A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.6 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.8 m/s from a height of 28.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
1) What is the speed of the blue ball when it reaches its max. height?
2) How long does it take the blue ball to reach its maximum height?
3) What is the maximum height the blue ball reaches?

me answer be ye Conduction

The correct answer is A. Sediment

Answer: A. Sediment

Eutrophication is a phenomena in which a water body gets enriched with minerals and nutrients frequently entered into water body due to run-off from the land, which facilitates dense growth of plants especially on the superficial layers of water body. Decaying organisms, fertilizers and pollution all add nutrients and minerals necessary for plant growth specifically remain floating on the superficial layers of water where plant growth takes place but sediments gets settled at the bottom of the water body, the minerals present in it remain settle at the bottom and hence, does not contribute to eutrophication.

The answer is A. Hope this helps! :)

Answer:

[tex]H = 273.4 m[/tex]

Explanation:

Let the falling object took "n" seconds to reach the ground and it travels H height

So we will have

[tex]H = \frac{1}{2}gn^2[/tex]

now we know that it covers one fourth of total height in last second

So we can say that it will cover 3H/4 distance in (n-1) seconds

so we will have

[tex]\frac{3H}{4} = \frac{1}{2}g(n-1)^2[/tex]

now from above two equations

[tex]\frac{4}{3} = (\frac{n}{n-1})^2[/tex]

[tex]1.155(n-1) = n[/tex]

[tex]n = 7.46 s[/tex]

Now we have

[tex]H = \frac{1}{2}(9.81)(7.46^2)[/tex]

[tex]H = 273.4 m[/tex]

Yes, the stuntman can jump from one building to the other building.

Further explanation:

The stuntman jump from one building inclined at some angle to the other traversing a parabolic path.

Given:

The velocity of stuntman is 5m/s.

The distance between the buildings is 4m.

The difference in the height of the buildings is 2.5m.

The angle of inclination is [tex]{15^ \circ }[/tex].

Concept used:

When stuntman jumps from the top of one building to the top of other building he start running first then he jump at a velocity of 5m/s inclined at an angle of [tex]{15^ \circ }[/tex] from the horizontal of first building.

The velocity of stuntman has two components [tex]{v_x}[/tex] and [tex]{v_y}[/tex], respectively in the X-direction and in the Y-direction.

The expression for the distance in horizontal direction is given as.

[tex]x = \left( {v\cos \theta } \right)t[/tex]

Rearrange the above expression for time.

[tex]t = \dfrac{x}{{\left( {v\cos \theta } \right)}}[/tex]                             …… (1)

Here, t is the time of flight, v is the velocity of object and [tex]\theta[/tex]  is the angle of inclination.

The expression for the distance in Y-direction is given by the second equation of motion.

[tex]y = ut + \frac{1}{2}\left( { - g} \right){t^2}[/tex]                        

Here, u is the velocity in Y-direction and (–g) is the acceleration due to gravity directed in the downward direction.

The expression for the component of velocity in Y-direction is given as.

[tex]u = v\sin \theta[/tex]

Substitute [tex]v\sin\theta[/tex] for u in the above expression.

[tex]y = \left( {v\sin \theta } \right)t + \frac{1}{2}\left( { - g} \right){t^2}[/tex]                       …… (2)

Substitute 5m/s for v, 4m for x and [tex]{15^ \circ }[/tex] for   in equation (1).

[tex]\begin{aligned}t&=\frac{{4\,{\text{m}}}}{{\left({\left( {5\,{\text{m/s}}}\right)\left( {\cos {{15}^ \circ }}\right)}\right)}}\\&=0.828\,{\text{s}}\\\end{aligned}[/tex]

Substitute [tex]0.828\,{\text{s}}[/tex] for t, 5m/s for v, [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for g and [tex]{15^ \circ }[/tex] for [tex]\theta[/tex] in equation (2).

