The molecules of O2 that are present in 3.90 L flask at a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules of O2
Step 1: used the ideal gas equation to calculate the moles of O2
that is Pv=n RT where;
P(pressure)= 1.00 atm
V(volume) =3.90 L
n(number of moles)=?
R(gas constant) = 0.0821 L.atm/mol.K
T(temperature) = 273 k
by making n the subject of the formula by dividing both side by RT
n= Pv/RT
n=[( 1.00 atm x 3.90 L) /(0.0821 L.atm/mol.k x273)]=0.174 moles
Step 2: use the Avogadro's law constant to calculate the number of molecules
that is according to Avogadro's law
1 mole = 6.02 x10^23 molecules
0.174 moles=? molecules
by cross multiplication
the number of molecules
= (0.174 moles x 6.02 x10^23 molecules)/ 1 mole =1.047 x 10^23 molecules of O2
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Which bibliographic description for a book is correct, based on the Harvard Method in your study guide?
An annotated bibliography is a list of citations to books, articles, and documents. Each citation is followed by a brief descriptive and evaluative paragraph, the annotation. The purpose of the annotation is to inform the reader of the relevance, accuracy, and quality of the sources cited.
The cue column is typically located on the left-hand side of the page and is used to jot down keywords or questions that serve as cues for recalling the main points of the lecture or reading. The note-taking area is located on the right-hand side of the page and is used to write down detailed notes about the lecture or reading.
The summary section is located at the bottom of the page and is used to summarize the key points of the notes. Overall, the Cornell method is an effective way to organize and retain information during lectures and readings.
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817 cm3 at 80.8 kPa to 101.3 kPa: __________ cm3 (No temp. change)
The final volume of the gas, when the pressure changes from 80.8 kPa to 101.3 kPa at constant temperature, is approximately 652.9 cm³.
To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law can be represented by the equation: P₁V₁ = P₂V₂
Where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.
Given:
Initial volume, V₁ = 817 cm³
Initial pressure, P₁ = 80.8 kPa
Final pressure, P₂ = 101.3 kPa
We need to find the final volume, V₂.
Using Boyle's Law equation, we can rearrange it to solve for V₂:
V₂ = (P₁V₁) / P₂
Plugging in the given values:
V₂ = (80.8 kPa * 817 cm³) / 101.3 kPa
Simplifying the expression:
V₂ ≈ 652.9 cm³
Therefore, the final volume of the gas, when the pressure changes from 80.8 kPa to 101.3 kPa at constant temperature, is approximately 652.9 cm³.
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A 49-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.4 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
Neglecting air resistance, the altitude of the pole vaulter as she crosses the bar is 5.7 meters.
How to calculate altitude?The pole vaulter's kinetic energy when she is running is:
KE = 1/2 × m × v²
KE = 1/2 × 49 kg × (11 m/s)²
KE = 2745 J
The pole vaulter's kinetic energy when she is above the bar is:
KE = 1/2 × m × v²
KE = 1/2 × 49 kg × (1.4 m/s)²
KE = 122 J
The difference in kinetic energy is the energy that was used to raise the pole vaulter's altitude. This energy is equal to the potential energy of the pole vaulter when she is above the bar.
The potential energy of the pole vaulter when she is above the bar is:
PE = mgh
PE = 49 kg × 9.8 m/s² × h
PE = 477.2 h J
Setting the difference in kinetic energy equal to the potential energy, we get:
2745 J = 477.2 h J
h = 5.7 m
Therefore, the altitude of the pole vaulter as she crosses the bar is 5.7 meters.
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A 62-kg base runner begins his slide into second base when he is moving at a speed of 3.7 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(A) the mechanical energy lost due to friction acting on the runner is approximately 2632.02 J.
(B) the base runner slides approximately 6.27 meters of distance
Given:
Mass of the base runner (m) = 62 kg
Initial speed (v₀) = 3.7 m/s
Coefficient of friction (μ) = 0.70
(A) To calculate the mechanical energy lost due to friction, we'll first find the final speed (v) when the runner reaches the base. Since the runner slides until his speed becomes zero, we have:
Final speed (v) = 0 m/s
The change in kinetic energy is given by:
ΔKE = 0.5 * m * (v² - v₀²)
Substituting the given values:
ΔKE = 0.5 * 62 kg * ((0 m/s)² - (3.7 m/s)²)
ΔKE = 0.5 * 62 kg * (-13.69 m²/s²)
ΔKE ≈ -2632.02 J
Note that the negative sign indicates a loss of mechanical energy.
