A bowling ball of mass m = 1.7 kg drops from a height h = 14.2 m. A semi-circular tube of radius r = 6.2 m rest centered on a scale. Write an expression for the reading of the scale when the bowling ball is at its lowest point, in terms of the variables in the problem statement.

The initial height H of the boy can be determined by adding the height of the slide h and the horizontal distance d the boy travels after leaving the track: H = h + d.

To determine the initial height H of the boy in terms of h and d, we can use the principle of conservation of energy. The total mechanical energy of the system remains constant throughout the motion.

At the top of the slide, the boy has gravitational potential energy given by mgh, where m is the mass of the boy, g is the acceleration due to gravity, and h is the height of the slide above the ground.

As the boy slides down the slide, there is no friction or other dissipative forces, so there is no change in mechanical energy. At the bottom of the track, the gravitational potential energy is converted entirely into kinetic energy.

Therefore, we can equate the initial potential energy to the final kinetic energy:

mgh = 1/2 m\(v^{2}\),

where v is the horizontal velocity of the boy when he leaves the track.

Since the boy leaves the track horizontally, the vertical component of his velocity is zero. Therefore, we can use the relationship between horizontal distance d and horizontal velocity v:

d = vt.

Solving these equations, we can express the initial height H in terms of h and d:

H = h + d.

So the initial height H of the boy can be determined by adding the height of the slide h and the horizontal distance d the boy travels after leaving the track.G

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The total energy of all the particles makes up thermal energy. This implies that larger objects with slower-moving particles at low temperatures can have more energy than smaller ones with higher temperatures.

Thermal energy is the sum of the energies of all the particles. This implies that larger objects at low temperatures can have more energy than smaller ones at higher temperatures due to slower-moving particles.

These molecules and atoms move more quickly when something is heated, providing it with more thermal energy. Compared to cold water, hot water has more thermal energy.

Therefore, molecules and atoms in colder objects move more slowly and have less thermal energy than those in warmer ones. The ability of particles to transfer energy increases with speed.

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Change in velocity, \(\Delta v\), is the initial velocity subtracted from the final velocity. \(\Delta v = v_{2} - v_{1} = 35 - 20 = 15kmh^{-1}\)

The change in the velocity of Maria's car is equal to 15 Km/h.

Velocity can be defined as a vector measurement of the rate of motion and direction of an object. The velocity of an object can be defined as the rate of change in the position of the body with respect to time.

Velocity can be defined as a vector parameter as it exhibits both magnitude and direction. A mathematical formula to calculate the velocity of the object can be represented as:

v = d / t

Where v is the velocity of the object, d is the distance, and t is the time taken by the object.

Although the S.I. unit for the measurement of the velocity is m/s and can also expressed in the terms of miles per hour (mph), and kilometers per hour(kph).

Given, the initial velocity of Maria's car, v₁ = 20 m/s

The final velocity of Maria's car, v₂ = 35 m/s

The change in the velocity of Maria's car, Δv = v₂ - v

Δv = = 35 - 20

Δv = 15 km/h

Therefore, the change in velocity is 15 Km/h.

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Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be \(723.91>SSD>620.2\)

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

\(SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}\)

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

\(SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}\)

\(SSD=723.9\ ft\)

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

\(SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}\)

\(SSD=620.2\ ft\)

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be \(723.91>SSD>620.2\)

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

\(SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}\)

\(SSD=910.5\ ft\)

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be \(723.91>SSD>620.2\)

(c), The SSD will be 910.5 ft.

Answer:

j120

Explanation:

qll energy for residential is 120 and that's what ruffly is always used for wiring

Answer:

805.48N/m

Explanation:

According to Hookes law

F = Ke

F is the force = mg

F = 2.4×9.8 = 23.52N

e is the extension = 2.92cm = 0.0292m

Force constant K = F/e

K = 23.52/0.0292

K = 805.48N/m

Hence the force constant of the spring is 805.48N/m

Answer:

5. An object of mass m = 2 kg, moving with velocity Vi1 = 12 m/s, collides head-on with a stationary object whose mass is m2 = 6 kg. The velocities of the objects after the collision are vj1 -6 m/s and Vr2 = 6 m/s.

Explanation:

We can use the conservation of momentum and kinetic energy to solve for the final velocities of the two objects.