[tex]\begin{aligned}y&=\left( {\left( {5\,{\text{m/s}}} \right)\sin \left( {{{15}^ \circ }} \right)} \right)\left( {0.828\,{\text{s}}} \right) + \frac{1}{2}\left( { - 9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right){\left( {0.828\,{\text{s}}} \right)^2}\\&=1.071 - 3.36 \\&=- 2.29\,{\text{m}}\\\end{aligned}[/tex]

Thus, the stuntman can jump from one building to another because its vertical distance is less than the difference in height of buildings.

Learn more:

1.  Motion under force https://brainly.com/question/6125929.

2.  Projection of ball https://brainly.com/question/11023695.

3. Conservation of momentum https://brainly.com/question/9484203.

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Force, motion, parabolic path, acceleration due to gravity, time of flight, angle of inclination, inclined plane, buildings, height difference, cliff, horizontal direction, vertical direction, 2.29 m, 2.3m, 0.828sec.

The question involves calculating whether a stuntman can jump from one rooftop to another using principles of projectile motion. By calculating the horizontal and vertical components of the stuntman's velocity, and the time and distance he will travel, we determine he will fall approximately 2.9 meters. Since the other roof is only 2.5 meters lower, the stuntman will successfully make the jump.

The student's question involves determining whether a stuntman can safely jump from one building to another, given his initial velocity, the angle of his jump, and the distance and height difference between the buildings. This is a classic physics problem involving projectile motion, where we consider the horizontal and vertical components of the motion separately. We can ignore air resistance and use the kinematic equations to solve this problem.

Firstly, we need to find the horizontal and vertical components of the initial velocity. The horizontal component, vx, is v × cos(θ), and the vertical component, vy, is v × sin(θ), where v is the initial velocity and θ is the angle of launch. For a jump of 5.0 m/s at 15°:

The time, t, it takes to travel horizontally across the 4.0 m gap is found using t = d / vx, where d is the horizontal distance:

t = 4.0 m / 4.83 m/s ≈ 0.83 s

Now, let's determine if he will drop less than 2.5 m in this time frame. We use the equation of motion for vertical displacement, Δy = vyt + ½gt², with g being the acceleration due to gravity (9.81 m/s²) to find the vertical drop:

Δy = (1.29 m/s × 0.83 s) + (0.5 × -9.81 m/s² × (0.83 s)²) ≈ -2.9 m

The stuntman will fall approximately 2.9 m, thus he will make it to the other roof, which is 2.5 m lower than his initial jump point.

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The problem involves applying conservation of momentum principles to a two-dimensional collision between an apple and an orange tossed by astronauts. The final direction of the apple after the collision can be determined using the velocity components and trigonometry, specifically the arctan function. To provide the direction, additional details like initial direction are needed.

The student is seeking help with a physics problem involving the concepts of momentum conservation and two-dimensional collisions. In this problem, two astronauts (in a hypothetical scenario) are tossing an apple and an orange to each other when the fruits collide in space. The provided information lets us analyze the collision using the principles of momentum to find the final direction of the apple after the collision.

To solve for the direction of the apple after the collision, we must apply the law of conservation of momentum in two dimensions because the collision sends the objects off in different directions. Since there are no external forces, the total momentum in each direction (x and y) should remain constant. We use the before-collision and after-collision momenta to form equations and solve for the final velocity components of the apple, allowing us to calculate the final direction using trigonometry.

However, to provide the exact direction, a diagram or more information indicating the initial directions of the fruits would be required. Assuming the initial throw was along the x-axis and the final velocity of the orange is given in the y-axis, you can apply the arctan function to the velocity components of the apple to find the direction in terms of angle from the x-axis.

The units can be best determined by expressing them both in terms of SI base units.

The Système International (SI) unit is named after the French word for it. The International System of Units (ISO) is the name of the metric system that is used as the industrial standard for measurements (SI). SI units are crucial for the growth of science and technology.

