The main waterline into a tall building has a pressure of 90 psia at 16 ft elevation below ground level. Howmuch extra pressure does a pump need to add to ensure a waterline pressure of 30 psia at the top floor 450 ft above ground?

Answer:

The speed of comet is 1442.77 m/s.

Explanation:

Given that,

Speed of comet's [tex]v_{c}= 5.3\times10^{4}\ m/s[/tex]

Distance from the sun [tex]r_{d}=9.8\times10^{10}\ m[/tex]

Distance [tex]r_{far}=3.6\times10^{12}\ m[/tex]

We need to calculate the speed of comet

Using conservation of angular momentum

[tex]L_{f}=L_{i}[/tex]

[tex]I\omega=I\omega[/tex]

Here. [tex]v = r\omega[/tex]

[tex]\omega=\dfrac{v}{r}[/tex]

[tex]mr_{far}^2\times\dfrac{v_{far}}{r_{far}}=mr_{d}^2\times\dfrac{v_{d}}{r_{d}}[/tex]

[tex]r_{far}\times v_{far}=r_{d}\times v_{d}[/tex]

Put the value into the formula

[tex]3.6\times10^{12}\times v_{far}=9.8\times10^{10}\times5.3\times10^{4}[/tex]

[tex]v_{far}=\dfrac{9.8\times10^{10}\times5.3\times10^{4}}{3.6\times10^{12}}[/tex]

[tex]v_{far}=1442.77\ m/s[/tex]

Hence, The speed of comet is 1442.77 m/s.

Final answer:

The speed of Halley's comet when it is 3.6 x 10^12 m away from the sun can be found using the conservation of angular momentum, resulting in a speed of 1.44 x 10^3 m/s.

Explanation:

To find the speed of Halley's comet at a distance of 3.6 × 1012 m from the sun, we will use the conservation of angular momentum. The orbital momentum of an object around the sun is given by L = mrv, where m is the mass of the comet, r is the radius of the orbit (distance from the sun), and v is the velocity of the comet along that radius.

We are given the comet's speed at a distance of 9.8 × 1010 m is 5.3 × 104 m/s. When the comet is at a distance of 3.6 × 1012 m from the sun, we want to find its new speed, v'. Because angular momentum is conserved, the initial and final angular momentum should be equal:

L initial = L final
m r v = m r' v'

Since the mass of the comet, m, cancels out, we are left with:

r v = r' v'

Plugging in the known values, we get:

9.8 × 1010 m × 5.3 × 104 m/s = 3.6 × 1012 m × v'

Therefore:

v' = (9.8 × 1010 × 5.3 × 104) / (3.6 × 1012)

v' = 1.44 × 103 m/s

The speed of Halley's comet at a distance of 3.6 × 1012 m from the sun is thus 1.44 × 103 m/s.

Answer: v = 5.79 * 10^10m/s.

Explanation: By using the work-energy theorem, we know that the work done on the electron by the potential difference equals the kinetic energy of the electrons.

Mathematically, we have that

qV = 1/2mv²

q= magnitude of an electronic charge = 1.609*10^-16c

V= potential difference = 95v

m = mass of an electronic charge = 9.11* 10^-31kg.

v = velocity of electron.

Let us substitute the parameters, we have that

1.609*10^-16 * 95 = (9.11*10^-31 * v²) /2

1.609*10^-16 * 95 * 2 = 9.11*10^-31 * v²

305.71 * 10^-16 = 9.11 * 10^-31 * v²

v² = 305.71 * 10^-16/ 9.11 * 10^-31

v² = 3.355 * 10^21.

v = √3.355 * 10^21

v = 5.79 * 10^10 m/s

Final answer:

Using the principle of energy conservation and the equation KE = qV = ½ mv^2, the velocity of the electrons in a CRT with an accelerating potential of 95.0 V is calculated to be approximately 5.8 x 10^6 m/s.

Explanation:

To determine how fast electrons will be moving when they hit the screen in a Cathode-Ray Tube (CRT) with an accelerating potential of 95.0 V, one can use the principle of energy conservation. The amount of kinetic energy (KE) gained by an electron when it is accelerated through a potential difference (V) is equal to the change in electrical potential energy (PE), which is given by the equation KE = qV, where q is the charge of the electron (1.602 x 10-19 Coulombs) and V is the potential difference.

