frac{118(1) + 74(-6)}{118 + 74}[/tex]

[tex]v = - 1.7 m/s[/tex]

Answer:

C) the reactants have more potential energy than the products.

Explanation:

Since there is a difference between the potential energy of the reactants and the products, that energy has to leave the formula and into its surroundings, meaning that there will be a liberation of energy, increasing the temperature of the surroundings and or in the closed system in which the reaction is taking place, that is an exhotermic reaction.

The reactants have more potential energy than the products due to which it releases heat energy.

There is a difference between the potential energy of the reactants and the products. That reactions which release thermal energy is also called exothermic reaction which only occurs when there is more potential energy in reactants than products.

So we can conclude that reactants have more potential energy than the products due to which it releases heat energy.

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Final answer:

Gauss' Law in differential form can be used to show that the electric field outside a long non-conducting hollow cylinder decreases as 1/r.

Explanation:

Gauss' Law in differential form can be used to show that the electric field outside a long non-conducting hollow cylinder decreases as 1/r. To do this, we need to consider a Gaussian surface in the form of a cylindrical shell with radius r and length L. The charge enclosed by this surface is the charge on a length L of the cylinder, which is given by Qenc = λL, where λ is the linear charge density. According to Gauss' Law in differential form, the electric field through the Gaussian surface is given by E⋅dA = λenc/ε₀, where E is the electric field, dA is the area vector of the Gaussian surface, λenc is the charge enclosed by the surface, and ε₀ is the permittivity of free space. Since the Gaussian surface is a cylindrical shell, the area vector is perpendicular to the electric field and its magnitude is equal to the area of the cylindrical shell, which is 2πrL. Substituting these values in the equation, we get E(2πrL) = λL/ε₀. Solving for E, we find that the electric field outside the cylinder is given by E = λ/(2πε₀r). Since λ is constant for a long hollow cylinder, we can replace it with the charge density ρ multiplied by the length of the cylinder, L. Therefore, the electric field outside the cylinder decreases as 1/r, where r is the distance from the center of the cylinder.

Answer:

The period of oscillation is 3.14 sec.

Explanation:

Given that,

Spring constant = 20 N/m

Mass = 5.0 kg

We need to calculate the time period

Using formula of time period

[tex]T =2\pi\sqrt{\dfrac{m}{k}}[/tex]

m= mass

k = spring constant

[tex]T =2\times3.14\sqrt{\dfrac{5.0}{20}}[/tex]

[tex]T =3.14\ sec[/tex]

Hence, The period of oscillation is 3.14 sec.

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

   Voltage, V = 6 V, Resistance, R = 550 Ω

   Current, I [tex]=\frac{6}{550}=0.011A=11mA[/tex]

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) [tex]\texttt{Power=}\frac{Energy}{Time}=\frac{0.066}{0.65\times 10^{-3}}=101.54W[/tex]    

Explanation:

The speed [tex]v[/tex] of a wave is given by the following equation:

[tex]v=\frac{\lambda}{T}[/tex] (1)

Where [tex]\lambda[/tex] is the wavelength and [tex]T[/tex] the period of the wave.

In addition we know there is an inverse relation between the period of a wave and its frequency [tex]f[/tex] :

[tex]f=\frac{1}{T}[/tex]  (2)

Substituting (2) in (1):

[tex]v=\lambda.f[/tex]  (3)

Now we can find the velocity of the wave with the known given values applying equation (3):

[tex]v=(10m)(2Hz)[/tex]  (4)

Finally:

[tex]v=20 \frac{m}{s}[/tex]  

Answer: [tex]1.079(10)^{-6}Earth-years[/tex]

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

This law states a relation between the orbital period [tex]T[/tex] of a body (the asteroid in this case) orbiting a greater body in space (the Sun in this case) with the size [tex]a[/tex] of its orbit:

[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex]    (1)

Where;

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]M=1.989(10)^{30}kg[/tex] is the mass of the Sun

[tex]a=2.47(Earth-radius)=2.47(6371000m)=15736370m[/tex]  is orbital radius of the orbit the asteroid describes around the Sun.

