What is the legal height the player is allowed to serve from a. Above the waist B. Below the waist C. On the waist D. above the chest

Complete Question

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Answer:

The kinetic energy for particle a is  [tex]K_a  =  \frac{9}{2}  mr^2 w^2[/tex]

The kinetic energy for particle b is   [tex]K_b  =  m w^2r^2 [/tex]

Explanation:

From the question we are told that

   The mass of particle a  is  [tex]m[/tex]  

   The mass of particle b is  [tex]2 m[/tex]  

Generally the kinetic energy for  particle a is mathematically represented as  

       [tex]K_a  =  \frac{1}{2} m v^2[/tex]

Here [tex]v_a[/tex] is the linear velocity of particle a which is mathematically represented as

         [tex]v_b  =  w *  3r[/tex]

So

      [tex]K  =  \frac{1}{2} m {w *  3r}^2[/tex]

=>   [tex]K_a  =  \frac{9}{2}  mr^2 w^2[/tex]

Generally the kinetic energy for  particle a is mathematically represented as  

    [tex]K_b  =  \frac{1}{2} 2m v_b^2[/tex]

Here [tex]v_b[/tex] is the linear velocity of particle a which is mathematically represented as

         [tex]v_b  =  w *  r[/tex]

So

      [tex]K_b  =  m (wr)^2[/tex]

=>    [tex]K_b  =  m w^2r^2[/tex]

The kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].

The given parameters;

The kinetic energy of the of the particle in terms of the angular velocity is calculated as follows;

[tex]K_b = \frac{1}{2} m (\omega r)^2\\\\K_b = \frac{1}{2} m \omega ^2 r^2[/tex]

where;

Thus, the kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].

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Answer:

1609.3

Explanation:

Answer:

Mass = 1450kg

Explanation:

P = M * V (where p is momentum, m is mass and v is velocity)

29000 = 20 * M

M = 29000 / 20

M= 1450 kg

Answer:

Explanation:

Given:

Distance S = 26.45m

Using the equation of motion

S = ut+1/2gt²

26.45 = 0+1/2(9.81)t²

26.45 = 4.905t²

t² = 26.45/4.905

t² = 5.392

t = 2.528s

Hence the it took the balloon 2.52s to reach the ground

Answer:

the third law

Explanation:

could u vote me brainliest plz? thx :)

Answer:

is this a mudder question

Answer:

Driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.

Explanation:

Let suppose that car is represented by a particle, then we proceed to construct its free body diagram and corresponding equations of equilibrium:

[tex]\Sigma F_{x'} = f - W\cdot \sin \theta = 0[/tex] (Eq. 1)

[tex]\Sigma F_{y'} = N-W\cdot \cos \theta = 0[/tex] (Eq. 2)

Where:

[tex]f[/tex] - Static friction force, measured in newtons.

[tex]N[/tex] - Normal force on the car from the ground, measured in newtons.

[tex]W[/tex] - Weight of the car, measured in newtons.

[tex]\theta[/tex] - Driveway inclination, measured in sexagesimal degrees.

By applying definitions of maximum static friction force and weight, we expand the system of equations presented above:

[tex]\mu_{s}\cdot N -m\cdot g\cdot \sin \theta = 0[/tex] (Eq. 1b)

[tex]N - m\cdot g \cdot \cos \theta = 0[/tex] (Eq. 2b)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]m[/tex] - Mass of the car, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

Then, we substitute normal force and simplify the resulting expression:

[tex]\mu_{s}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0[/tex]

[tex]\mu_{s} = \tan \theta[/tex]

[tex]\theta = \tan^{-1}\mu_{s}[/tex] (Eq. 3)

Based on such result, we can conclude that if driveway inclination is greater than value reported, then car shall not be safe in case of parking. Our reference angle is: ([tex]\mu_{s} = 0.15[/tex])

[tex]\theta = \tan^{-1} 0.15[/tex]

[tex]\theta \approx 8.531^{\circ}[/tex]

By direct comparison, we find that driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.

Answer:

the main character

Explanation:

i think

Answer:

The total displacement of the object is 20 m

Explanation:

Velocity vs Time Graph

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The distance covered by the car can be calculated as the area behind the graph.

The total area can be calculated in two parts: One triangle from t=0 sec to t=2 sec, and one rectangle from t=2 sec to t=3 sec.

The area of a triangle is

A1= B*H/2, where B is the length of the base, and H is the length of the Height. The base is B=2 seconds, and the height is H = 10 m/s, thus:

A1= (2 sec)*(10 m/sec)/2=10 m

The area of a rectangle is

A2=W*L, where W is the width, L is the length. W=1 sec, L=10 m/s, thus:

A2=(1 sec)*(10 m/sec) = 10 m

Total displacement= 10 m + 10 m = 20 m

The total displacement of the object is 20 m

Answer D

Explanation I did the quiz

Answer:

Alkali metals are the chemical elements found in Group 1 of the periodic table. The alkali metals include: lithium, sodium, potassium, rubidium, cesium, and francium.

Explanation:

Answer:

alkali metals

Explanation:

Answer: is  13

Explanation:

Answer:

Explanation:

16=m(2)*2

16=m4

m=16÷4

m=4

Answer:

friction, should be correct i think!

Answer:

Fx  = 4.92 [N]

Fy = 0.868 [N]

Explanation:

Let's take the 10 degrees as a measure from the horizontal component to the vector.