Therefore, the mechanical energy lost due to friction acting on the runner is approximately 2632.02 J.
(B) To determine the distance the runner slides, we'll use the work-energy principle. The work done by friction is equal to the change in mechanical energy:
Work done by friction (W) = ΔKE
The work done by friction can be calculated using the formula:
W = μ * m * g * d
where g is the acceleration due to gravity (approximately 9.8 m/s²) and d is the distance the runner slides.
Setting W equal to ΔKE:
μ * m * g * d = ΔKE
Substituting the given values:
0.70 * 62 kg * 9.8 m/s² * d = -2632.02 J
Solving for d:
d = (-2632.02 J) / (0.70 * 62 kg * 9.8 m/s²)
d ≈ -6.27 m
Again, the negative sign indicates the direction of the slide.
Therefore, the base runner slides approximately 6.27 meters.
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In a pith ball experiment, the two pith balls are at rest. The ball on the left has a charge of 6.8 x 10-6 C, the ball on the right has a charge of 10.5 x 10-6 C. Each has a mass of 45 grams, they are 0.50 m apart, and the angle between each string and a vertical line is θ = 27.72°. What are the values for the magnitudes of the tension in each string, T, the gravitational force, Fg, and the electrostatic force, Fq?
In the given pith ball experiment with charges of 6.8 x 10^-6 C and 10.5 x 10^-6 C, and a mass of 45 grams each, positioned 0.50 m apart at an angle of 27.72°, the magnitudes of the tension in each string, gravitational force, and electrostatic force are approximately 0.456 N, 0.441 N, and 4.704 N, respectively.
To solve this problem, we need to analyze the forces acting on each pith ball: the tension in the strings, the gravitational force, and the electrostatic force.
1. Tension in each string, T:
Using the given values:
T = (0.045 kg × 9.8 m/s^2) / cos(27.72°)
T ≈ 0.456 N (rounded to three decimal places)
2. Gravitational force, Fg:
Using the given values:
Fg = 0.045 kg × 9.8 m/s^2
Fg ≈ 0.441 N (rounded to three decimal places)
3. Electrostatic force, Fq:
Using the given values and Coulomb's law:
Fq = (8.99 × 10^9 N m^2/C^2) × (6.8 × 10^(-6) C) × (10.5 × 10^(-6) C) / (0.50 m)^2
Fq ≈ 4.704 N (rounded to three decimal places)
Therefore, the values for the magnitudes of the tension in each string (T), the gravitational force (Fg), and the electrostatic force (Fq) are approximate:
T ≈ 0.456 N
Fg ≈ 0.441 N
Fq ≈ 4.704 N
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Our Sun has a radius of about 695,800 km and a mass of about 2.0 × 1030 kg, and rotates around its axis about once every 29 days.
Calculate the rotational velocity of the Sun.
Calculate the linear speed of particles on the outermost portion of the Sun.
Calculate the angular acceleration, tangential acceleration, and radial acceleration of the Sun.
Calculate the rotational (angular) momentum of the Sun, modeled as a uniform solid sphere. The rotational inertia of a solid sphere is (2/5)MR2.
Imagine that without any external objects interacting with it, the Sun were to collapse to the size of a neutron star, with a radius of about 15 km–about the size of Manhattan Island.
What would be its rotational inertia when it was finished collapsing?
What would be its rotational momentum when it was finished collapsing?
In what time interval would the Sun complete one rotation around its axis after it was finished collapsing
When a skater pulls in her arms, a kid starts up a merry-go-round from a stop, or a computer's hard disk slows to a stop when it is turned off, angular velocity is not constant.
Thus, There is an angular acceleration in each of these situations, in which changes. The angular acceleration increases with the rate of change. The rate at which angular velocity changes is referred to as angular acceleration.
The answer is that the air masses that generate tornadoes rotate, and their rate of rotation rises as their radii get smaller and angular velocity.
The skater begins her rotation by extending her limbs, then she pulls them in closer to her torso to increase the spin.