The formula for kinetic energy is KE = ½ mv^2, with 'm' representing the mass of the electron (9.109 x 10^-31 kg) and 'v' being the velocity of the electron. Setting these equations equal gives us qV = ½ mv^2. Solving for 'v' and plugging in the values for q, V, and m, we can find the velocity of the electrons when they hit the screen.

Calculation:

qV = ½ mv^2

2qV = mv^2

v = √(2qV/m)

v = √(2 * 1.602 x 10^-19 C * 95.0 V / 9.10^9 x 10^-31 kg)

v ≈ 5.8 x 10^6 m/s

Thus, the electrons would be traveling at approximately 5.8 x 10^6 m/s when they hit the screen.

We will make a graph to better understand the displacement of the individual. From the trigonometric properties we will find the required distances.

[tex]d_1 = \frac{450}{\sqrt{2}} = 318.198[/tex]

[tex]d_2 = \frac{400}{\sqrt{2}} = 282.843[/tex]

[tex]D = 500+d_1+d_2 = 1101.041[/tex]

Displacement ,

[tex]x = \sqrt{1101.041^2+(318.198-282.843)^2}}[/tex]

[tex]x = 1101.608m[/tex]

The angle would be

[tex]\theta = cos^{-1} (\frac{1101.041}{1101.608})[/tex]

[tex]\theta = E 1.838\° N[/tex]

Therefore the displacement was of 1101.608m to a angle of 1.838° from East to North.

The sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.

This is a problem of physics specifically in the field of mechanics, about displacement and vectors. To find the total displacement, the directions in which the sailor traveled need to be considered. A north direction has an angle of 90 degrees, northeast is 45 degrees, and northwest is 135 degrees.

Using the Pythagorean theorem and some trigonometry, we can calculate the displacement in the x and y directions. The total displacement in the x direction (East-West) is 550*cos(45) - 500*cos(45) = 35.35 meters (approximately). The total displacement in the y direction (North-South) is 600 + 550*sin(45) + 500*sin(45) = 1116.57 meters (approximately).

The total displacement would then be the square root of (35.35 ^2 + 1116.57 ^2) = 1118.16 meters (approximately).

To find the angle of the displacement, we can use the arctangent function (atan function in most calculators). The angle is atan (1116.57 / 35.35) = 88.13 degrees (approximately).

Therefore, the sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.

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Answer

given,

mass of the block, m = 0.5 Kg

displacement, x = 30 cm = 0.3 m

Spring constant, k = 2 N/m

a) Angular frequency

   [tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]

   [tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]

  [tex]\omega = 2\ rad/s[/tex]

b) Period of oscillation

   [tex]T=\dfrac{2\pi}{\omega}[/tex]

   [tex]T=\dfrac{2\pi}{2}[/tex]

           T = 3.14 s

c) Position at  t = 2

 x = -A cos ω t

A = 0.3   ω = 2 rad/s

x = -0.3 cos (2 x 2)

x = 0.196 m

d) velocity at t= 2

 v = A ω sin ω t

 v = 0.3 x 2 x sin 4

 v = -0.454 m/s

e) acceleration at t= 2

   a = A ω² cos ω t

   a = 0.3 x 2² cos 4

   a = -0.784 m/s²

Answer:

934701926.438 Hz

Explanation:

Mass of molecule

[tex]m=3\times 16\times 1.67\times 10^{-27}\ kg[/tex]

Moment of inertia is given by

[tex]I=\dfrac{1}{12}ml^2\\\Rightarrow I=\dfrac{1}{12}\times 3\times 16\times 1.67\times 10^{-27}\times (2.5\times 10^{-10})^2\\\Rightarrow I=4.175\times 10^{-46}\ kgm^2[/tex]

E = Electric field = 8000 N/C

p = Dipole moment = [tex]1.8\times 10^{-30}\ Cm[/tex]

l = Length of rod = [tex]2.5\times 10^{-10}\ m[/tex]

Frequency of oscillations is given by

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{1.8\times 10^{-30}\times 8000}{4.175\times 10^{-46}}}\\\Rightarrow f=934701926.438\ Hz[/tex]

The frequency of oscillations is 934701926.438 Hz

Answer:

The mass of coal is 108 kg.

