If an insulated (power side) wire rubbed through a part of the insulation and the wire conductor touched the steel body of a vehicle, the type of failure would be called a(an)

Answer:

If the truck were lifted twice as high, it would have twice as much potential energy

Explanation:

Potential energy = mgh

where;

m is mass of the truck (kg)

g is acceleration due to gravity (m/s²)

h is the height above the floor.

Initial Potential Energy, E₁ = mgh₁

When the truck was lifted twice as high, New height (h₂) = 2h₁

New potential Energy, E₂ =mgh₂ = mg(2h₁)

E₂ = 2mgh₁

Recall, E₁ = mgh₁, then substitute in E₁  into E₂

E₂ = 2(mgh₁)

E₂ = 2E₁

Therefore, If the truck were lifted twice as high, it would have twice as much potential energy

Explanation:

Power is the ratio of energy to time.

Here we need to consider kinetic energy,

Mass, m = 102 kg

Initial velocity = 0 m/s

Final velocity = 16.2 m/s

Time, t = 2.4 s

Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J

Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J

Change in energy = Final kinetic energy - Initial kinetic energy

Change in energy = 13384.44 - 0

Change in energy = 13384.44 J

Power = 13384.44  ÷ 2.4 = 5576.85 W = 7.48 hp

Power of cheetah is 5576.85 W = 7.48 hp

The cat used 10 units of potential energy to pounce but only 4 units were converted meaningfully into kinetic energy of the pounce. The remaining 6 units were dissipated as heat.

When the cat pounces on a mouse, the energy needed for this action is less than the total potential energy consumed. The cat's body uses 10 units of potential energy, but the pounce only required 4 units of kinetic energy. The difference between these two values signifies the amount of energy dissipated as heat. Therefore, we can perform the subtraction: 10 units (total potential energy) - 4 units (kinetic energy used) = 6 units. Thus, "6 units of energy were dissipated as heat."

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Answer:

total magnification = 400 X

Explanation:

given data

ocular lens = 10 X

objective lens = 40 X

to find out

total magnification

solution

we know that total magnification is express as

total magnification = Objective magnification ×  ocular magnification  .................1

put here value we get

total magnification =  10 × 40

total magnification = 400 X

The total magnification for a microscope with a 10X ocular lens and a 40X objective lens is 400X.

When using a microscope, the total magnification can be found by multiplying the magnification of the ocular lens (also known as the eyepiece) by the magnification of the objective lens. Thus, in your case, the total magnification of the specimen being viewed with a 10X ocular lens and a 40X objective lens would be 400X. This is because 10 times 40 equals 400. Therefore, you are observing the specimen at 400 times its actual size.

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Answer: FALSE

Explanation:Ball End Mills are hemispherical tip used to cut and and shape/ milling rounded objects, such as the metal bearing grooves, slotting and pocketing.

Ball End Mill also called Ball nose end mills, the scallop height will not increase as a result of the increase in diameter. The diameter is the distance between one end of the round tool to another when taken from the middle.

Answer:

398.22 kg

Explanation:

[tex]m_1[/tex] = Mass of canoe 1 = 320 kg

[tex]m_2[/tex] = Mass of canoe 2

[tex]v_1[/tex] = Velocity of canoe 1 = 0.56 m/s

[tex]v_2[/tex] = Velocity of canoe 2 = 0.45 m/s

In this system the linear momentum is conserved

[tex]m_1v_1=m_2v_2\\\Rightarrow m_2=\dfrac{m_1v_1}{v_2}\\\Rightarrow m_2=\dfrac{320\times 0.56}{0.45}\\\Rightarrow m_2=398.22\ kg[/tex]

The mass of the second canoe is 398.22 kg

Answer:

398.22 kg

Explanation:

mass of I canoe, m1 = 320 kg

initial velocity of both the canoe = 0

final velocity of I canoe, v1 = 0.56 m/s

final velocity of II canoe, v2 = - 0.45 m/s

Let the mass of second canoe is m2.