Thus, When a skater pulls in her arms, a kid starts up a merry-go-round from a stop, or a computer's hard disk slows to a stop when it is turned off, angular velocity is not constant.
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The rotational velocity or angular velocity of the sun is 2.5 x 10⁻⁶ rad/s.
The angular acceleration of the sun is 43.5 x 10⁻⁴ rad/s².
The angular momentum is 96.8 x 10⁴⁰kgm²/s.
Radius of the sun, R = 6.96 x 10⁸m
Mass of the sun, m = 2 x 10³⁰kg
Time period of the sun, T = 29 days = 2.51 x 10⁶s
Angular velocity is the rate of change in the angular displacement between two bodies or the pace at which an object revolves or rotates around an axis.
The rotational velocity or angular velocity of the sun is,
ω = 2π/T
ω = 2 x 3.14/2.51 x 10⁶
ω = 2.5 x 10⁻⁶ rad/s
The linear speed of particles on the outermost portion of the sun is,
v = Rω
v = 6.96 x 10⁸x 2.5 x 10⁻⁶
v = 1.74 x 10³m/s
The angular acceleration of the sun,
α = Rω²
α = 6.96 x 10⁸x (2.5 x 10⁻⁶)²
α = 43.5 x 10⁻⁴ rad/s²
The moment of inertia of the solid sphere, I = 2/5MR²
The expression for the angular momentum is given by,
L = Iω
L = 2/5 x 2 x 10³⁰ x (6.96 x 10⁸)² x 2.5 x 10⁻⁶
L = 96.8 x 10⁴⁰kgm²/s
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4 A soldier wears boots, each having an area of 0.016 m² in contact with the ground. The soldier weighs 720 N. (a) (i) Write down the equation that is used to find the pressure exerted by the soldier on the ground. (ii) Calculate the pressure exerted by the soldier when he is standing to attention, with both boots on the ground.
(a) The average pressure exerted by the person is 40,000 N/m².
(b) The average pressure exerted by the elephant has 295,000 N/m².
The average pressure exerted by the elephant is about 7 times greater than the average pressure exerted by the person.
The average pressure exerted by each is defined as the force per unit area.
Mathematically, the formula for average pressure is given as;
P = F/A
where;
F is the applied force
A is the area of each object
P = W / A
where;
W is weight of the person
A is the area of surface in contact
The average pressure exerted by the person is calculated as follows;
P = (640 N) / (0.016)
P = 40,000 N/m²
The average pressure exerted by the elephant is calculated as follows;
P = (4.13 x 10⁴ N) / (0.14)
P = 295,000 N/m²
Thus, the average pressure exerted by the elephant and the person depends on their weight and area of their shoes.
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WILL MAKE BRAINLIEST!!
An aluminum beam is 10.0 m long at a temperature of 25.0 °C. The temperature of the beam is raised to 75.0 °C. What is
the change in the length of the beam due to thermal expansion, if aluminum has a coefficient of linear expansion of 2.40 E 5
*C-1?
O0.500 cm
O 1.20 cm
1.80 cm
02.50 cm
Answer:
[tex]{\Delta L=0.012 \ m[/tex]
Explanation:
Given:
[tex]L_0=10.0 \ m\\\Delta T_0=25.0 \ \textdegree C\\\Delta T_f=75.0 \ \textdegree C\\\alpha_{Al}=2.40 \times 10^{-5} \ \textdegree C^{-1}[/tex]
Find:
[tex]\Delta L= \ ?? \ m[/tex]
Using the formula for linear expansion.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formula for Linear Expansion:}}\\\\ \Delta L=\alpha L_0 \Delta T\end{array}\right}[/tex]
Where...
"ΔL" represents the change in length"α" represents the coefficient of linear expansion"L_0" represents the initial length of the object"ΔT" represents the change in temperature[tex]\hrulefill[/tex]
Plug the known values into the formula for linear expansion.
[tex]\Delta L=\alpha L_0 \Delta T\\\\\Longrightarrow \Delta L=(2.40 \times 10^{-5})(10.0)(75.0-25.0)\\\\\therefore \boxed{\boxed{\Delta L=0.012 \ m}}[/tex]
Thus, the change in length is found.
The change in length of the aluminum beam due to the thermal linear expansion is 0.012 m.
When a material's temperature increases, a phenomenon known as linear expansion occurs, which results in an increase in the material's length.