Explanation:

Given that,

Energy of coal = 1200 kg

Mass of aluminum = 1.0 kg

Energy required for single soda can = 15 g of Al

Energy required for 6 pack of cans = [tex]6\times15=90\ g\ of\ Al[/tex]

We need to calculate the mass of coal

Using formula of mass

[tex]\text{mass of coal}=\dfrac{\text{Energy of coal}\times\text{Energy required for 6 pack of cans}}{\text{Mass of aluminum}}[/tex]

Put the value into the formula

[tex]m=\dfrac{1200\times90\times10^{-3}}{1.0}[/tex]

Put the value into the formula

[tex]m=108\ Kg[/tex]

Hence, The mass of coal is 108 kg.

Answer:

d = 3.53 m

Explanation:

The Coulomb Force is given as

[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]

The x-component of the force is equal to

[tex]F_x = F\cos(\theta) = F\frac{x}{\sqrt{x^2 + y^2}} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{(5^2 + d^2)}\frac{d}{\sqrt{5^2 + d^2}} = \frac{1}{4\pi\epsilon_0}\frac{dq_1q_2}{(5^2 + d^2)^{3/2}}[/tex]

This is basically a function of (d). So, the maximum value of this function is the point where its derivative with respect to d is equal to zero.

[tex]\frac{dF_x}{dd} = \frac{kq_1q_2}{(d^2 + 5^2)^{3/2}} - \frac{3d^2kq_1q_2}{(d^2 + 5^2)^{5/2}} = 0\\3d^2 = d^2 + 5^2\\2d^2 = 25\\d = 3.53~m[/tex]

The value of d for which the x component of the force on the charge is the greatest, in accordance to Coulomb's Law, is when d equals 5 meters. At this distance, the x and y components of the force are equal, thus maximizing the x component.

The scenario you described involves the concept of Coulomb's Law which states the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Looking at your question, the value of d for which the x component of the force on 9×10−18 C is the greatest would be when d equals 5 meters. The charges would then form an equilateral triangle with the origin, meaning the x component and y component of the force would be equal, hence maximizing the x component.

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Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

Answer:

The two balls will hit the ground at the same time neglecting air resistance.

Explanation:

The vertical motion of the two ball are independent of each other. Also, the balls are falling from the same height in the same time. The ball projected horizontal neglecting air resistance is traveling with a constant velocity ( the same distance in the equal time) has only force of gravity acting on it. They will therefore hit the ground at the same time because they are acted upon by the same acceleration in the same time from the same height.

The vertical and horizontal motions of an object are independent. Hence, despite different trajectories, a ball dropped vertically and a ball launched horizontally from the same height will hit the ground simultaneously, as exemplified by Galileo's thought experiment.

In the realm of Physics, an interesting question has been asked: Which ball hits the ground first - a ball that is dropped vertically or a ball that is launched horizontally? According to the principle of independence of motion, the horizontal motion and the vertical motion of an object are independent of each other. It implies that the horizontal velocity of a body has no effect on its vertical velocity. Thus, the ball dropped directly and the ball launched horizontally would strike the ground at the same time. Both balls are subjected to the same gravitational acceleration (g), without any initial vertical velocity.

This concept is a perfect illustration of Galileo's thought experiment, whereby he proved that the time for the balls to hit the ground is determined by their vertical motion alone and that horizontal motion does not affect the descent time.

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Answer:

[tex]3.8469943828\times 10^{26}\ W[/tex]

Explanation:

[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]

T = Surface temperature of the Sun = 5778 K

r = Radius of the Sun = 696000 km

From Stefan-Boltzmann law

[tex]F=\sigma T^4\\\Rightarrow F=5.67\times 10^{-8}\times 5778^4\\\Rightarrow F=63196526.546\ W/m^2K[/tex]

Power is given by

[tex]P=F4\pi r^2\\\Rightarrow P=63196526.546\times 4\pi\times (696000000)^2\\\Rightarrow P=3.8469943828\times 10^{26}\ W[/tex]

The power output of the Sun is [tex]3.8469943828\times 10^{26}\ W[/tex]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is [tex]3.2\cdot 10^7 N/C[/tex] (radially inward)

