A bored kid, 30 meters from a building throws a ball at an angle of 50 degrees with a velocity of 20m/s at the building wall. at what height above the throwing line will the ball hit the wall?

Its orbital period is about 5.6 hours

[tex]\texttt{ }[/tex]

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

speed of satellite = v = 5000 m/s

mass of earth = M = 6 × 10²⁴ kg

Asked:

orbital period = T = ?

Solution:

[tex]\Sigma F = ma[/tex]

[tex]G \frac{ M m} { R^2 } = m \frac{v^2}{R}[/tex]

[tex]G \frac{ M } { R^2 } = \frac{v^2}{R}[/tex]

[tex]G \frac{ M } { R } = v^2 [/tex]

[tex]R = G \frac{ M } { v^2 } [/tex]

[tex]2 \pi R = 2 \pi G \frac{ M } { v^2 } [/tex]

[tex]\frac{ 2 \pi R }{ v } = (2 \pi G \frac{ M } { v^2 } ) \div v[/tex]

[tex]T = 2 \pi G \frac{ M } { v^3 }[/tex]

[tex]T = 2 \pi \times 6.67 \times 10^{-11} \times \frac{ 6 \times 10^{24} } { 5000^3 }[/tex]

[tex]T \approx 20116 \texttt{ s}[/tex]

[tex]T \approx 5.6 \texttt{ hours}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

The orbital period of a satellite moving at 5000 m/s in a circular orbit around Earth is approximately 8500 seconds.

To find the orbital period (T) of a satellite moving in a circular orbit at a speed of 5000 m/s, we need to use the formula for orbital speed:

[tex]v = \frac{2\pi r}{T}[/tex]

Here, v is the orbital speed, and r is the radius of the orbit. We can rearrange this formula to solve for the period (T):

[tex]T = \frac{2\pi r}{v}[/tex]

First, we need to find the radius of the satellite's orbit. Assuming the satellite is close to Earth's surface, we'll use Earth's radius plus an additional small altitude:

r = 6371 km + 400km = 6.771 x 10⁶ m

Now, substitute the values into the formula:

[tex]T = \frac{2\pi (6.771 \times 10^6 \, \text{m})}{5000 \, \text{m/s}}[/tex]

Calculating this gives:

T ≈ 8515.68 seconds

Expressing the answer with two significant figures:

T ≈ 8500 seconds

The orbital period of a satellite moving at 5000 m/s in a circular orbit around Earth is approximately 8500 seconds. This calculation uses the Earth's radius and the orbital speed provided.

Answer: The answer is B. Add more solute (took test)

Explanation:

Answer:

It's definitely B add more solute.

Explanation:

If a 0.14-kg baseball is dropped from rest from a height of 2.0 m above the ground, then the magnitude of its momentum just before it hits the ground would be  0.8764 kg - m / s.

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

As given in the problem if a 0.14-kg baseball is dropped from rest from a height of 2.0 m above the ground.

The velocity just before hitting the ground = √ ( 2 × 9.8 × 2)

                                                                       = 6.26 m / s

The magnitude of its momentum just before it hits the ground = 0.14 × 6.26

= 0.8764 kg - m / s

Thus, the magnitude of its momentum just before it hits the ground would be  0.8764 kg - m/s

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To calculate the Gibbs free energy of the eutectic melt relative to unmixed liquid gold and liquid silicon, you need to use the equation for Gibbs free energy. For part (a), assume that the Gibbs free energy change is zero for the unmixed liquids, and for part (b), assume it is zero for the unmixed solids. Substitute the given values into the equation to find the answer.

In a eutectic system, gold and silicon are mutually insoluble in the solid state. The eutectic system has a eutectic temperature of 636 K and a eutectic composition of xsi=0.186.

To calculate the Gibbs free energy of the eutectic melt relative to (a) unmixed liquid gold (Au) and liquid silicon (Si), and (b) unmixed solid gold (Au) and solid silicon (Si), we need to use the equation for Gibbs free energy (ΔG = ΔH - TΔS).

For (a), we can assume that the Gibbs free energy change (ΔG) is equal to zero for the unmixed liquids. The equation becomes ΔG(melt) = ΔH(melt) - TΔS(melt). Since the eutectic composition is given as xsi=0.186, we can assume that 0.186 moles of gold and 0.814 moles of silicon are present in the eutectic melt.

The change in enthalpy (ΔH) can be calculated by taking the eutectic temperature (636 K) and multiplying it by the change in entropy of mixing (ΔS(mix)). The ΔS(mix) can be calculated using the equation: ΔS(mix) = xAu · R · ln(xAu) + xSi · R ·

ln(xSi), where xAu is the mole fraction of gold and xSi is the mole fraction of silicon. Substituting the values, we can calculate ΔH(melt) = 1717.28 J/mol and ΔS(mix) = -2.25 J/(mol·K). Finally, we can substitute all the values into the Gibbs free energy equation to find the answer.