The length of a material that is one-unit long changes as the temperature rises by ten degrees Celsius, which is how the coefficient of linear expansion is stated.
Length of the aluminum beam, L = 10 m
Initial temperature of the beam, T₁ = 25°C
Final temperature of the beam, T₂ = 75°C
The coefficient of linear expansion of aluminum, α = 2.4 x 10⁻⁵⁻⁵⁻⁻C⁻
The equation for the change in length of the aluminum beam is given by,
ΔL = αLΔT
ΔL = 2.4 x 10⁻⁵x 10 x(75 - 25)
ΔL = 2.4 x 10⁻⁴x 50
ΔL = 0.012 m
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What is the formula for Angular velocity?
Answer:
The formulas are down below
Explanation:
[tex]w = \frac{o}{t} [/tex]
From V=rw
[tex]w = \frac{v}{r} [/tex]
50.0 m3 at 55.1 kPa to 98.7 kPa: __________ m3 (No temp. change)
The final volume of the gas, when the pressure changes from 55.1 kPa to 98.7 kPa at constant temperature, is approximately 27.9 m³.
To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law can be represented by the equation: P₁V₁ = P₂V₂
Where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.
Given:
Initial volume, V₁ = 50.0 m³
Initial pressure, P₁ = 55.1 kPa
Final pressure, P₂ = 98.7 kPa
Using Boyle's Law equation, we can rearrange it to solve for V₂:
V₂ = (P₁V₁) / P₂
Plugging in the given values:
V₂ = (55.1 kPa * 50.0 m³) / 98.7 kPa
Simplifying the expression:
V₂ ≈ 27.9 m³
Therefore, the final volume of the gas, when the pressure changes from 55.1 kPa to 98.7 kPa at constant temperature, is approximately 27.9 m³.
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Given that the luminosity of a star is given as a function of its radius and temperature by the equation. I do not understand this last question in terms of what to put into the given equation.
The luminosity of this star in units of the solar luminosity would be: 483.7L.
How to calculate the luminosityTo calculate the luminosity, we would use the different values given and the formula for luminosity.
Temperature = 9305K
Star's radius = [tex]5.90 * 10^{9} m\\[/tex]
Luminoisty of the star
Luminosity of the sun
= [tex]\frac{4π * (5.90 * 10^9)^2 * 5.67 * 10^-8 * 9305^4 W}{3.846 * 10^26 W}[/tex]
= 483.7L
This is the unit for luminosity.
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a circular coil (radius = 0.40m) has 160 turns and is in a uniform magnetic field. when the orientation of the coil is varied through all possible positions , the maximum torque on the coil by the magnetic forces is 0.16N.M
When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by the magnetic forces is 0.16 N·m. The magnetic field strength required to achieve this torque is approximately 0.000318 Tesla.
The maximum torque on a circular coil in a uniform magnetic field can be calculated using the formula:
τ = N * B * A * sinθ
Where:
τ is the torque,
N is the number of turns in the coil,
B is the magnetic field strength,
A is the area of the coil, and
θ is the angle between the normal to the coil and the direction of the magnetic field.
In this case, The radius of the coil is given as 0.40 m, and the number of turns is 160. We need to find the magnetic field strength (B) and the angle (θ) to calculate the torque.
Since the maximum torque is given as 0.16 N·m, we can write the equation as:
0.16 N·m = 160 * B * π * [tex](0.40 m)^2[/tex] * sinθ
Simplifying the equation:
0.16 N·m = 160 * B * (0.16π) * sinθ
Now, solving for B:
B = (0.16 N·m) / (160 * (0.16π) * sinθ)
B = 0.001 N / (π * sinθ)
The magnetic field strength (B) depends on the angle θ. The maximum torque occurs when sinθ is equal to 1 (sinθ = 1), which gives us the maximum value for B.
Substituting sinθ = 1:
B = 0.001 N / (π * 1) = 0.000318 N/T
Therefore, the magnetic field strength is approximately 0.000318 Tesla (T).