C) The electric field at 6.00 cm from the surface is [tex]1.2\cdot 10^7 N/C[/tex] (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states  that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

[tex]\int EdS = \frac{q}{\epsilon_0}[/tex]

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

[tex]\epsilon_0[/tex] is the vacuum permittivity

By taking a sphere centered in the origin,

[tex]\int E dS = E \cdot 4\pi r^2[/tex]

where [tex]4\pi r^2[/tex] is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

[tex]R=9.0 cm = 0.09 m[/tex] (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

[tex]q=0[/tex]

And therefore,

[tex]E4\pi r^2 = 0\\\rightarrow E = 0[/tex]

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

[tex]r=R=0.18 m[/tex]

The charge contained within the Gaussian sphere is therefore

[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]

Therefore, the electric field is

[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C[/tex]

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

[tex]E=3.2\cdot 10^7 N/C[/tex]

C)

For this part again, we apply Gauss Law:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]

Therefore, the electric field in this case is

[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C[/tex]

And again, this is radially inward, so according to the sign convention asked in the problem,

[tex]E=1.2\cdot 10^7 N/C[/tex]

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To find the electric field just inside the paint layer, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.

A) Inside the paint layer:

Since the paint is spread in a very thin uniform layer over the surface of the plastic sphere, we can consider a Gaussian surface just inside the paint layer, with a radius slightly smaller than the sphere's radius (let's call it r). The charge enclosed within this Gaussian surface is the charge on the sphere, which is -29.0 μC.

Gauss's law equation can be written as:

Φ = Q_enclosed / ε₀

The electric field just inside the paint layer (E₁) is given by:

E₁ * A = Q_enclosed / ε₀

E₁ * 4πr² = -29.0 μC / ε₀

E₁ = (-29.0 μC / ε₀) / (4πr²)

The direction of the electric field just inside the paint layer will be radially outward, opposite to the direction of the charge.

B) Just outside the paint layer:

The electric field just outside the paint layer (E₂) will be the same as the electric field inside the sphere when it was uncharged. This is because the charge on the paint layer is spread on the outer surface of the sphere.

So, E₂ = E₁ (since it's radially outward)

C) 6.00 cm outside the surface of the paint layer:

To find the electric field 6.00 cm outside the surface of the paint layer, we need to calculate the electric field due to the charge on the sphere.

E₃ = k * Q / r²

where k is the Coulomb's constant (k = 9.0 x 10^9 N m²/C²), Q is the charge on the sphere (-29.0 μC), and r is the distance from the center of the sphere to the point where we want to find the electric field (6.00 cm = 0.06 m + radius of the sphere).

Remember that the electric field due to a charged object decreases with the square of the distance.

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Answer:

The boiling point of Acetone is 329K (in 3 significant figures)

Explanation:

Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)

Answer: using the formula 0°C + 273.15 = 273.15K the boiling point in units of kelvin to significant figures is 329.15k.

Explanation: The boiling point of a substance ( acetone) is the temperature at which the vapour pressure of the liquid substance equals the pressure surrounding it. The boiling point of acetone serves as it's characteristic physical properties. This is measured in degree Celsius (°C ) which can be converted to units of Fahrenheit or kelvin. To convert degree Celsius to kelvin this formula is used: 0°C + 273.15 = 273.15K . Given that acetone has boiling point of 56°C,from the formula 0°C is substituted for 56°C. This gives us:

56°C + 273.15= 319.15k.

Also,measurements given in Kelvin will always be larger numbers than in Celsius and the Kelvin temperature scale does not use the degree (°) symbol because Kelvin is an absolute scale, based on absolute zero, while the zero on the Celsius scale is based on the properties of water. I hope this helps. Thanks

Answer:

[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

Explanation:

Electric Potential of Point Charges

The electric potential from a point charge Q at a distance r from the charge is

[tex]\displaystyle V=k\frac{Q}{r}[/tex]

Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.

We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is

[tex]d=\sqrt2 a[/tex]

where a is the length of the side.

The distance from any corner to the center is half that diagonal, thus

[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]

The total potential in the center is  

[tex]V_t=V_1+V_2+V_3+V_4[/tex]

Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.