Conservation of momentum:

m1v1i + m2v2i = m1v1f + m2v2f

where m1 and v1 are the mass and velocity of object 1 before the collision, and m2 and v2 are the mass and velocity of object 2 before the collision.

Plugging in the values:

(2 kg)(12 m/s) + (6 kg)(0 m/s) = (2 kg)(v1f) + (6 kg)(v2f)

Simplifying:

24 kg m/s = 2 kg v1f + 6 kg v2f

Conservation of kinetic energy:

(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2

Plugging in the values:

(1/2)(2 kg)(12 m/s)^2 + (1/2)(6 kg)(0 m/s)^2 = (1/2)(2 kg)(v1f)^2 + (1/2)(6 kg)(v2f)^2

Simplifying:

144 J = 1 kg v1f^2 + 3 kg v2f^2

Now we have two equations with two unknowns (v1f and v2f). Solving for v1f in terms of v2f in the first equation:

v1f = (24 kg m/s - 6 kg v2f)/2 kg = 12 m/s - 3v2f

Plugging this into the second equation:

144 J = 1 kg (12 m/s - 3v2f)^2 + 3 kg v2f^2

Simplifying and solving for v2f:

144 J = 1 kg (144 m^2/s^2 - 72 v2f + 9 v2f^2) + 3 kg v2f^2

144 J = 144 J - 72 kg m/s v2f + 9 kg m^2/s^2 v2f^2 + 3 kg v2f^2

6 kg v2f^2 - 72 kg m/s v2f + 144 J = 0

Dividing by 6 kg:

v2f^2 - 12 kg m/s v2f + 24 J/kg = 0

Using the quadratic formula:

v2f = [12 kg m/s ± sqrt((12 kg m/s)^2 - 4(1)(24 J/kg))]/(2)

v2f = [12 kg m/s ± sqrt(96) m/s]/2

v2f = 6 kg m/s ± 2sqrt(6) m/s

v2f ≈ 9.90 m/s or v2f ≈ 2.10 m/s

Plugging these values into the equation we found for v1f:

v1f = 12 m/s - 3v2f

v1f ≈ -16.70 m/s or v1f ≈ 38.70 m/s

Since the negative velocity doesn't make physical sense, the final velocities of the two objects are:

v1f ≈ 38.70 m/s and v2f ≈ 2.10 m/s

Answer:

5.0 s

Explanation:

The equation for time is:

\(t = \frac{d}{s} \)

We plug in the given values into the equation

\(t=\frac{16\ m}{3.2\ m/s^}\)

\(t=5.0 \ s\)

Answer:

ratio variables

Ratio scale of measurement

Weight, height and distance are all examples of ratio variables. Data in the ratio scale can be added, subtracted, divided and multiplied. Ratio scales also differ from interval scales in that the scale has a 'true zero'. The number zero means that the data has no value point

Answer:

~4%

Explanation:

% = |(7.6 - 7.9)|/7.9

= 0.3/7.9 ≈ 0.04 = 4%

Answer:

partecls

Explanation:

because they are to small to see with plain eyes

Graphical analysis of motion gives the graphical the relationships between position, velocity, and acceleration, versus time

2. Please find attached the required position versus time and acceleration versus time graphs

3. Please find attached the required velocity versus time and position versus time graphs

The reasons the attached graphs are correct are given as follows:

2. The given coordinates of the vertex points obtainable from the graph is first written as follows:

\(\begin{array}{|c|c|c|c|c|}\underline {Time, \ t}&\underline {Velocity, \ v}&Acceleration, a&\underline {Distance , \ s =v\cdot \Delta t+\dfrac{1}{2} \cdot a \cdot (\Delta t)^2}&Position\\0&-1&0&0&0\\2&1&1&0&0\\4&-1&-1&0&0\\6&1&1&0&0\\10&1&0&4&4\\12&0&-0.5&1&5\end{array}\)The distance covered between time intervals of time, s, is given as follows;

\(s = v \cdot \Delta t + (1/2) \cdot a \cdot ( \Delta t)^2\)

The position, Pₙ = s₁ + s₂ + s₃ + ... + sₙ

Other values on the graph obtained by calculation on a spreadsheet are;