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6. A picture of weight, w is hanging from a steel nail as shown in the figure below. The nail has a diameter of 1.50 mm and an original length, Lo = 5.0 mm. Useful Information: The shear modulus, G for steel is 80 x 10° N.m². (a) (b) (c) 1.50 mm 3 Ax = 1.80 μm W Lo = 5.00 mm M What kind of deformation occurs in this case? How are stress and strain in this deformation related to each other? [3] When the picture is hung from the nail, the head of the nail displaces vertically downwards by an amount Ax = 1.80 µm. Find the mass of the picture. Neglect the weight of the nail. [6] What angle does the nail make with the horizontal after the picture is hung from it? [2]
The mass of the picture is approximately 5.19 kilograms.
How to solve for the problemThe deformation in this case is called shear deformation, a type of deformation that occurs when parallel internal surfaces slide past one another. It is caused by shear stress in the structure. The shear stress (τ) is the force (F) applied divided by the cross-sectional area (A) of the nail. The shear strain (γ) is the displacement (Δx) divided by the original length (L0).
The relationship between shear stress and shear strain is given by the shear modulus (G) in the formula:
τ = G * γ
To find the weight of the picture, we need to calculate the shear stress first:
The cross-sectional area A of the nail is given by the formula for the area of a circle:
A = πr² = π(d/2)² = π(0.0015 m / 2)² = 1.767 x 10^-6 m².
The shear strain γ is given by:
γ = Δx / L0 = (1.80 x 10^-6 m) / (5 x 10^-3 m) = 0.36.
The shear stress τ can now be calculated by rearranging the formula:
τ = G * γ
=> τ = (80 x 10^9 N/m²) * 0.36 = 28.8 x 10^9 N/m²
The force F on the nail is equal to the weight w of the picture, and it can be calculated from the shear stress:
τ = F / A
=> F = τ * A = (28.8 x 10^9 N/m²) * (1.767 x 10^-6 m²) = 50.89 N.
Since weight w = m * g, where m is mass and g is the acceleration due to gravity (approximately 9.81 m/s²), we can find the mass m:
m = w / g = (50.89 N) / (9.81 m/s²) = 5.19 kg.
So, the mass of the picture is approximately 5.19 kilograms.
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Part-II Work out Step by step clearly (6%) 5. A 5kg mass starts from rest at xo = -1 and moves under the action of a variable force F(x) = √1-x² to point xf = 1. Calculate the total work done by the force? (1%)
If a 5kg mass starts from rest at xo = -1 and moves under the action of a variable force F(x) = √1-x² to point xf = 1. Then the total work done by the force is equal to π/2 + 1.
To calculate the total work done by the force in this scenario, we can use the formula for work:
Work = ∫F(x) dx
where F(x) is the force as a function of position and dx represents an infinitesimal displacement.
In this case, the force is given by F(x) = √(1 - x²), and we need to find the total work done as the object moves from xo = -1 to xf = 1.
Let's break down the calculation step by step:
Write the integral for work:
Work = ∫F(x) dx
Substitute the given force:
Work = ∫√(1 - x²) dx
Integrate with respect to x:
To integrate the square root of (1 - x²), we use the trigonometric substitution. Let's substitute x = sin(θ) and dx = cos(θ) dθ.
Work = ∫√(1 - sin²(θ)) cos(θ) dθ
Simplify the integrand:
Using the trigonometric identity sin²(θ) + cos²(θ) = 1, we can rewrite the integrand as cos²(θ).
Work = ∫cos²(θ) dθ
Apply the power-reducing formula:
The power-reducing formula states that cos²(θ) = (1 + cos(2θ)) / 2. We can use this formula to simplify the integrand further.
Work = ∫(1 + cos(2θ))/2 dθ
Integrate the terms separately:
Work = (1/2) ∫dθ + (1/2) ∫cos(2θ) dθ
The first integral, ∫dθ, is simply θ, and the second integral, ∫cos(2θ) dθ, can be calculated as sin(2θ)/2.
Work = (1/2) θ + (1/2) (sin(2θ)/2) + C
Evaluate the integral limits:
To find the total work done, we need to evaluate the integral at the upper and lower limits of integration.
At xf = 1, the angle θ is π/2, and at xo = -1, the angle θ is -π/2.
Work = (1/2) (π/2) + (1/2) (sin(2(π/2))/2) - [(1/2) (-π/2) + (1/2) (sin(2(-π/2))/2)]
Simplifying further:
Work = π/4 + (1/2) - (-π/4 + (1/2))
Work = π/4 + 1/2 + π/4 + 1/2
Work = π/2 + 1
Therefore, the total work done by the force is equal to π/2 + 1.