Use conservation of momentum

momentum before push = momentum after push

0 + 0 = m1 x v1 + m2 x v2

0 = 320 x 0.56 - m2 x 0.45

m2 = 398.22 kg

Thus, the mass of second canoe is 398.22 kg.

Final answer:

a) The measured lifetime of the pi meson as observed by an Earth observer is longer than its proper lifetime due to time dilation. b) The average distance the meson travels before decaying can be calculated using the velocity and measured lifetime. c) If time dilation did not occur, the distance the meson would travel can be calculated using the velocity and proper lifetime.

Explanation:

a) According to time dilation, the average lifetime of the pi meson as measured by an observer on Earth is longer than its proper lifetime. To calculate it, we use the formula for time dilation: t' = t / γ, where t' is the measured lifetime, t is the proper lifetime, and γ is the Lorentz factor. Given that the proper lifetime is 2.6 × 10-8 s and the speed of the meson is 0.88c, we can calculate γ using the formula γ = 1 / √(1 - v2/c2). After calculating γ, we can substitute it into the time dilation formula to find the measured lifetime on Earth.

b) To calculate the average distance the meson travels before decaying, we use the formula d = v · t', where d is the distance, v is the velocity, and t' is the measured lifetime. We can substitute the values given in part a) to find the distance traveled by the meson.

c) If time dilation did not occur, the distance the meson would travel can be calculated using the formula d = v · t, where d is the distance, v is the velocity, and t is the proper lifetime. By substituting the values given in part a) into this formula, we can find the distance the meson would travel in the absence of time dilation.

Answer:

a) [tex]\mu_s=0.65[/tex]

b) [tex]\mu_k=0.54[/tex]

Explanation:

Static friction is when the body is at rest or is about to move while kinetic friction is when the body is already in motion. According to Newton's second law:

[tex]\sum F_y:N=mg\\\sum F_x:F_f=F_x[/tex]

a)  In this case, the static friction must be equal to the horizontal force to set the clock in motion:

[tex]F_f=\mu_sN=\mu_smg\\\mu_smg=F_x\\\mu_s=\frac{F_x}{mg}\\\mu_s=\frac{670}{106kg(9.8\frac{m}{s^2})}\\\mu_s=0.65[/tex]

b) In this case, the kinetic friction is equal to the horizontal force that keep the clock moving with constant velocity:

[tex]\mu_kmg=F_x'\\\mu_k=\frac{F_x'}{mg}\\\mu_k=\frac{557}{106kg(9.8\frac{m}{s^2})}\\\mu_k=0.54[/tex]

Answer:

True

Explanation:

Answer:

The answer is TRUE

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Answer:

mass of the meter stick=0.063 kg

or

mass of the meter stick=63.3 g

Explanation:

Given data

m₁=41.0g=0.041kg

r₁=(39.2 - 23)cm

r₂=(49.7 - 39.2)cm

g=9.8 m/s²

To find

m₂(mass of the meter stick)

Solution

The clockwise and counter-clockwise torques must be equal if the meter stick   is in rotational equilibrium

[tex]Torque_{cw}=Torque_{cw}\\F_{1}r_{1}=F_{2}r_{2}\\ m_{1}gr_{1}=m_{2}gr_{2}\\(0.041kg)(9.8m/s^{2} )(0.392m-0.23m)=m_{2}(9.8m/s^{2})(0.497m-0.392m)\\0.0651N.m=1.029m_{2}\\m_{2}=0.063 kg\\or\\m_{2}=63.3g[/tex]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            [tex]V^{2} = U^{2} + 2gh[/tex]

Substituting the values,

                             [tex]V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)[/tex]

                             [tex]V^{2} = 24.5 m/s[/tex]

                             [tex]V = \sqrt{24.5} \ m/s[/tex]

                             [tex]V = 4.95 \ m/s[/tex]

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            [tex]V^{2} = U^{2} + 2gh[/tex]

Substituting the values,

                            [tex]0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)[/tex]

                            [tex]0 = U^{2} - 16.072 m/s[/tex]

                            [tex]U^{2} = 16.072 m/s[/tex]

                            [tex]U = \sqrt{16.072} \ m/s[/tex]

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            [tex]F = \frac{mv - mu}{t}[/tex]

                          [tex]F.t = m(v - u)[/tex]

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

Answer:

The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are due to water's partial charges.

