A block with velocity v>0 slides along the floor (with no friction). It hits an ideal spring at time t=0 (configuration #1). The spring starts to compress until the block comes to a (momentary) stop (configuration #2). (Figure 1) (Later, the spring will of course expand, pushing the block back). Here we show you some plots relating to the motion of the block and spring. You will need to identify what these plots represent. In each plot, the point we label as "1" refers to configuration #1 (when the block first comes in contact with the spring). The point we label "2" refers to configuration #2 (which is the moment the block comes to rest, with the spring fully compressed). Here, "force" refers to the x-component of the force of the spring on the block and "position" (and "velocity") refer to the x-components of the position (and velocity) of the block. In all cases, consider the origin to be (0,0); that is, the x-axis represents y=0 and the y-axis represents x=0.

Part A

Look first a t graph A. (Figure 2)

Which of the choices given could this graph represent?

1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position

Part B

Now look at graph B. (Figure 3)

Which of the choices given could this graph represent?

1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position

Part C

Next look at graph C. (Figure 4)

Which of the choices given could this graph represent?

1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position

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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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:

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 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:

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.

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:

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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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

Answer:

r = 3721.04 m          

Explanation:

Given that,

Weight of human, F = 650 N

Charge on two humans, [tex]q_1=q_2=1\ C[/tex]

We need to find the distance between charges if the electric attraction between them to equal their 650 N weight. It is given by :

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

[tex]r=\sqrt{\dfrac{kq^2}{F}}[/tex]

[tex]r=\sqrt{\dfrac{9\times 10^9\times 1^2}{650}}[/tex]

r = 3721.04 m

So, the distance between charges is 3721.04 m if the electric attraction between them to equal their 650 N weight. Hence, this is the required solution.

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: 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:

a. its constitution and mixed government.

Explanation:

Polybius tell us about the Roman mixed constitution as a fundamental in the Roman victory and the Carthaginian defeat in the Punic war, conceiving the mixed constitution as the best, since this constitution was at its peak, which implies that among the elements of the constitution, the aristocratic component was the dominant one.

Answer:

The maximum no. of electrons- [tex]2.25\times 10^{22}[/tex]

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

[tex]I = \frac{Q}{t}[/tex]

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

[tex]Q = 1.0\times 3600 = 3600\ C[/tex]

Maximum number of electrons, n is given by:

[tex]n = \frac{Q}{e}[/tex]

where

e = charge on an electron = [tex]1.6\times 10^{- 19}\ C[/tex]

Thus

[tex]n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}[/tex]

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

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.

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:

0.06

Explanation:

The thickness of a human hair is about 60 micrometers. When converted to millimeters, this measures to 0.06 millimeters.

The thickness of a human hair is about 60 micrometers (um). To convert this to millimeters (mm), you need to understand that 1 millimeter is equal to 1000 micrometers. Therefore, you can convert 60 um to mm by dividing 60 by 1000.

This gives an answer of 0.06 mm. So, the thickness of a human hair is 0.06 millimeters.

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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:

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.

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:

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.

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:

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.

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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:

False

Explanation:

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

Answer:

The average diameter of a single alveolus is 0.0222 cm.

Explanation:

Volume of the lung ,V= 1.9 L

[tex]1 L = 1000 cm^3[/tex]

[tex]1.9 L=1.9\times 1000 cm^3=1900 cm^3[/tex]

Number of alveoli in a human lung = [tex]330\times 10^6[/tex]

Volume of single alveoli =v

[tex]v\times 330\times 10^6=V[/tex]

[tex]v=\frac{1900 cm^3}{330\times 10^6}[/tex]

[tex]v=5.7575\times 10^{-6} cm^3[/tex]

The alveoli are spherical.

Radius of an alveolus = r

Volume of the sphere = [tex]\frac{4}{3}\pi r^3[/tex]

[tex]v=\frac{4}{3}\pi r^3[/tex]

[tex]5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3[/tex]

[tex]r=0.0111 cm[/tex]

Diameter of the alveolus =d

d = 2r = 2 × 0.0111 cm = 0.0222 cm

The average diameter of a single alveolus is 0.0222 cm.

The question seeks to find the average diameter of a single alveolus in the human lung, using information about the total volume of the lung and the number of alveoli. By using the formula for the volume of a sphere, one can calculate the volume of a single alveolus and derive its radius and thereby its diameter.

The problem is essentially asking to find the average diameter of a single alveolus, given we know the total volume of the lung and the number of alveoli. Using the formula for the volume of a sphere, V=4/3πr³, where V is the volume and r is the radius, we can find the volume of a single alveolus by dividing the total volume of the lungs (1.9 liters, which equals to 1.9 x 10^9 cubic millimeters) by the number of alveoli (approximately 330 million). We can then calculate the radius by rearranging the sphere volume formula to r = ((3*V)/4π)^1/3. The diameter would be double the radius.

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The total amount of energy per hour is [tex]2.039\cdot 10^{16} kJ[/tex]

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

[tex]E_1 = 2073 kJ[/tex]

The total surface area of the United States is

[tex]A=9.834\cdot 10^6 km^2[/tex]

And converting into squared metres,

[tex]A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2[/tex]

Therefore, the total energy reaching the entire United States per hour is given by:

[tex]E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ[/tex]

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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:

[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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