[tex]V_t=4\times V[/tex]

[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]

Operating

[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

Thus:

[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]

Answer:

time takes for the bullet to strike the ground is 0.5711 second

Explanation:

given data

height h = 1.60 m

initial speed u = 1058 m/s

solution

we get here time that is express here as

time = [tex]\sqrt{\frac{2h}{g}}[/tex]     ......................1

put here value and we will get here time that is

time = [tex]\sqrt{\frac{2*1.60}{9.81}}[/tex]

time = 0.5711 second

so time takes for the bullet to strike the ground is 0.5711 second

Final answer:

The time it takes for a bullet to strike the ground when fired horizontally from 1.60 m is calculated using the equation t = √(2h/g). Substitute 1.60 m for h and 9.81 m/s² for g to obtain t ≈ 0.571 seconds.

Explanation:

Calculating the Time for a Bullet to Strike the Ground

To calculate the time it takes for a bullet to strike the ground when fired from a horizontal position 1.60 meters above the ground with an initial speed of 1058 m/s, we can use the equations of motion for uniformly accelerated motion (in this case, the acceleration due to gravity). Since the bullet is fired horizontally, there is no initial vertical velocity component, so the vertical motion can be treated as a free-fall problem.

Using the formula for the time of free fall (t) from a height (h):
t = √(2h/g), where g is the acceleration due to gravity (approximately 9.81 m/s²).

Plugging in the given height of 1.60 m and solving for t gives us:
t = √(2 * 1.60 m / 9.81 m/s²) = √(0.3261 s²) ≈ 0.571 seconds.

Therefore, the bullet will take approximately 0.571 seconds to hit the ground.

Answer:

a)  y = A x² , b)   A = - ½ g / v₀², c)    v₀ = 15.46 m / s

Explanation:

For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same

X axis

         x = v₀ₓ t

         t = x / v₀ₓ

Y Axis  

         y = [tex]v_{oy}[/tex] t - ½ g t²

As it shoots horizontally the vertical speed is zero

        y = - ½ g t²

We replace

         y = - ½ g (x / v₀ₓ)²

The initial speed is all horizontal

        v₀ₓ = v₀

         y = - ½ g / v₀²    x²

        y = A x²

b) the expression for the constant is

           A = - ½ g / v₀²

c) we look for the initial speed

           v₀² = - ½ g x² / y

   As the object falls below the exit point its height is negative

          v₀ = √ (- ½  9.8   3²/ (-0.21))

          v₀ = 15.46 m / s

Answer:

0.0167 m

Explanation:

f = Frequency = 18 GHz

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

When the speed of the wave is divided by the freqeuncy we get the wavelength

Wavelength is given by

[tex]\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{18\times 10^9}\\\Rightarrow \lambda=0.0167\ m[/tex]

The wavelength of the radio waves is 0.0167 m

Answer:

W = 60.62 ft-lbs

Explanation:

given,

Horizontal force = 7 lb

distance of push, d = 10 ft

angle of ramp, θ = 30°

Work done on the box = ?

We know,

Work done is equal to force into displacement.

W = F.d cos θ

W = 7 x 10 x cos 30°

W = 70 x 0.8660

W = 60.62 ft-lbs

Hence, work done on the box is equal to W = 60.62 ft-lbs

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under  gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I  am  sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

Explanation:

The given data is as follows.

     height (h) = 98.0 m,     speed (v) = 73.0 m/s,

Formula of height in vertical direction is as follows.

          h = [tex]\frac{gt^{2}}{2}[/tex],

or,      t = [tex]\sqrt{\frac{2h}{g}}[/tex]

Now, formula for the required distance (d) is as follows.

       d = vt

          = [tex]v \sqrt{\frac{2h}{g}}[/tex]  

        = [tex]73.0 m/s \sqrt{\frac{2 \times 98.0 m}{9.8 m/s^{2}}}[/tex]  

          = 326.5 m

Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.