Explanation:

The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor =  3.53 V

The potential drop across the conductor is  3.54 V

The rate of flow of electric charge along a conductor per unit time is known as current.

Mathematically it is expressed as,

As we know that

Power,   [tex]P = VI[/tex]

Where P is power, V is  potential drop and  I is  current.

 Also,  

It is given that,  energy = 742 J and time = 30s

Power [tex]=\frac{742}{30} =24.74watt[/tex]

Substituting the value of power in equation [tex]P=VI[/tex]

      [tex]V=\frac{P}{I} =\frac{24.74}{7}=3.54V[/tex]

Thus,  the potential drop across the conductor is  3.54 V

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Answer:

The standard free energy is the difference between that of products and reactants. The value of G has a greater influence on the reaction been considered.

Explanation:

The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system. The spontaneity of a reaction is explained by the standard gibb's free energy.

Use this hints for any reaction involving the Gibb's free, Enthalpy and entropy.

For each factor of 10 change in Keq, there is a corresponding change of approximately 5.7 kJ/mol in standard free energy change, ΔG°', at standard biochemical conditions (298.15 K).

For every factor of 10 difference in the equilibrium constant, Keq, there is a corresponding change in the standard free energy change, ΔG°', as calculated using the relationship between ΔG°' and Keq:

ΔG°' = -RT ln Keq, where R is the universal gas constant and T is the temperature in Kelvin.

Given that R = 8.314 J/mol·K and at standard biochemical conditions T = 298.15 K, a tenfold change in Keq results in a change of (8.314 J/mol·K x 298.15 K x ln(10)), which equates to approximately 5.7 kJ/mol. Therefore, for example, increasing Keq from 102 to 103 would decrease ΔG°' by 5.7 kJ/mol, making the reaction more product-favored under standard conditions. Conversely, decreasing Keq from 10-3 to 10-4 would increase ΔG°' by 5.7 kJ/mol, making the reaction less product-favored under standard conditions.

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

[tex]F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\   F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2}  }{(6.50*10^{-10} )^{2}  } \\F=5.46*10^{-10}N[/tex]

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

[tex]a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}[/tex]

For(b) Acceleration of proton

[tex]a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}[/tex]

Acceleration of the body is rate of change of the increasing velocity with respect to the time.  is The acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and  acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.

A proton and an electron are released when they are 6.50×10⁻¹⁰ m apart.

The the distance of proton and electron is 6.50×10⁻¹⁰ m.

Now it is known that,

        [tex]m_e=9.109\times10^{-31}[/tex]

        [tex]m_p=1.673\times10^{-27}[/tex]

        [tex]Q_e=-1.602\times10^{-19}[/tex]

        [tex]Q_e=1.602\times10^{-19}[/tex]

Acceleration of the body is rate of change of the increasing velocity with respect to the time. Acceleration of a body is the ratio of the force to the mass of the body.

The coulomb force between the two charges can be given as,

[tex]F=k\dfrac{q_1q_2}{r^2} [/tex]

Here k is the coulombs constant, r is the distance and q is the charge of proton and electron.Put the values,

[tex]F=9\times10^9\times\dfrac{(1.602\times10^{-19})^2}{(6.5\times10^{-10})^2} [/tex]

[tex]F=5.46\times10^{-10}[/tex]

Now it is known that the acceleration of a body is the ratio of the force to the mass of the body.