It is composed of 7 base units, from which 22 derivative units are generated. Either a standard multiple or a fractional quantity can be used to express SI units. Prefix multipliers with powers of 10 in the range from [tex]10^{-24}[/tex]  To [tex]10^{24}[/tex]  Are used to define these numbers.

Now, according to the question :

J=kg×m²/s²

J/m²=kg/s²

and, N=kg×m/s²

⇒m×s²/(kg×m/s²)

=[tex]s^{4}[/tex]/kg

Hence, after comparing both of them, we can say that they are not equivalent.

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Accepting the null hypothesis when the experimental hypothesis is true is called type error.

The type error object denotes a failure to perform an operation.When a value is not of the anticipated type.

When an operand or argument provided to a function is incompatible with the type required by that operator or function,the condition of type error is found.

Accepting the null hypothesis when the experimental hypothesis is true is called type error.

Hence,type error is the correct answer for the blank.

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The two balls meet at approximately 272.5 meters above the surface of the tower after around 5.1 seconds.

To determine when and at what height the two balls meet, we'll analyze their motions individually and then together.

Step 1: Motion of the dropped ball

For the ball dropped from the roof at a height of 400 meters, the equation of motion under gravity is:

h₁ = 400 - 0.5 * g * t²

where:

h₁ is the distance traveled by the dropped ball in meters.

g is the acceleration due to gravity (9.8 m/s²).

t is the time in seconds.

Putting in the values, h₁ = 400 - 4.9 t²

Step 2: Motion of the thrown ball

For the ball thrown upwards with an initial velocity of 50 m/s, the equation of motion is:

h₂ = 50t - 0.5 * g * t²

where:

h is the distance traveled by the thrown ball in meters.

Putting in the values, h₂ = 50t - 4.9 t²

Step 3: Equating the distances

The balls meet when their total distance is 400 meters, so:

h₁ + h₂ = 400

Substituting the equations for h₁ and h₂, 400 - 4.9 t² + 50t - 4.9 t² = 400

Simplifying, -9.8 t² + 50t = 0

Factoring out t, t(50 - 9.8t) = 0

Thus, t = 0 (initial point, not useful) or t = 50 / 9.8 ≈ 5.1 seconds.

Step 4: Calculating the height

Using t ≈ 5.1 seconds in any height equation:

h₁ = 400 - 4.9 * (5.1)²

h₁ = 400 - 127.5

h₁ ≈ 272.5 meters

Therefore, the two balls meet approximately 272.5 meters above the surface of the tower after about 5.1 seconds.

To vaporize 3.0 L of liquid oxygen with a density of 1.14 kg/L and a heat of vaporization of 213 kJ/kg, 728460 joules of energy would be absorbed.

The student is asking about the amount of energy required to vaporize a certain volume of liquid oxygen. Given that the density of liquid oxygen at its boiling point is 1.14 kg/L, and its heat of vaporization is 213 kJ/kg, we can calculate the energy needed for vaporization using these two properties.

First, calculate the mass of 3.0 L of liquid oxygen:

Mass = Density times Volume = 1.14 kg/L times 3.0 L = 3.42 kg

Then, calculate the energy required for vaporization:

Energy = Mass times Heat of Vaporization = 3.42 kg times 213 kJ/kg = 728.46 kJ

Since 1 kJ = 1000 J, we can convert the energy to joules:

Energy in joules = 728.46 kJ times 1000 J/kJ = 728460 J

Therefore, 728460 joules of energy would be absorbed by 3.0 L of liquid oxygen as it vaporizes.

Answer:

Acceleration, [tex]a=4.83\ m/s^2[/tex]              

Explanation:

It is given that,

Mass of the meteor, m = 521 kg

Force acting on the meteor, F = 2520 N

Let a is the acceleration of the meteor. It can be calculated using the Newton's second law of motion. According to this law the force acting on an object is equal to the product of mass and acceleration with which it is moving. Mathematically, it is given by :

[tex]F=m\times a[/tex]

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{2520\ N}{521\ kg}[/tex]

[tex]a=4.83\ m/s^2[/tex]

So, the acceleration of the meteor is [tex]4.83\ m/s^2[/tex]. Hence, this is the required solution.