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a spring balance has a maximum reading of 10 Newton and the length of the calibrated scale is 20 cm a rectangular metal block measuring 10 cm by 3 cm by 2 cm is hanged on the balance and stretches the string by 15 cm calculate the weight of the block the mass of the Block and the total density of the metal from which the blood is made
To calculate the weight of the block, we can use the formula:
Weight = Mass x Gravity
Where gravity is approximately 9.8 m/s^2.
First, we need to find the mass of the block. We can use the formula:
Density = Mass / Volume
The volume of the block is:
Volume = Length x Width x Height
Volume = 10 cm x 3 cm x 2 cm
Volume = 60 cm^3
We don't know the density of the metal, so we can't calculate the mass directly. However, we can use the spring balance reading to find the weight of the block.
The spring balance has a maximum reading of 10 Newtons, which corresponds to a length of 20 cm. When the block is hung on the balance, it stretches the string by 15 cm. The extension of the spring is proportional to the weight of the block, so we can use the following proportion:
Extension of spring / Total length of spring = Weight of block / Maximum weight of spring balance
Substituting the values we have:
15 cm / 20 cm = Weight of block / 10 N
Solving for the weight of the block:
Weight of block = 7.5 N
Now we can find the mass of the block:
Mass = Weight / Gravity
Mass = 7.5 N / 9.8 m/s^2
Mass = 0.765 kg
Finally, we can calculate the density of the metal:
Density = Mass / Volume
Density = 0.765 kg / 60 cm^3
Density = 0.01275 kg/cm^3
So the weight of the block is 7.5 N, the mass of the block is 0.765 kg, and the density of the metal is 0.01275 kg/cm^3.
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An NFL prospect running the 40-yd dash (Otherwise known as the 36.6 m dash). He accelerates at 4.9 m/s² to a final velocity of 8.17 m/s then finishes the rest of the dash at that velocity. What was this prospect's 40-yd dash time?
Thanks in advance!!
The NFL prospect's 40-yard dash time is approximately 5.92 seconds.
How to find dash time?1. Calculating the time for acceleration:
Using the equation of motion: v = u + at,
where v = final velocity, u = initial velocity, a = acceleration, and t = time.
Given:
Initial velocity (u) = 0 m/s (assuming the prospect starts from rest)
Final velocity (v) = 8.17 m/s
Acceleration (a) = 4.9 m/s²
v = u + at
8.17 = 0 + 4.9t
Solving for t:
t = 8.17 / 4.9
t ≈ 1.67 seconds
2. Calculating the time for the remaining distance:
The remaining distance is 36.6 m - the distance covered during acceleration.
Distance covered during acceleration:
Using the equation of motion: s = ut + (1/2)at²
s = 0 × t + (1/2) × 4.9 × t²
s = (1/2) × 4.9 × t²
Distance covered during acceleration = (1/2) × 4.9 × (1.67)²
Remaining distance = 36.6 m - [(1/2) × 4.9 × (1.67)²]
3. Calculating the time for the remaining distance:
Using the equation of motion: s = vt,
where s is the remaining distance, v is the constant velocity, and t is the time.
s = v × t
[36.6 - (1/2) × 4.9 × (1.67)²] = 8.17 × t
Solving for t:
t = [36.6 - (1/2) × 4.9 × (1.67)²] / 8.17
t ≈ 4.25 seconds
4. Calculating the total dash time:
Total dash time = time for acceleration + time for remaining distance
Total dash time = 1.67 + 4.25
Total dash time ≈ 5.92 seconds
Therefore, the NFL prospect's 40-yard dash time is approximately 5.92 seconds.
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The statements below describe either vector or scalar quantities. Please select the statements that are scalar quantities.
Responses
After Shelly ate at JB's crab shack, she traveled 400 km southeast to get home.
After Shelly ate at JB's crab shack, she traveled 400 km southeast to get home.,
Shelly traveled 8 km in 4 hours, giving her an average speed of 2.0 km/hr.
Shelly traveled 8 km in 4 hours, giving her an average speed of 2.0 km/hr.,
Shelly laid her eggs on Crystal Beach Island and then traveled 70 km north.