Now, if we want to find the period, we have to express equation (1) as written below and substitute all the values:

[tex]T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}[/tex]   (2)

[tex]T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.989(10)^{30}kg)}(15736370m)^{3}}[/tex]   (3)

[tex]T=34.0428s[/tex]   (4) This is the period in seconds, but we were asked to find it in Earth years. So, we have to make the conversion:

[tex]T=34.0428s.\frac{1Earth-hour}{3600s}.\frac{1Earth-day}{24Earth-hour}.\frac{1Earth-year}{365Earth-day}[/tex]  

Finally:

[tex]T=1.079(10)^{-6}Earth-years[/tex]  This is the period of the asteroid around the Sun in Earth years.

The asteroid's orbital period is about 3.88 years

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

Orbital Radius of Earth = R

Orbital Radius of Asteroid = R' = 2.47 R

Orbital Period of Earth = T = 1 year

Unknown:

Orbital Period of Asteroid = T' = ?

Solution:

Firstly , we will use this following formula to find the orbital period:

[tex]F = ma[/tex]

[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]

[tex]G M = \omega^2 R^3[/tex]

[tex]\frac{GM}{R^3} = \omega^2[/tex]

[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]

[tex]T = 2\pi \sqrt {\frac{R^3}{GM}}[/tex]

[tex]\texttt{ }[/tex]

Next , we could find the asteroid's orbital period by using above formula:

[tex]T' : T = 2\pi\sqrt{ \frac{(R')^3}{GM}} : 2\pi\sqrt{ \frac{R^3}{GM}}[/tex]

[tex]T' : T = \sqrt{(R')^3} : \sqrt{R^3}[/tex]

[tex]T' : 1 = \sqrt{(2.47R)^3} : \sqrt{R^3}[/tex]

[tex]T' : 1 = \sqrt{2.47^3}\sqrt{R^3} : \sqrt{R^3}[/tex]

[tex]T' : 1 = \sqrt{2.47^3} : 1[/tex]

[tex]T' = \sqrt{2.47^3}[/tex]

[tex]T' \approx 3.88 \texttt{ years}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer:

initial: 1654.6 J, final: 0 J, change: -1654.6 J

Explanation:

The length of the slide is

L = 8.80 ft = 2.68 m

So the height of the child when he is at the top of the slide is (with respect to the ground)

[tex]h = L sin \theta = (2.68 m)sin 25.0^{\circ}=1.13 m[/tex]

The potential energy of the child at the top is given by:

[tex]U = mgh[/tex]

where

m = 63.0 kg is the mass of the child

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.13 m

Substituting,

[tex]U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J[/tex]

At the bottom instead, the height is zero:

h = 0

So the potential energy is also zero: U = 0 J.

This means that the change in potential energy as the child slides down is

[tex]\Delta U = 0 J - (1654.6 J) = -1654.6 J[/tex]

The potential energy at the top is 700.9 J. At the bottom is 0 J. Change is -700.9 J (loss).

The length of the slide [tex]\( L = 8.80 \, \text{ft} \)[/tex] and the angle with the horizontal [tex]\( \theta = 25.0^\circ \)[/tex], we can use the sine function to find the vertical height h from the top to the bottom of the slide:

[tex]\[ h = L \sin \theta \][/tex]

[tex]\[ h = 8.80 \, \text{ft} \times \sin 25.0^\circ \][/tex]

First, calculate [tex]\( \sin 25.0^\circ \)[/tex]:

[tex]\[ \sin 25.0^\circ \approx 0.4226 \][/tex]

Now, calculate h:

[tex]\[ h = 8.80 \times 0.4226 \approx 3.719 \, \text{ft} \][/tex]

Potential energy PE is given by:

[tex]\[ PE = mgh \][/tex]

where [tex]\( m = 63.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \)[/tex] , and h is the height

Convert h from feet to meters (1 ft = 0.3048 m):