\(\begin{array}{|l|cl|}Time \ (s)&&Position \ (m)\\0.5&&-0.375\\1&&-0.5\\1.5&&-0.375\\2.5&&0.375\\3&&0.5\\3.5&&0.375\\4.5&&-0.375\\5&&-0.5\\5.5&&-0.375\\6.5&&0.5\\7&&1\\7.5&&1.5\\8&&2\\8.5&&2.5\\9&&3\\10&&4\\10.5&&4.4375\\11&&4.75\\11.5&&4.9375\\12&&5\end{array}\right]\)

Please find attached the acceleration versus time graph

3. The coordinates of the points are presented as follows:

\(\begin{array}{|c|cc|}\underline{Time\ (s)}&&\underline{Acceleration\ (m/s^2)}\\0&&0.5\\2&&0.5\\2&&0\\4&&0\\4&&-0.5\\6&&-0.5\\6&&0\\8&&0\\8&&0.5\\10&&0.5\\10&&-0.5\\12&&-0.5\end{array}\right]\)

The distance covered between time intervals, s, is given as follows;

\(s = v \cdot \Delta t + (1/2) \cdot a \cdot ( \Delta t)^2\)

The position, Pₙ = s₁ + s₂ + s₃ + ... + sₙ

Using a spreadsheet application, more detailed values of the position can be found as shown in the graph created with MS Excel

0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10, 10.5, 11, 11.5, 12

0, 0.06, 0.25, 0.56, 1.00, 1.50, 2, 2.5, 3, 3.5, 3.81, 4.00, 4.06, 4.06, 4.06, 4.06, 4.06, 4.13, 4.31, 4.63, 5.06, 5.50, 5.81, 6, 6.06

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Answer:

Explanation:

Properties of the sun

Introduction to Stars A. The Sun 1. ...

Layers of the sun (Atmosphere) a. Photosphere: Layer that emits the radiation we see. ...

Layers of the sun (Interior) a. ...

Solar Wind: a. ...

The sun in X-Rays: a. ...

Luminosity: The measure of the energy put out by the sun. ...

Luminosity continued: c. ...

Luminosity continued: e.

The heat of the Sun's surface is so large that neither solid nor liquid can reside there. The heat is the physical property of the sun.

The Sun is a billion star at the core of our universe, a hot. It is an incandescent ball of hydrogen and helium.

The Sun is millions of kilometers away from the land, and life is not possible without its energy.

The major properties of the sun are as follows;

1. The mass of the sun is 1.98892 x 10³⁰ kg.

2 The density of the sun is 1.622 x 10⁵ kg/m³

3. The surface gravity of the sun is 27.94 g.

4. The diameter is 1,391,000 kilometers.

5. The volume of the Sun is 1.412 x 1018 km³

The heat of the Sun's surface is so hot that neither solid nor liquid can reside there instead, the component materials are mostly gaseous atoms with a few molecules. As a result, no permanent surface exists.

Hence heat, density, and mass are the major properties of the sun.

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Answer:

Yes, the horizontal Velocity is Constant

Explanation:

This is according to Newtons first law where an object in motion stays in motion until a force acts upon it. if there are no forces acting on a projectile in the horizontal direction then the velocity is constant.

Answer:

The bag of sand

Explanation:

I think it is the bag of sand because according to the definition of impulse, impulse is the average force acting on a particule when an external force is being acted on it.

The number and wattage of lamps required to illuminate the workshop would be approximately 8 lamps and 70 watts respectively.

To estimate the number and wattage of lamps required to illuminate a workshop space of 60x1.5 meters, we can follow these steps:

Calculate the area of the workshop:

Determine the total lumens required:

Adjust for the coefficient of utilization and luminous efficiency:

Adjust for space-height ratio and candle power depreciation:

Determine the number of lamps required:

Determine the wattage of each lamp:

Therefore, approximately 8 lamps with a wattage of 70 watts each would be required to illuminate the workshop space.

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Answer:

Did you forget to attach the image?

Explanation:

So far, we're not sure that you even HAVE a bulb or any switches.

If you had them, you probably would have included a diagram of the circuit you built with them. THEN we could look at the diagram and answer this question.

Right now, we can't.  We haven't seen such a diagram.