The efficiency of the machine is defined as

[tex]\eta = \frac{W_{out}}{W_{in}}[/tex]

Here

Work out is the work output and Work in is the work input

To find the Work in we have then

[tex]W_{in} = \frac{W_{out}}{\eta}[/tex]

[tex]W_{in} = \frac{mgh}{\eta}[/tex]

Replacing with our values

[tex]W_{in} = \frac{(58)(9.8)(3)}{73\%}[/tex]

[tex]W_{in} = 2335.89J[/tex]

The work done by the applied force is

W = Fd

Here,

F = Force

d = Distnace

Rearranging to find F,

[tex]F = \frac{W}{d}[/tex]

[tex]F = \frac{2335.89J }{18}[/tex]

F = 129.77N

Therefore the force exerted on the machine after rounding off to two significant figures is 130N

The force exerted on the pulley system when a rope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m with an efficiency of 73 percent is about 129.42 Newtons.

To solve this problem, we need to understand the concept of machine efficiency and work. The efficiency of a machine is the ratio of output work to input work.

In this case, the efficiency of the pulley system is given as 73%. Meaning output work is 73% of the input work. The force exerted on the machine is equivalent to the input work divided by the distance pulled.

The output work (W_out) can be determined using the formula W_out = mass * gravity * height = 58 kg * 9.8 m/s^2 * 3.0 m = 1701.6 Joules.

Then, we can find the input work (W_in) using the efficiency formula: W_in = W_out / efficiency = 1701.6 J / 0.73 = 2329.59 Joules.

Finally, we can find the force exerted on the pulley system (F_in) by dividing the input work by the distance pulled: F_in = W_in / distance = 2329.59 J / 18.0 m = 129.42 Newtons.

So, the force exerted on the machine would be approximately 129.42 Newtons.

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Answer:

Explanation:

Given

Astronaut in space gives an equal push to baseball and bowling ball.

Since the mass of the bowling ball is more than the baseball so inertia associated with a bowling ball is more as compared to baseball

Force applied on baseball and bowling ball is the same so the acceleration of baseball will be more because the mass of baseball is less.

[tex]Force=mass\times acceleration[/tex]

If both are traveling with the same speed then momentum associated with them is given by-product of mass and velocity

Since the mass of the bowling ball is more, therefore, the momentum of bowling ball is more as compared to baseball

Answer:

The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

[tex]F= qE[/tex]

[tex]mg=qE[/tex]

[tex]V\times\rho\times g=qE[/tex]

[tex]q=\dfrac{V\times\rho}{E}[/tex]

[tex]q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}[/tex]

Put the value into the formula

[tex]q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}[/tex]

[tex]q=3.106\times10^{-16}\ C[/tex]

We need to calculate the quantization of charge

Using formula of quantization

[tex]n = \dfrac{q}{e}[/tex]

Put the value into the formula

[tex]n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}[/tex]

[tex]n=1941.25[/tex]

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].

Yes, quantization of charge is obeyed within experimental error.

Answer:

3.11 x 10^-16 C

Explanation:

radius of drop, r = 1.6 micro metre = 1.6 x 10^-6 m

Electric field, E = 46 N/C

density of oil, d = 0.085 g/cm³ = 85 kg/m³

Let the charge on the oil drop is q.

So, the weight of the drop is balanced by the electric force on the drop.

mass of drop x g = charge of the drop x electric field

m x g = q E

Volume x density x g = q E

[tex]\frac{4}{3} \pi\times r^{3}\times d\times g = q \times E[/tex]

[tex]\frac{4}{3} \pi\times 1.6^{3}\times 10^{-18}\times 85\times 9.8 = q \times 46[/tex]

q = 3.11 x 10^-16 C

Thus, the charge on the oil drop is 3.11 x 10^-16 C.  

To find out how far the firefighting crew should position their cannon from the building, we use the equations of projectile motion separately for the vertical and the horizontal components. Since there are two viable times that satisfy the conditions of motion, there are two potential positions of the cannon. This question combines the principles of physics, which are specifically related to projectile motion.

This question is about the physics of projectile motion, specifically under gravity. To find out how far from the building they should position their cannon, we have to deal with the vertical and horizontal components of the projectile motion separately.

First, we determine the time it takes for the water to reach the desired height. We get this by using the equation of motion: H = V*sin(theta)*t - 0.5*g*t^2, where H is the height above the ground level (10m), V is speed (25.0 m/s), sin(theta) is the vertical component of initial velocity, g is acceleration due to gravity (approximated as 10 m/s^2 for ease of calculation), and t is time. We can rearrange this equation to find t.