Thus taking the components in the X & y axes respectively:

Fx = 5*cos(10) = 4.92 [N]

Fy = 5*sin(10) = 0.868 [N]

They both are two different elements.SO the answer is Element

Answer:

b)72

Explanation

work done= force* distance

work done=40*1.8

72J

Answer:

Answers in solutions.

Explanation:

Question 6:

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

Density = mass ÷ volume = [tex]\frac{1930}{100}[/tex] = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to [tex]\frac{29.8}{2}[/tex] = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

Question 7:

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

Question 8:

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

[tex]\frac{0.02 * 1}{0.001}[/tex] =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

[tex]\frac{0.02 * 1,000,000}{1}[/tex] = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

Question 9:

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

Question 10:

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; [tex]\frac{20 * 1}{1}[/tex] = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

Answer:

Epidermis, in botany, outermost, protoderm-derived layer of cells covering the stem, root, leaf, flower, fruit, and seed parts of a plant. The epidermis and its waxy cuticle provide a protective barrier against mechanical injury, water loss, and infection.

hope this helps

Answer:

The balloon needs a vertical velocity of approximately 22.14 m/s to reach the window

Explanation:

The given parameters are;

The destination of the designed water balloon we shoot = The 3rd story window

The height of the 3rd story window which the water balloon should reach = 25 m

Therefore, we have, the equation of motion of the water balloon given as follows;

v² = u² - 2 × g × s

Where;

[tex]u_y[/tex] = The initial vertical velocity of the balloon

[tex]v_y[/tex] = The final vertical velocity of the balloon = 0 m/s

g = The acceleration due to gravity = 9.8 m/s

s = The height the balloon is intended reach at the final velocity becomes 0 m/s = The height of the 3rd story window

∴ s = 25 m

Substituting the values, we have;

0² = [tex]u_y[/tex]² - 2 × 9.8 × 25

[tex]u_y[/tex]² = 2 × 9.8 × 25 = 490

[tex]u_y[/tex] = √490 = 7·√10

The initial vertical velocity of the balloon = [tex]u_y[/tex] = 7·√10 m/s

Therefore, the balloon needs a vertical velocity of 7·√10 m/s which is approximately 22.14 m/s to reach the window.

Answer:

i would want to be a dog or a cat

Explanation:

there just funny

Technology can be the four things listed below...

Answer:

mechanical

electronic

medical

industrail and manufacturing

Answer:

Unlike a fact witness, an expert is entitled to compensation for participation in the case. A key distinction between a fact witness and an expert witness is that an expert witness may provide an opinion. Fact witnesses must limit their testimony to facts in regard to evidence they may have observed or been involved.

Explanation:

there

Answer:

Unlike a fact witness, an expert is entitled to compensation for participation in the case. A key distinction between a fact witness and an expert witness is that an expert witness may provide an opinion. Fact witnesses must limit their testimony to facts in regard to evidence they may have observed or been involved.

Explanation:

E2020

Answer:

At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.

Explanation:

We first need to find a relation between the Kelvin scale and the Fahranheit scale. We'll use the Celsius scale to relate them.

The Kelvin and Celsius scales are related by the formula:

K = °C + 273.15

Solving for °C:

°C = K - 273.15

Besides, the Kelvin and Celsius scales are related by:

°C = 5 ⁄ 9(°F - 32)

Now we find a temperature, say X, where both scales coincide. Equating both formulas:

X - 273.15=5 ⁄ 9(X - 32)

Multiply by 9:

9X - 2,458.35 = 5X - 160

Simplifying:

4X = 2,458.35 - 160=2,298.35

Solving:

X =2,298.35 / 4 = 574.59

At 574.59 Kelvin, the Fahrenheit temperature will be 574.59 °F.

Answer:

science is the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

Explanation:

Option D) 1.5 m/s² is the correct answer.

The magnitude of the acceleration of the car is 1.5 m/s².

given the data in the question;

record of the observation;

Time(s)      Speed(m/s)

3.0                4.0

5.0                7.0

6.0                8.5

Since its undergoing constant acceleration, we can make use any two from the three data recorded.

So, lets consider motion between t₁ = 3.0s and t₂ = 5.0s

Initial velocity at t₁ ; u = 4.0 m/s

final velocity at t₂; v = 7.0 m/s

Now, Time Elapsed; [tex]t = t_2 - t_1 = 7.0s - 5.0s = 2.0s[/tex]

To determine the magnitude of acceleration "a", we use the first equation of motion which says;

[tex]v = u + at[/tex]

we make "a" the subject of the formula

[tex]a =\frac{v - u}{t}[/tex]

so we substitute in our values;

[tex]a = \frac{(7.0m/s) - (4.0m/s)}{2.0s}[/tex]

[tex]a = \frac{3.0m/s}{2.0s}[/tex]

[tex]a = 1.5 m/s^2[/tex]

Therefore, the magnitude of the acceleration of the car is 1.5 m/s².

Option D) 1.5 m/s² is the correct answer.

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Answer:Electrones

Explanation:Because its building up static

Answer:

B) Electrons

Explanation:

When William slides down the slide, the rubbing action between William and the slide causes electrons to be transferred to William and stored as a negative charge.

When he touches a piece of metal, the negative charge (i.e the build up of excess electrons), is released (discharged) into the metal causing what we know as a static shock.