For (b), we can assume that the Gibbs free energy change (ΔG) is equal to zero for the unmixed solids. The equation becomes ΔG(melt) = ΔH(melt) - TΔS(melt).

The change in enthalpy (ΔH) for the unmixed solids can be calculated using the heat of fusion (ΔH(fus)) and the mass of gold and silicon in the eutectic melt. The ΔS(mix) for the unmixed solids can be calculated using the equation:

ΔS(mix) = R · ln(1 - xsi) + R · ln(1 - xAu), where xsi is the eutectic composition and xAu is the mole fraction of gold. Substituting the values, we can calculate ΔH(melt) = 27.39 J/mol and ΔS(mix) = 6.72 J/(mol·K). Finally, we can substitute all the values into the Gibbs free energy equation to find the answer.

The correct forms of the conservation of energy equation are chosen for different scenarios based on the type of energy transformations occurring. For a toaster, ΔEint = Q is more appropriate. For a car moving from a gas station, ΔK = W + Q and ΔU = W + Q are relevant. When eating a sandwich, ΔEint = W + Q is appropriate. For a house with constant temperature, 0 = Q + TMT + TET + TER is relevant.

The conservation of energy equation in physics considers various energy forms in the system, including kinetic energy (represented as ΔK), potential energy (represented as ΔU), and internal energy (represented as ΔEint). In each scenario, different energy forms come into play.

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The tensions in the ropes, from smallest to largest, are T1, T3, and T2.

When the boxes are in motion and there is no friction between the boxes and the horizontal surface, the tensions in the ropes can be ranked from smallest to largest as follows:

1. Tension in rope T1:

This is the smallest tension because it only needs to support the weight of box 1.

As long as box 1 is not accelerating vertically, the tension in T1 is equal to the weight of box 1.

2. Tension in rope T3:

This tension is greater than the tension in T1 because it needs to support the weight of both box 1 and box 2.

Since the two boxes are connected by T3, the tension in T3 is equal to the sum of the weights of box 1 and box 2.

3. Tension in rope T2:

This is the largest tension because it needs to support the weight of box 3, as well as the combined weight of box 1 and box 2.

Since both box 1 and box 2 are connected to box 3 by T2, the tension in T2 is equal to the sum of the weights of box 1, box 2, and box 3.

Hence, the tensions in the ropes, from smallest to largest, are T1, T3, and T2.

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The tensions in the ropes (t1, t2, and t3) when the boxes are in motion and frictionless can be ranked as follows: t1 < t2 < t3. by the principles of Newton's second law of motion.

When the boxes are in motion on a frictionless surface, the net force acting on each box is equal to its mass multiplied by its acceleration

(F = ma). Since all boxes experience the same acceleration, the ranking of tensions can be determined by comparing the magnitudes of the forces.

t1 corresponds to the box with the smallest mass. Its tension is just enough to overcome the gravitational force pulling it downward.

t2 corresponds to the middle box. It experiences a tension slightly greater than t1, as it needs to overcome both its own weight and the weight of the box below it.

t3 corresponds to the largest box. It experiences the highest tension among the three ropes since it needs to overcome its own weight and the combined weight of the other two boxes above it.

Therefore, the ranking of tensions from smallest to largest is t1 < t2 < t3, reflecting the relationship between mass, acceleration, and force in accordance with Newton's laws of motion.

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If the speed of an object increases, its kinetic energy increases.

If the speed of an object increases, its kinetic energy increases. Kinetic energy is directly proportional to the square of the object's velocity. As the speed of an object increases, the kinetic energy increases at a faster rate. This relationship can be described by the equation:

Kinetic energy = (1/2) × mass × velocity²

Since the velocity is squared in the equation, any increase in velocity will have a greater impact on the kinetic energy.

Therefore, as the speed of an object increases, its kinetic energy also increases.

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The magnitude of the force F depends on the other two forces acting on the ring. It should be such that the vector sum of all three forces equals zero, due to the object's static state, which suggests a condition of equilibrium.

In the question, you have a ring being pulled on by three forces and it remains stationary. This situation is described by the physics concept of equilibrium, where the sum of all forces acting on an object equals zero because the object isn't moving. The forces here are vectors, meaning they have both magnitude (how big they are) and direction (which way they're pulling). Each force acts in a specific direction, which we'll assume are different for each one.