[tex]\[ h = 3.719 \, \text{ft} \times 0.3048 \, \text{m/ft} = 1.133 \, \text{m} \][/tex]

The potential energy at the top:

[tex]\[ PE_{\text{top}} = 63.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.133 \, \text{m} \][/tex]

[tex]\[ PE_{\text{top}} \approx 63.0 \times 9.81 \times 1.133 \approx 700.9 \, \text{J} \][/tex]

Since the child is at the bottom of the slide, where h = 0, the potential energy at the bottom [tex](\( PE_{\text{bottom}} \))[/tex] is 0 Joules.

The change in potential energy

The change in potential energy [tex](\( \Delta PE \))[/tex] is:

[tex]\[ \Delta PE = PE_{\text{bottom}} - PE_{\text{top}} \][/tex]

[tex]\[ \Delta PE = 0 - 700.9 \, \text{J} \][/tex]

[tex]\[ \Delta PE = -700.9 \, \text{J} \][/tex]

Answer:

Power required, P = 175 watts

Explanation:

It is given that,

Weight of a student, F = mg = 200 N

The student climbs a flight of stairs of height, h = 7 m

Time taken, t = 8 s

We have to find the power required to perform this task. Work done per unit time is called the power required. Mathematically, it is given by :

[tex]P=\dfrac{W}{t}[/tex]

W = work done

t = time taken

[tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{200\ N\times 7\ m}{8\ s}[/tex]

P = 175 watts

Hence, the power required to complete this task is 175 watts.

[tex]a=5000\dfrac{mm}{s}=5\dfrac{m}{s}[/tex]

[tex]f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s[/tex]

[tex]a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}

[/tex]

Hope this helps.

r3t40

The maximum displacement or amplitude of the oscillating machine, given the acceleration of 5,000 mm/s² and a frequency of 10 Hz, is calculated using principles of simple harmonic motion as approximately 13 mm.

The question you asked involves the principles of Physics, in particular, the topics of simple harmonic motion and amplitude are applied. The machine's oscillation modelled as a simple spring-mass system operates under the principles of simple harmonic motion. The maximum displacement the machine undergoes, also known as the amplitude, can be calculated using the given data of acceleration and frequency.

Acceleration in a simple harmonic system is given by the formula a = -ω²x, where ω represents angular frequency, x represents displacement, and a represents acceleration. The maximum displacement, or amplitude, can be described as A = a_max/ω².

To find ω, we use the given frequency and recall that ω=2πf, so substituting 10 Hz for f, we get ω=20π rad/sec. To find the acceleration, we need to convert 5000 mm/s² to m/s² which is 5 m/s².

Substituting these values back into the amplitude equation A = a_max/ω², we get A = 5/400π² which calculates approximately as 0.013 m or 13 mm.

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(a) 0.5 kg m/s

Before the collision, only the first body is moving, so only the first body contributes to the total momentum.

The first body has

m = 400 g = 0.4 kg (mass)

v = 1.25 m/s (velocity, towards east direction)

So its momentum is

[tex]p_x=mv = (0.4 kg)(1.25 m/s)=0.5 kg m/s[/tex]

And since the body is moving along the east direction, this is also the easterly component of the total momentum before the collision.

(b) Zero

Before the collision, we have:

- The first body moving along the east direction --> so its northerly component is zero

- The second body at rest --> this means that it does not contribute to the momentum, since it is zero

This means that the northerly component of the total momentum before the collision is zero.

(c) 0.5 m/s at 53.1 degrees south of east

The law of conservation of momentum states that each component of the total momentum must be conserved.