Answer;

-Reference point

The x represents the reference point on a motion map.

Explanation;

-Motion maps are another way to represent the motion of an object.  (other representations are graphical and mathematical models).

-Motion maps are a visual way to represent an object’s motion at various time

-. A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point.

-The relationships between motion and reference point is that When an object changes position over time relative to a reference point, the object is in motion. This means that there relationship is they are similar.

The x represents a reference point on a motion map. The correct option is A.

Another technique to depict an object's motion is motion maps. Models in mathematics and graphics are alternative forms of representation. A visual representation of an object's motion through time is called a motion map.

A reference point is a location or object that is used as a point of comparison to ascertain whether something is moving. When an object shifts in relation to a fixed point, it is said to be in motion.

The relationship between motion and reference point is that an object is in motion when its position with respect to a reference point varies over time. This indicates that they have a similar relationship.

Therefore, the value x represents a reference point on a motion map. The correct option is A.

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Answer:

E₃ = energy in third energy level.

Explanation:

The energy of an electron is third orbit is given by :

[tex]E=-\dfrac{13.6Z^2}{n^2}[/tex]

Where

Z is the atomic number

n is principal quantum number and n = 1,2,3....

i.e. the energy of an electron is inversely proportional to the principal quantum number.

E₁,E₂,E₃.......... shows the energy of an electron in first, second and third energy level respectively.

Answer: 3150 J

Explanation:

Work is the result when a force acts on an object and moves it by some distance and it is calculated in Joules.  

So,

Distance covered by the box = 35m

Force applied on the box = 90N

Formula:

The force and the distance covered was in the same direction, so the angle between them is 0°. The work can be found using the formula:

W = Fd cosθ

W = Fd cos0  

W = Fd(1)

W= (90) * (35)

W = 3150 J

Answer:

c. a difference in electric potential

Explanation:

Answer:

a = 0

Explanation:

When a baseball is thrown straight upward, it moves upward under the action of gravity. We know that at the highest point, the speed of the object is equal to 0. The rate of change of velocity is called acceleration of an object.

Since, the speed is 0 at the height point, so, its acceleration at its greatest height is 0. Hence, this is the required solution.

When a baseball is thrown straight upward, its acceleration at its greatest height is equal to: c. 0 [tex]m/s^2[/tex]

In Physics, the acceleration of an object or body is calculated by subtracting its initial velocity from the final velocity and dividing by the time.

Mathematically, acceleration is given by the formula;

[tex]Acceleration = \frac{V\; - \;U}{t}[/tex]

Where:  

A baseball that is thrown thrown straight upward, experiences motion due to the acceleration of gravity.

Also, the final speed of an object is equal to zero (0) at the greatest height because it came to a stop while the initial velocity is equal to zero (0), since it would accelerate from rest.

Deductively, the rate of change of velocity with respect to time is equal to zero (0) and this is the value of baseball's acceleration.

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Answer:

In The Atmosphere

Explanation:

edg2021

The mass of the aluminum cylinder is 4,050 kg, obtained by multiplying the given density of aluminum with the given volume of the cylinder.

The mass (m) of the aluminum cylinder can be calculated using the formula for density (ρ), which is Density = Mass/Volume. In this case, since the density of the aluminum (ϱ) is 2.70 * 103 kilograms per cubic meter (kg/m3) and the volume (V) of the cylinder is 1.50 m3, we can rearrange the formula to find mass such that Mass = Density * Volume. Substituting the given values, we get m = 2.70 * 103 kg/m3 * 1.50 m3 = 4,050 kg. Therefore, the mass of the aluminum cylinder is 4,050 kg.

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