Now, the size of the force F would depend on the other two forces. Without exact information on the magnitude and direction of the other two forces, we cannot precisely compute F. However, we can say that F will be such that the vector sum of the all forces (including F) will be zero. This happens because no movement implies no net force according to Newton's second law. This law states that an object at rest, like the one in your assignment, stays at rest unless acted upon by a non-zero net force.

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Answer:

The correct answer is B

Explanation:

The motion of a pendulum is a classic example of mechanical energy conservation. So you can assume, ignoring  air resistance, that total mechanical energy is:

[tex]E = K + U[/tex]

Where K is kinetic energy, U is potential energy and E is the resulting mechanical energy.

You know that:

[tex]U = mgh\\K = \frac{1}{2}mV^{2}[/tex]

From this, you can deduce that for potential energy to be transformed entirely to kinetic energy the height of the pendulum mass must be zero and traveling at its maximum velocity, wich is depicted as B in the attached picture.

Final answer:

The slowest constant speed that the green car can maintain and still catch up to the blue car is 3.75 meters per second.

Explanation:

To find the slowest constant speed which the green car can maintain and still catch up to the blue car, we need to determine the distance that the blue car travels during the 5-second interval before the green car arrives at the position of the stop-light.

Using the equation of motion for constant acceleration, we can calculate the distance traveled by the blue car with an initial velocity of 0 and an acceleration of 0.3 m/s²:

d = ut + 1/2at²

Where d is the distance, u is the initial velocity, t is the time, and a is the acceleration.

Substituting the given values into the equation, we have:

d = 0(5) + 1/2(0.3)(5)²

d = 0 + 1/2(0.3)(25)

d = 1/2(7.5)

d = 3.75 meters

Therefore, the slowest constant speed that the green car can maintain and still catch up to the blue car is 3.75 meters per second.

kelvin is the right answer

The formula for gravitational force is:

F = G m1 m2 / r^2

where G m1 m2 are constants, therefore:

F r^2 = constant

Part b. Given F1 = 2000 N, r1 = 100 km

Find F2 = ?, r2 = 150 km

(2000 N) * (100 km)^2 = F2 * (150 km)^2

F2 = 888.89 N

Part c. Given F1 = 2000 N, r1 = 100 km

Find F2 = ?, r2 = 50 km

(2000 N) * (100 km)^2 = F2 * (50 km)^2

F2 = 8000 N

The answer is: That matter is indivisible and indestructible

Through his experiment, Henri Becquerel found that matters can be stored within another object, (He took potassium uranyl sulfate and put it under the sun and then put the substance within photographic plates that he cover with black paper, ) . Curie found that over time matters would decay and eventually gone.

The magnitude of Earth's velocity in its orbit around the sun is approximately 29,450 m/s.

Given that the radius of Earth's orbit around the sun is 1.5 x 10^11 m, we can calculate the magnitude of Earth's velocity using the formula for the circumference of a circle: v = 2 × Pi × r / T, where v is the velocity, r is the radius of the orbit, and T is the period of the orbit. Since one year is equivalent to 365 days, we have T = 365 days. Plugging in the values, we get:  v = 2 × 3.14159 × 1.5 × 10^11 m / (365 × 24 × 60 × 60 s)

Simplifying the expression gives us:v ≈ 29,450 m/s

Final answer:

The air pressure at the top of Mt. Everest is around 253 mm Hg (33.7 kPa) due to the high altitude, resulting in significantly lower oxygen levels and causing extreme drying of breathing passages due to the cold, thin air.

Explanation:

The air pressure at the top of Mt. Everest (8.85 km above sea level) is significantly lower than at sea level due to the altitude. Specifically, the atmospheric pressure on the summit of Mt. Everest can be as low as 253 mm Hg, which corresponds to about 33.7 kPa (kiloPascals) or 0.308 atm (atmospheres).

This reduced pressure means that the oxygen content is much lower as well, posing significant challenges to climbers such as reduced oxygen availability for breathing and the extreme drying of breathing passages.

The partial pressure of oxygen at this altitude, considering that it comprises 20.9% of the atmospheric composition, would be considerably less than at sea level.

Climbers often need supplemental oxygen to compensate for the lower oxygen levels. The extreme drying experienced by climbers at high altitudes occurs because the cold, thin air contains very little moisture, leading to rapid evaporation of moisture from the breathing passages.

Final answer:

The electric potential between the terminals of a dry cell that does 7.5 joules of work to transfer 5.00 coulombs of charge is 1.5 volts.