- Along the easterly direction:

[tex]p_x = p_{1x} + p_{2x}[/tex]

where

[tex]p_x = 0.5 kg m/s[/tex] is the easterly component of the total momentum

[tex]p_{1x} = m v cos \theta = (0.4 kg)(1.00 m/s) cos 36.9^{\circ} =0.32 kg m/s[/tex] is the easterly component of the momentum of the first body after the collision

[tex]p_{2x}[/tex] is the easterly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision

Solving the equation we find

[tex]p_{2x} = p_x - p_{1x} = 0.5 kg m/s - 0.32 kg m/s = 0.18 kg m/s[/tex]

- Along the northerly direction:

[tex]p_y = p_{1y} + p_{2y}[/tex]

where

[tex]p_y = 0 kg m/s[/tex] is the northerly component of the total momentum

[tex]p_{1y} = m v sin \theta = (0.4 kg)(1.00 m/s) sin 36.9^{\circ} =0.24 kg m/s[/tex] is the northerly component of the momentum of the first body after the collision

[tex]p_{2y}[/tex] is the northerly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision

Solving the equation we find

[tex]p_{2y} = p_y - p_{1y} = 0 - 0.24 kg m/s = -0.24 kg m/s[/tex]

So now we find the momentum of the 600 g body after the collision:

[tex]p_2=\sqrt{p_{2x}^2 + p_{2y}^2}=\sqrt{(0.18)^2+(-0.24)^2}=0.3 kg m/s[/tex]

and so its final speed is

[tex]v=\frac{p_2}{m}=\frac{0.3 kg m/s}{0.6 kg}=0.5 m/s[/tex]

and the direction is

[tex]\theta=tan^{-1} (\frac{p_{2y}}{p_{2x}})=tan^{-1} (\frac{-0.24}{0.18})=-53.1^{\circ}[/tex]

so 53.1 degrees in the south-east direction.

Answer:

Range, R = 5090.1 meters

Explanation:

It is given that,

Velocity of cannon, v = 240 m/s

A cannon is shot at a 30 degree angle above the horizontal. The range of this shot is given by the formula as follows :

[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]

g = acceleration due to gravity

[tex]R=\dfrac{(240\ m/s)^2\ sin2(30)}{9.8\ m/s^2}[/tex]

R = 5090.1 meters

So, the range of this shot is 5090.1 meters. Hence, this is the required solution.

Answer:

0.042 m^3

Explanation:

side, a = 0.5 m, m = 100 kg

density of fluid, d = 1.2 g/cm^3 = 1200 kg/m^3

Volume of cube, V = side^3 = (0.5)^3 = 0.125 m^3

density of solid = mass of solid / volume of solid cube

density of solid, D = 100 /  0.125 = 800 kg/m^3

Let v be the volume of cube immersed in fluid.

According to the principle of flotation

Weight of cube = Buoyant force acting on the cube

V x D x g = v x d x g

0.125 x 800 = v x 1200

v = 0.083 m^3

Volume above the level of fluid = V - v = 0.125 - 0.083 = 0.042 m^3

Answer:

  162 km

Explanation:

A diagram can be helpful.

Using the law of cosines, we can find the magnitude of the distance (c) to satisfy ...

  c^2 = a^2 +b^2 -2ab·cos(C)

where C is the internal angle of the triangle of vectors and resultant. Its value is ...

  180° -39.8° -59.9° = 80.3°

Filling in a=76 and b=156, we get ...

  c^2 = 76^2 +156^2 -2·76·156·cos(80.3°) ≈ 26116.78

  c ≈ √26116.78 ≈ 161.607

The magnitude of the total displacement is about 162 km.

_____

Please note that in the attached diagram North is to the right and East is up. That alteration of directions does not change the angles or the magnitude of the result.

Answer:

Magnitude of total displacement = 162.87 km

Explanation:

Let east be x axis and north be y axis.

The first is 76 km at 39.8◦ west of north.

Displacement 1 = 76 km at 39.8◦ west of north =  76 km at 129.8◦ north of east.

Displacement 1 = 76 cos129.8 i + 76 sin 129.8 j = -48.65 i + 58.39 j

The second is 156 km at 59.9◦ east of north.

Displacement 2 = 156 km at 59.9◦ east of north =  156 km at 31.1◦ north of east.