[tex]a_e=\dfrac{F}{m_e} [/tex]

[tex]a_e=\dfrac{5.46\times10^{-10}}{9.109\times10^{-31}} [/tex]

[tex]a_e=5.992\times10^{20}[/tex]

[tex]a_p=\dfrac{F}{m_p} [/tex]

[tex]a_p=\dfrac{5.46\times10^{-10}}{1.673\times10^{-27}} [/tex]

[tex]a_e=3.625\times10^{17}[/tex]

Thus the acceleration of the electron is [tex]5.992\times10^{20}[/tex] meter per second squared and  acceleration of the proton is [tex]3.265\times10^{17}[/tex] meter per second squared.

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Complete Question:

The complete question is on the all the uploaded image

Answer:

Explanation:

This solution were gotten by examining the diagrams and figuring out the rules that are broken

For A

we see that the two charges are positive so rule 1 is broken,on the same diagram we see that there is no radial symmetry close to the charge so rule 3 is broken then looking again at the diagram A we can see that the electric field is not tangent to any electric line at any point hence rule 5 is broken

For B:

We see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken

secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule  3 is broken

For C:

We see that the number of line is not proportional to the magnitude of the charge hence rule 2 is broken.

For D:

we see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken

secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule  3 is broken

For E:

We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.

For F

Looking at the diagram we see that none of the rules are broken

For G:

looking  at the diagram we see that the line are not uniformly distributed near the charge hence rule  3 is broken

taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken

For H:

We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.

taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken

Looking again at the diagram we see that the at any point on the electric field line that the electric field itself is not tangent to the electric field line hence rule 5 is broken

Incomplete question.The complete question is here

A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10m wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s.What is her acceleration on the rough ice?

Answer:

acceleration = -4.103 m/s²

Explanation:

Given data

Initial velocity Vi=8.30 m/s

Final velocity Vf=5.20 m/s

Initial distance xi=0 m.......(We choose xi=0 the start point of acceleration motion )

Final distance xf=5.10 m

To find

Acceleration

Solution

From the kinetic equation

[tex](v_{f})^{2}=(v_{i} )^{2}+2a(x_{f} -x_{i} )\\  (5.20 m/s)^{2}=(8.30m/s)^{2}+2a(5.10m-0m)\\a=\frac{(5.20m/s)^{2}-(8.30m/s)^{2}  }{2*(5.10m)}\\a=-4.103 m/s^{2}[/tex]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

[tex]f'=f(\dfrac{v+v_{p}}{v-v_{s}})[/tex]

Where, f = frequency

v = speed of sound

[tex]v_{p}[/tex] = speed of passenger

[tex]v_{s}[/tex] = speed of source

Put the value into the formula

[tex]f'=262\times(\dfrac{344+18}{344-30})[/tex]

[tex]f'=302.05\ Hz[/tex]

Hence, The frequency is 302.05 Hz.

Answer:

0.02896 kg/s

Explanation:

[tex]A_1[/tex] = Initial displacement = 0.5 m

[tex]A21[/tex] = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

[tex]x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)[/tex]

At maximum displacement

[tex]cos(\omega t+\phi)=1[/tex]

[tex]\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s[/tex]

The magnitude of the damping coefficient is 0.02896 kg/s

Final answer:

To find the damping constant b for a damped harmonic motion scenario with a given change in amplitude over time, you employ the decay of amplitude formula in damped harmonic motion, then rearrange and solve for b.

Explanation:

To calculate the magnitude of the damping constant b for a 45.0 g hard-boiled egg moving at the end of a spring with a force constant of 25.0 N/m given a decrease in amplitude from 0.500 m to 0.100 m over 5.00 seconds, we apply the principle of damped harmonic motion. The damping force is given by Fx= −bvx, where b is the damping coefficient we wish to find. From the information provided, it's understood that this is a case of exponentially decaying amplitude in harmonic motion.

The equation that relates the exponential decay of amplitude in damped harmonic motion is A(t) = A0e−(bt/2m), where A0 is the initial amplitude, A(t) is the amplitude at time t, and m is the mass of the object. By re-arranging this equation to solve for b, and substituting the given values A0=0.500 m, A(t)=0.100 m, t=5.00 s, and m=0.045 kg (since 45.0 g=0.045 kg), we can calculate b.