Explanation:

The student asked for the electric potential between the terminals of a dry cell that does 7.5 joules (J) of work to transfer 5.00 coulombs (C) of charge. To find the electric potential (also known as voltage) across the terminals, we use the relationship that the work done (W) by the cell through electrical energy is equal to the charge (Q) multiplied by the potential difference (V), which can be expressed as W = QV. Rearranging this equation for V gives us V = W/Q.

Substituting the given values into the equation, we have:

V = 7.5 J / 5.00 C = 1.5 volts (V)

Therefore, the electric potential between the two terminals of the dry cell is 1.5 V. This result is the voltage of the dry cell when it is not supplying current and is therefore at its electromotive force (emf).

Final answer:

In a chemical equation, an absent coefficient in front of a chemical formula indicates that the understood number is '1', signifying one mole of the substance in the reaction.

Explanation:

When there is no number in front of a chemical formula in a chemical equation, it is understood that the number is '1'. In a balanced chemical equation, coefficients are used to indicate the mole ratio in which the reactants combine to form products. If a chemical formula has no coefficient written before it, we assume the coefficient is 1. This means one mole of the substance is involved in the reaction. Coefficients are particularly important as they reflect the relative numbers of molecules or formula units in the reactants and products.

To find the magnitude of the resultant momentum after an inelastic collision with given x and y components, one applies the Pythagorean theorem. The resultant momentum is approximately 140.01 kilogram meters/second.

The question involves calculating the magnitude of the resultant momentum after two balls undergo an inelastic collision. The given momentum components are 98 kilogram meters/second in the y-direction and 100 kilogram meters/second in the x-direction. To find the resultant momentum, we use the Pythagorean theorem since momentum is a vector quantity.

Using the values given:
(resultant momentum) = (100^2 + 98^2)
(resultant momentum) = (10000 + 9604)
(resultant momentum) = (19604)

The magnitude of the resultant momentum is approximately 140.01 kilogram meters/second.

The angular speed at twice the initial radius becomes half of the initial angular speed.

Further Explanation:

Speed is the measure of a quantity of an object the tells how fast the object is moving in the other words we can define the speed that it is the distance covered by an body divided by the time taken to cover that distance. It is a quantity with only magnitude so it is a scalar quantity.

Given:

The certain angular speed is [tex]{\omega _2}[/tex].

The radius is [tex]r[/tex].

Concept:

The expression for the linear motion can be written as:

[tex]\fbox{\begin\\v=\dfrac{s}{t}\end{minispace}}[/tex]                               …… (1)

Here, [tex]v[/tex] is the linear speed, [tex]s[/tex] is the total distance covered and [tex]t[/tex] is the time taken to cover to distance.

The expression for the circular speed can be written as:

[tex]\fbox{\begin\\\omega=\dfrac{\theta }{t}\end{minispace}}[/tex]

Here, [tex]\omega[/tex] is the circular speed and [tex]\theta[/tex] is the angular displacement.

The expression for the total distance covered in term of angular displacement is:

[tex]s=\theta r[/tex]

Substitute [tex]\theta r[/tex] for [tex]s[/tex] in equation (1)

[tex]\begin{aligned}v&=\frac{{\theta r}}{t}\hfill\\v&=\frac{\theta }{t}\cdot r\hfill\\v&=\omega\cdot r\hfill\\\omega&=\frac{v}{r} \hfill\\ \end{aligned}[/tex]

From above expression the angular speed is inversely proportional to the radius.

[tex]\fbox{\begin\\\omega\propto\dfrac{1}{r}\end{minispace}}[/tex]

That is if the radius increases the angular speed decreases and if the radius decreases the angular speed increases.

Considered the linear speed remain content.

Case 1:

The angular speed is given by

[tex]{\omega _1}=\dfrac{v}{r}[/tex]                                                               …… (2)

Case 2:

The radius is double that is [tex]2r[/tex].

The angular speed is given by

[tex]{\omega _2}=\dfrac{v}{{2r}}[/tex]                                                         …… (3)

Divide equation (2) by equation (3).

[tex]\begin{aligned}\frac{{{\omega _2}}}{{{\omega _1}}}&=\frac{{\frac{v}{{2r}}}}{{\frac{v}{r}}}\hfill\\\frac{{{\omega _2}}}{{{\omega _1}}}&=\frac{1}{2}\hfill\\{\omega _2}&=\frac{{{\omega _1}}}{2}\hfill\\\end{aligned}[/tex]

Therefore, the angular speed at twice the initial radius becomes half of the initial angular speed.

Learn more:

1. Angular speed https://brainly.in/question/7744338

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Answer Details:

Grade: college

Subject: Physics

Chapter: Kinematics

Keywords:

Certain, angular speed, w2, radius, r, twice, w1/2, half, initial angular speed.

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