Displacement 2 = 156 cos31.1 i + 156 sin 31.1 j = 133.58 i + 80.58 j

Total displacement = Displacement 1 + Displacement 2

Total displacement = -48.65 i + 58.39 j + 133.58 i + 80.58 j = 84.93 i + 138.97 j

[tex]\texttt{Magnitude of total displacement =}\sqrt{84.93^2+138.97^2}=162.87km[/tex]

Magnitude of total displacement = 162.87 km

Answer:

The value of A will increase as you increase the charges.

Explanation:

A is actually kq1*q2. So if you increase the charges, A will also increase. The big increase comes if you make r smaller.

Increased charges is a direct variation. As the charge increases the value of A increases. As the charges decrease, the value of A decreases.

r is an inverse variation. As r increases the force decreases. As r decreases the force increases. R does not affect a.

Answer:

Increase the radius of curvature of surface.

Explanation:

Radius of curvature is the measurement that is two times the focal length, for a given lens. It lies on either side of the lens.

Focal length is the distance which is half of the radius of curvature. Radius of curvature is a measure of the radius of the circle . Focal length is the distance between the center of curvature of the lens and the point where all the rays are brought to a focus for a distant object.

The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C with 6 negatively charged electrons on it.

If the charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electron [tex]10^{-19[/tex] C.  and Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons.

Given that :

Number of electrons = 6

Given charge = −1.60× [tex]10^{-19}[/tex]C

The total charge can be calculated as:

Total charge = Number of electrons × Fundamental charge

Total charge  = 6 × −1.60× [tex]10^{-19}[/tex]C

                       = 9.6 ×[tex]10^{-19}[/tex]

The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C.

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Using the findings of Millikan's oil drop experiment, the charge on an oil droplet with 6 negatively charged electrons would be -9.60×10^−19 Coulombs.

According to the data from Millikan’s oil drop experiment, the charge of a single electron is identified to be −1.60×10−19 C. It was found that the charge on any negatively charged oil droplet is always a whole-number multiple of this fundamental charge. Therefore, if Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons on it, he would calculate the charge as

-1.60×10−19 C (charge of a single electron) multiplied by 6 (number of electrons) resulting in a total charge of -9.60×10^−19 C.

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(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

[tex]\tau = F r[/tex]

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

[tex]\tau = (0.795 N)(30.9 m)=24.6 N m[/tex]

(b) [tex]0.035 rad/s^2[/tex]

The equivalent of Newton's second law for a rotational motion is

[tex]\tau = I \alpha[/tex]

where

[tex]\tau[/tex] is the torque

I is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

[tex]I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2[/tex]

And so we can solve the previous equation to find the angular acceleration:

[tex]\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2[/tex]

(c) [tex]1.08 m/s^2[/tex]

The linear acceleration (tangential acceleration) in a rotational motion is given by

[tex]a=\alpha r[/tex]

where in this problem we have

[tex]\alpha = 0.035 rad/s^2[/tex] is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

[tex]a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2[/tex]

Explanation:

It is given that, a mass attached to a spring oscillates and completes 53 full cycles in 28 s. Frequency of the system is given by the number of oscillations or vibrations per second. Here, the frequency of this system is given by :

[tex]f=\dfrac{no\ of\ oscillations}{time}[/tex]

[tex]f=\dfrac{53}{28}[/tex]

f = 1.89 Hertz

The relationship between the frequency and the time period of the spring is inverse i.e.

[tex]T=\dfrac{1}{f}[/tex]

[tex]T=\dfrac{1}{1.89\ Hz}[/tex]

T = 0.52 seconds

Hence, this is the required solution.

Answer:

The cost of energy is 0.29$.