Next, we use the time obtained and the horizontal component of initial velocity (V*cos(theta)) to find the horizontal distance made by using the equation: s = V*cos(theta)*t.

The hint here is that there are two potential time values, a smaller one and a larger one, leading to two potential distances from the building that the cannon could be positioned. Both times satisfy the conditions of the motion.

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The firefighting crew should place their cannon either 8.0 meters or 56.1 meters from the building to reach a blaze 10.0 meters above ground.

To determine how far the firefighting crew should position their cannon, we need to analyze the projectile motion of the water. We'll break this down into x-direction (horizontal) and y-direction (vertical) components.

Step-by-Step Solution

1. Break down initial velocity into components

Initial speed, v₀ = 25.0 m/s

Angle, θ = 53.0°

v₀x = v₀ x cos(θ) = 25.0 x cos(53.0°) ≈ 15.0 m/s

v₀y = v₀ x sin(θ) = 25.0 x sin(53.0°) ≈ 20.0 m/s

2. Write the equations of motion

Vertical (y-direction): y = v₀y x t - 0.5 x g x t²

Given y = 10.0 m, v₀y = 20.0 m/s, g = 9.8 m/s²

Using the quadratic formula, we find the time (t) it takes for the water to reach 10.0 m:

10.0 = 20.0 x t - 0.5 x 9.8 x t²

4.9t² - 20.0t + 10.0 = 0

Solving for t, we get t ≈ 0.54 s or t ≈ 3.74 s

Both times are valid because water can take two paths: one on the way up (0.54 s) and one on the way down (3.74 s).

3. Calculate horizontal distance

Distance (x) = v₀x x t

For t ≈ 0.54 s: x = 15.0 x 0.54 ≈ 8.0 m

For t ≈ 3.74 s: x = 15.0 x 3.74 ≈ 56.1 m

Therefore, the cannon can be placed either 8.0 meters or 56.1 meters from the building.

Final answer:

To calculate the speed of a muon before it decays, the distance the muon travels and the time it takes are used in the formula L = v x t. For a distance of 9.5 cm and a time of 2.20 microseconds, the speed is found to be approximately 43,182 meters per second.

Explanation:

The question is seeking to calculate the speed of a muon before it decays. Assuming the mean lifetime of a muon is 2.20 microseconds (us) and the muon makes a track 9.5 cm long in this time, we can calculate the muon's speed. Let's consider the formula:

L = v x t

where L is the distance traveled, v is the velocity, and t is the time. Given L = 9.5 cm (or 0.095 m) and t = 2.20 x 10-6 s, we plug in the values:

v = L/t

v = 0.095 m / 2.20 x 10-6 s

v = 4.31818 x 104 m/s

The muon's speed was approximately 43,182 m/s.

Springfield has the smallest slope.

Answer: Option B.

Explanation:

Springfield is a city in the United States of America. This city is in the states of Massachusetts. It is besides the river Connecticut river which adds to the beauty of this state.

Springfield is the nick name of the city "The queen of the Ozarks". The other name of this city is "the cultural center of the Ozarks". It has a beautiful scenery and this scenic beauty adds to the tourist attraction to a lot of people all around the world.

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = [tex]20\times 1=20\ m[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s[/tex]

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

The final speed of the motorist as she reaches the light is 5.72 m/s

To calculate the speed of the motorist as she reaches the light, we need to first find the deceleration of the motorist.

Formula:

Where:

Given:

Therefore,

Then,

Where:

we use the formula below to calculate the deceleration of the motorist.

Where:

Given:

Substitute these values into equation 2

Solve for a

Finally, we use the formula below to get the speed of the motorist as she reaches the light.