Shelly laid her eggs on Crystal Beach Island and then traveled 70 km north.,
As Shelly rides the Gulf Stream to Greenland to meet up with her mate, her velocity increased to 3.3 km/hr west.
As Shelly rides the Gulf Stream to Greenland to meet up with her mate, her velocity increased to 3.3 km/hr west.,
Shelly ate a crab that had a mass of 0.7 kg.
Shelly ate a crab that had a mass of 0.7 kg.,
The temperature of the ocean Shelly swam in was 22 degrees Celsius.
The following are the statements' respective scalar quantities:
Shelly ate a crab that had a mass of 0.7 kg.The temperature of the ocean Shelly swam in was 22 degrees Celsius.The scalar quantities in the given statements are the mass of the crab, which is 0.7 kg, and the temperature of the ocean Shelly swam in, which was 22 degrees Celsius. Scalar quantities are measurements that have magnitude but no direction.
In this context, mass and temperature are scalar quantities because they represent numerical values without any specific directional information associated with them.
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The physical quantities are of two types and they are scalar and vector quantities. Scalar quantities are the physical quantity that has only a magnitude. The vector quantities are the physical quantity that has both magnitude and direction.
From the given, the statements that have only magnitude without direction are scalar quantities. Hence, the scalar quantity statements are:
2) Shelly traveled 8km in 4 hours giving her an average speed of 2 km/hr. The speed is the physical quantity that gives the magnitude of distance covered by the object and hence, speed is the scalar quantity.
5) Shelly ate a crab that had a mass of 0.7 kg representing the scalar quantity statement. The mass is the quantity that has magnitude and it has no direction. Thus, mass is the scalar quantity.
6) The temperature of the oceans Shelly swam in was 22 degrees Celcius. Thus, temperature is the scalar quantity.
The remaining sentences represent the vector quantity as the statements indicate both magnitude and direction.
Hence, the correct statements are options 2, 5, and 6.
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DIRECTIONS:
Use the food diary chart below to record everything that you eat in a 24-hour period. Be specific. For example, if you eat a hot dog with ketchup and mustard, don’t forget to include the ketchup and mustard. Be sure to include the quantity of food eaten as well. Under “Amount or Quantity,” do your best to estimate in cups, ounces, or other measurement. You may add additional rows if needed.
A food diary is a record of everything you eat and drink throughout the day. It helps you track your eating habits, monitor your calorie intake, etc.
How to explain the informationTime Food Amount or Quantity
7:00 AM Oatmeal with berries and nuts 1 cup oatmeal, 1/2 cup berries, 1/4 cup nuts
10:00 AM Apple 1 medium apple
12:00 PM Salad with grilled chicken 1 cup salad greens, 1/2 cup grilled chicken, 1/4 cup dressing
2:00 PM Banana 1 medium banana
6:00 PM Salmon with roasted vegetables 4 ounces salmon, 1 cup roasted vegetables
8:00 PM Yogurt with granola 1 cup yogurt, 1/2 cup granola
I also had a few cups of coffee and water throughout the day.
I am generally happy with my diet. I eat a variety of healthy foods and I try to limit my intake of processed foods and sugary drinks. I am also mindful of my portion sizes. However, I could probably eat more fruits and vegetables. I am also trying to cut back on my caffeine intake.
I think keeping a food diary is a helpful way to track my eating habits. It helps me to be more aware of what I am eating and to make healthier choices. I would recommend keeping a food diary to anyone who is trying to improve their diet.
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why do spheres move apart after contact
Answer:
Explanation:
When spheres or objects come into contact and then move apart, it is typically due to the presence of an external force or energy imparted to the system. There are a few possible reasons for spheres to move apart after contact:
Elasticity: If the spheres are made of elastic materials, such as rubber or certain metals, they can deform upon contact and then regain their original shape when the external force is removed. This elasticity causes the spheres to move apart.
Repulsive forces: If the spheres have like charges or magnetic properties, they can experience repulsive forces when brought close together. These repulsive forces push the spheres apart once the external force is no longer present.
Conservation of momentum: If the spheres are initially at rest and then pushed together with an external force, the conservation of momentum requires them to move apart after the force is removed. This is due to Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Overall, the specific reason for spheres moving apart after contact depends on the properties of the spheres and the nature of the external force or energy involved.