Explanation:

Given that,

Voltage V= 12 V

Charge Q= 160 A-h

We need to calculate the energy

Using formula of energy

[tex]E = Pt[/tex]

[tex]E =VIt[/tex]

Where, E = energy

I= current

V = Potential

t = time

Put the value into the formula

[tex]E = 12\times160\times3600[/tex]

[tex]E =6.9\times10^{6} J[/tex]

The cost of this energy at 0.15/kwh

[tex]Total\ cost = Energy\times 0.15 kwh[/tex]

Here, Energy = power x time

In units,

[tex]J=watt\times s[/tex]

But ,  Energy in 1 hour is:

[tex]E = 1920\ J[/tex]

So, The cost of this energy is

[tex] Total\ cost =1920\dfrac{\times0.15}{1000}[/tex]

[tex]total\ cost = 0.29\$[/tex]

Hence, The cost of energy is 0.29$.

For a battery rated at 12V and 160 A-h, the total energy stored is about 6912000 Joules or 1.92 kWh. The cost of this energy, at a rate of $0.15 per kWh, would therefore be $0.288.

To find out the amount of energy stored in the battery, we first understand that the battery's rated capacity is given as Ampere-hours (A-h). It is a measure of the electric charge that a battery can deliver over time. We can find the energy stored in this battery by multiplying the charge by the battery's voltage. With a rating of 12 V and 160 A-h, the total energy would be calculated as Energy (E) = Voltage (V) x Charge (Q), where Q (the charge) is 160 A-h, which must be converted to seconds by multiplying by 3600 (the number of seconds in one hour). So, Q = 160 A-hr x 3600 = 576000 Coulombs. Once we substitute these values into the formula, the total energy stored would be E = 12 V * 576000 C = 6912000 J.

Now, to find the cost of this energy, we need to convert this energy (which is in Joules) to kWh by dividing by 3.6 million (as 1 kWh = 3.6 million Joules), leading to E = 6912000 J / 3600000 = 1.92 kWh. The cost of this energy would then be calculated by multiplying the energy by the cost per unit of energy, here given as $0.15 per kWh. Therefore, Cost = 1.92 kWh * $0.15/kWh = $0.288.

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Answer:

Minimum elastic modulus of fiber = 455.64 GPa

Explanation:

Contents of composite material = Epoxy and Unidirectional fibers

Elastic modulus of epoxy = 3.5 GPa

Elastic modulus of composite material = 320 GPa

Volume fraction of fiber = 70 %

Volume fraction of epoxy = 100 - 70 = 30%

Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320

0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95

Elastic modulus of fiber = 455.64 GPa

Minimum elastic modulus of fiber = 455.64 GPa

Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, [tex]a=-44\ ft/s^2[/tex]

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}[/tex]

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

Answer:

The distance covered by car before stopping is 61.13 ft.

Explanation:

Given data:

Initial Speed of car is, [tex]u=50 \;\rm mi/h = 50 \times 1.467 =73.35 \;\rm ft/s[/tex].

Deceleration of car is, [tex]a=-44\;\rm ft/s^{2}[/tex]. (Negative sign shows negative acceleration)

Applying the second kinematic equation of motion as,

[tex]v^{2}=u^{2}+2as[/tex]

Here, s is the distance covered and v is the final speed. Since, car stops finally, v = 0.

Solving as,

[tex]0^{2}=73.35^{2}+2(-44) \times s\\88s =73.35^{2}\\s \approx 61.1 \;\rm ft[/tex]

Thus, distance covered by car before stopping is 61.1 ft.

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Answer:

Explanation:

When two equal and opposite force acting at a line of action, they form a couple. It always gives a turning effect to the body.

The turning force is said to be torque .

Torque is the product of either force and the perpendicular distance between the forces.

Torque = force x perpendicular distance

Its SI unit is Nm. It is a vector quantity.

Torque is a physical quantity that measures the effectiveness of a force in changing or accelerating a rotation. The equation for the magnitude of torque is t = rF sin θ.

Torque is a physical quantity that measures the effectiveness of a force in changing or accelerating a rotation. It incorporates the magnitude, direction, and point of application of the force. The equation for the magnitude of torque is given by t = rF sin θ, where t is torque, r is the distance from the pivot point to the point of force application, F is the magnitude of the force, and θ is the angle between the force and the lever arm.

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