Explanation:

To determine the velocity and pressure head at position B in a horizontally converging pipe, we can use the principle of conservation of mass and Bernoulli's equation.

According to the principle of conservation of mass, the mass flow rate remains constant throughout the pipe. Therefore, we can write:

A₁V₁ = A₂V₂

where A₁ and A₂ are the cross-sectional areas at positions A and B, respectively, and V₁ and V₂ are the velocities at positions A and B, respectively.

Given:

A₁ = (π/4)(d₁)² = (π/4)(200 cm)² = 31416 cm²

A₂ = (π/4)(d₂)² = (π/4)(150 cm)² = 17671 cm²

V₁ = 2 m/s

We can calculate V₂ using the equation:

V₂ = (A₁V₁) / A₂

Substituting the values:

V₂ = (31416 cm² * 2 m/s) / 17671 cm² ≈ 3.54 m/s

Therefore, the velocity at position B is approximately 3.54 m/s.

Next, to determine the pressure head at position B, we can use Bernoulli's equation:

P₁ + (1/2)ρV₁² + ρgh₁ = P₂ + (1/2)ρV₂² + ρgh₂

Assuming the datum is at position B, where the pressure head (h₂) is zero, the equation simplifies to:

P₁ + (1/2)ρV₁² + ρgh₁ = P₂ + (1/2)ρV₂²

Given:

g = 10 m/s² (acceleration due to gravity)

Z = 0 m (datum)

ρ = density of the liquid (not given)

Since the density (ρ) of the liquid is not provided, we cannot determine the absolute pressure at position B or calculate the pressure head. The information given is insufficient to determine the pressure head at position B.

In summary:

- The velocity at position B is approximately 3.54 m/s.

- The pressure head at position B cannot be determined with the given information.

Answer:

The speed of the bus is 40 km/hr so this means the bus is travelling at a speed of 40 km per hour.

Answer:

Ions are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells. When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.

Explanation: CAN I RECEIVE A BRAINLIEST PLEASE

Answer:

peat them with my dislocated legs

Explanation:

Gravitational potential energy

= mgh

= (85kg)(9.81N/kg)(8848m)

= 7,377,904.8J.

a) The value of t u = 140/t`b.

b) The y position of the cannonball at the time t is  55.5 mc.

c) The initial speed of the projectile is 52.4 m/s.

Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.

a. To find the value of t:Here, we have to use the formula of distance

i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`

Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is

(2H) = 2 × 55.5

= 111 m`2H/g

= 111/9.8`

= 11.32653 s

From equation (1),u×t = 140u = 140/t

Therefore, `u = 140/t`b.

b)To find the y position of the cannonball at the time t:

Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,

Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².

We have already calculated the time taken by the cannonball to hit the ground in the previous step.`

y = 140 - (1/2) × 9.8 × t²`

On substituting the value of t as `t = 11.32653`,

we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.

c)  To find the initial speed of the projectile:

To calculate the initial speed of the projectile, we need to use the formula of range of projectile

.i.e.,R = u²sin2θ/g

Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)

u² = R × g/sin2θ

   = 140 × 9.8/sin0°  

   = 2744m²/s²u  

   = \(\sqrt(2744m^2/s^2)\)

   = 52.4 m/s

Hence, the initial speed of the projectile is 52.4 m/s.

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Answer:

In picture one because string telephones are best heard when there is more tension on the string

Explanation:

Answer:

\(v=\sqrt{26}~m/s\)

Explanation:

Parametric Equation of the Velocity

Given the position of the particle at any time t as

\(r(t) = (x(t),y(t))\)

The instantaneous velocity is the first derivative of the position:

\(v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))\)

The speed can be calculated as the magnitude of the velocity:

\(v=\sqrt{v_x^2+v_y^2}\)

We are given the coordinates of the position of a particle as:

\(x=5t-3t^2\)

\(y=5t\)

The coordinates of the velocity are:

\(v_x(t)=(5t-3t^2)'=5-6t\)

\(v_y(t)=(5t)'=5\)

Evaluating at t=1 s:

\(v_x(1)=5-6(1)=-1\)

\(v_y(1)=5\)

The velocity is:

\(v=\sqrt{(-1)^2+5^2}\)

\(v=\sqrt{1+25}\)

\(\mathbf{v=\sqrt{26}~m/s}\)