The rearranged equation for b would be b = 2m[ln(A0/A(t))]/t. Substituting the values gives b = 2(0.045)[ln(0.500/0.100)]/5.00, which after calculation yields the magnitude of b.

Answer:

0.7549kg

Explanation:

The mass of the slice + mass of the remaining cake = total mass of cake.

mass of remaining cake = total mass of cake - the mass of the slice

total mass=0.870kg

mass of slice = 0.1151kg

mass of remaining cake = 0.870 - 0.1151

mass of remaining cake=0.7549kg

The magnitude of the force of friction is 6.1 N

Explanation:

The work done by a force when moving an object is given by the equation:

[tex]W=Fd cos \theta[/tex]

where :

F is the magnitude of the force

d is the displacement

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

In this problem, we have:

W = -55.2 J (work done by the force of friction)

d = 8.98 m (displacement of the sled)

[tex]\theta=180^{\circ}[/tex] (because the force of friction acts opposite to the direction of motion)

By solving the equation for F, we find the magnitude of the force of friction on the sled:

[tex]F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N[/tex]

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Answer:

6.1

Explanation:

Acellus

Answer

A)   Positively charged two insulating rod are brought closed to an object  they repel each other. It means the Object is positively charged. Because similar charge repel each other.

   The correct answer is Option A.

B) we know force between to charges is calculated using Formula

         [tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex].......(1)

   form the given condition in the question

          [tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r}{2})^2}[/tex]

          [tex]F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r^2}{4})}[/tex]

          [tex]F' = \dfrac{4}{16}\dfrac{kQ_1Q_2}{r^2}[/tex]

from equation (1)

          [tex]F' = \dfrac{F}{4}[/tex]

hence, the correct answer is Option C.

The object is positively charged if repelled by a positively charged rod. The force between two charged objects reduces to F/16 when both charges are one-fourth and the distance is halved.

Charged Objects and Electrostatic Forces

If the object suspended by a string is repelled away from a positively charged insulating rod, we can conclude that the object is positively charged (A). This is because like charges repel each other, according to electrostatic principles.

Regarding the force between two charged objects, Coulomb's law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. If the charge on each object is reduced to one-fourth of its original value and the distance between them is reduced to d/2, the new force is calculated as:

New charge = Original charge / 4

New distance = d / 2

New force = (1/4 × 1/4) / (1/2 × 1/2)^2 = 1/16 / 1/4 = F/16 (A)

Answer:

93328 kg

Explanation:

[tex]\rho[/tex] = Density of mud = 1900 kg/m³

Volume of the cuboid is given by

[tex]V=2300\times 750\times 1.4\\\Rightarrow V=2415000\ m^3[/tex]

Volume is also given by

[tex]V=Al\\\Rightarrow l=\dfrac{V}{A}\\\Rightarrow l=\dfrac{2415000}{520\times 520}\\\Rightarrow l=8.9312\ m[/tex]

The length of the cuboid is 0.89312 m

For the small volume

[tex]V_a=al\\\Rightarrow V_a=5.5\times 8.9312\\\Rightarrow V_a=49.12\ m^3[/tex]

Mass is given by

[tex]m=\rho V_a\\\Rightarrow m=1900\times 49.12\\\Rightarrow m=93328\ kg[/tex]

The mass of the mud is 93328 kg

Explanation:

Diameter of Sun = 100 x Diameter of Earth.