Given:

Substitute these values into equation 3

Hence, The final speed of the motorist as she reaches the light is 5.72 m/s

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Answer:

a)    vₓ = 6,457 m / s ,  v_{y}  = 0.518 m / s , b)    v = 6.478 m / s,      θ = 4.9°

Explanation:

a) This is a kinematic problem, let's use trigonometry to find the components of acceleration

           sin 31 = [tex]a_{y}[/tex] / a

           cos 31 = aₓ = a

           a_{y} = a sin31

           aₓ = a cos 31

Now let's use the kinematic equation for each axis

X axis

         vₓ = v₀ₓ + aₓ (t-t₀)

         vₓ = v₀ₓ + a cos 31 (t-t₀)

         vₓ = 2.6 + 0.45 cos 31  (20-10)  

         vₓ = 6,457 m / s

Y Axis

       v_{y} = v_{oy} + a_{y} t

       v_{y} = v_{oy} + a_{y}  sin31 (t-to)

       v_{y} = -1.8 + 0.45 sin31   (20-10)

       v_{y}  = 0.518 m / s

b) let's use Pythagoras' theorem to find the magnitude of velocity

       v = √ (vₓ² + v_{y}²)

       v = √ (6,457² + 0.518²)

       v = √ (41.96)

       v = 6.478 m / s

We use trigonometry for direction

       tan θ = v_{y} / vₓ

       θ = tan⁻¹  v_{y} / vₓ

       θ = tan⁻¹ 0.518 / 6.457

       θ = 4.9°

c) let's look for the vector at the initial time

        v₁ = √ (2.6² + 1.8²)

        v₁ = 3.16 m / s

        θ₁ = tan⁻¹ (-1.8 / 2.6)

        θ₁ = -34.7

We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.

          a = (v₂ -v₁) / (t₂-t₁)

Answer:

[tex]6.88\times 10^{-6}\ C/m^2[/tex]

Explanation:

Q = Total charge = [tex]6.3\times 10^{-8}\ C[/tex]

[tex]\rho[/tex] = Surface charge density

r = Radius = 2.7 cm

A = Surface area = [tex]4\pi r^2[/tex]

Charge is given by

[tex]Q=A\rho\\\Rightarrow Q=4\pi r^2\times \rho\\\Rightarrow \rho=\dfrac{Q}{4\pi r^2}\\\Rightarrow \rho=\dfrac{6.3\times 10^{-8}}{4\pi (2.7\times 10^{-2})^2}\\\Rightarrow \rho=0.00000687706544224\\\Rightarrow \rho=6.88\times 10^{-6}\ C/m^2[/tex]

The surface charge density is [tex]6.88\times 10^{-6}\ C/m^2[/tex]

Answer:

Explanation: see attachment

Answer:

a)-296 m/s

b)-176 m

c) 198 m

Explanation:

given data:

[tex]v_{oy}\\[/tex]=-20 m/s

[tex]v_{x}[/tex]= 15 m/s

t = 6 s

To find

a)[tex]H_{1}[/tex]

b)[tex]H_{2}[/tex]

c) d

d) find the velocity of observer at rest in basket and rest on ground

a) according to kinematic equation the displacement is given by:

[tex]H_{1}[/tex]=[tex]v_{oy}t\\[/tex]-[tex]\frac{1}{2} gt^{2}[/tex]..............(1)

putting values in 1

[tex]H_{1}[/tex]=-296 m/s

negative sign is due to direction and also we choose to take off point to be the origin

b) while the stone was travelling [tex]H_{1}[/tex] for 6 s so the ballon was also travelling displacement  [tex]H_{b}[/tex] with [tex]v_{oy}\\[/tex]

[tex]H_{b}[/tex] =[tex]v_{oy}t\\[/tex]

     =-6*20

     =-120 m

the height above the earth is given by

[tex]H_{2}[/tex]=[tex]H_{1}[/tex]-[tex]H_{2}[/tex]= -176 m

c) the horizontal displacement is given by

x=[tex]v_{x}t\\[/tex]=15*6=90 m

so the distance between balloon and stone is

d=[tex]\sqrt{H_{1}^{2}+x^{2} }[/tex]=198 m

d) the velocity component of the balloon :

1) relative to person rest in the balloon are given by

The vertical component:

        [tex]v_{yr} =v_{by} +v_{py}[/tex]=[tex](-gt)-0=9.8*6=-58.8 m/s[/tex]

The horizontal component:

         [tex]v_{xr} =v_{bx} +v_{px}= 15-0=15m/s[/tex]

2) relative to person rest in the earth are given by

The vertical component:

        [tex]v_{yr} =v_{by} -v_{py}=(-gt)-20=9.8*6-20=-78.8 m/s[/tex]

The horizontal component:

        [tex]v_{xr} =v_{bx} -v_{px}= 15-0=15m/s[/tex]