We are creating a model where Earth is same as golf ball,

Diameter of Earth in model = Diameter of golf ball

Diameter of Earth in model = 1.68 inches

Diameter of Sun in model =  100 x Diameter of Earth in model

Diameter of Sun in model =  100 x 1.68

Diameter of Sun in model =  168 inches = 14 feet

Diameter of Sun in model = 168 inches = 14 feet = 4.27 m

Answer:

Force acting on the driver will be 776 N

Explanation:

We have given mass of the driver m = 97 kg

It starts from rest so initial velocity u = 0 m /sec

And reaches to a velocity of 40 m /sec in 5 sec

So final velocity v = 40 m /sec

And time taken t = 5 sec

From first equation of motion v = u +at

So [tex]40=0+a\times 5[/tex]

[tex]a=8m/sec^2[/tex]

Now we have to find the force acting the driver

From newtons law we know that F = ma

So force F = 97×8 = 776 N

So force acting on the driver will be 776 N

The question involves the application of Newton's law to a ball's position over time, launched from a cannon at an angle. The x and y-axis positions can be calculated using horizontal and vertical principles of motion respectively. For the ball's distance (r) to increase throughout the flight, the launch angle needs to be 90°.

The subject of this question relates to the application of Newton's second law in two dimensions, specifically in cases of projectile motion. The position of the ball fired from a cannon is determined by the initial velocity of the ball, the launch angle θ, and the acceleration due to gravity.

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Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]{\omega}=\dfrac{2\pi }{0.2}\ rad/s[/tex]

[tex]{\omega}=31.41\ rad/s[/tex]

We know that

[tex]{\omega}^2=m\ K[/tex]

K=Spring constant

[tex]K=\dfrac{\omega^2}{m}[/tex]

[tex]K=\dfrac{31.41^2}{0.12}\ N/m[/tex]

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

Final answer:

To find the magnitude of the maximum force acting on the body in simple harmonic motion, we use the equation F = -kx. To find the spring constant, we can rearrange the equation F = -kx.

Explanation:

To find the magnitude of the maximum force acting on the body in simple harmonic motion, we can use the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. In this case, we are given the amplitude A = 8.5 cm and the mass m = 0.12 kg. The displacement x at the maximum amplitude is equal to the amplitude, so x = A.

Plugging in the values, we get F = -(k)(A).

(a) To find the magnitude of the maximum force, we need to find the spring constant k. We can use the equation T = 2π√(m/k), where T is the period. Plugging in the values, T = 0.20 s and m = 0.12 kg, we can solve for k.

(b) To find the spring constant, we can rearrange the equation F = -kx to solve for k. Plugging in the values, F = 2.00 mg and x = A, we can solve for k.

Answer:

The time taken by the car with a useful power output is 4.03 seconds.

Explanation:

Given that,

Mass of the car, m = 875 kg

Initial speed of the car, u = 0

Final speed of the car, v = 17 m/s  

Power of the car, P = 42 hp = 31332 W

The power output of the car is given by the work done per unit time. It is given by :

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

Initial kinetic energy of the car will be 0 as it is at rest.

Final kinetic energy of the car is :

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

[tex]K=\dfrac{1}{2}\times 875\times (17)^2[/tex]

K = 126437.5

As per the work energy theorem, the change in kinetic energy of the object is equal to the work done.

So,

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

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

[tex]t=\dfrac{126437.5}{31332}[/tex]

t = 4.03 seconds

So, the time taken by the car with a useful power output is 4.03 seconds. Hence, this is the required solution.

Answer:

Modulation

Explanation:

Modulation is the the process whereby a waveform has one or more of it's properties varied. Amplitude, frequency and phase are all properties of a waveform which can be varied hence Modulation is the process of sending data over a signal by varying either its amplitude, frequency or phase.

Answer:

u= 187.61 ft/s

Explanation:

Given that

g= - 32 ft/s²

The maximum height ,h= 550 ft

Lets take the initial velocity = u ft/s

We know that

v²=u² + 2 g s

v=final speed ,u=initial speed ,s=height

When the object reach at the maximum height then the final speed of the object will become zero.

That is why

u²= 2 x 32 x 550

u²= 35200

u= 187.61 ft/s

That is why the initial speed will be 187.61 ft/s

Answer:

Explanation:

a = - 32 ft/s²

h = 550 ft

Let the initial velocity is u. The velocity at maximum height is zero.

use third equation of motion

v² = u² + 2 a h

0 = u² -  2 x 32 x 550

u² = 35200

u = 187.62 ft/s