Dr. Jones performed an experiment to monitor the effects of sunlight exposure on stem density in aquatic plants. In the study, Dr. Jones measured the mass and volume of stems grown in 5 levels of sun exposure. The data is represented below.
Sun exposure Stem mass (g) Stem volume (mL)
30 275 1100
45 415 1215
60 563 1425
75 815 1610
90 954 1742
a. Convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters (mº).
b. Calculate the density of the samples using the equation d = m/v. d = density m = mass (kg) v = volume (m)
c. Convert the density values to scientific notation.

Answers

Answer 1

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

\(1\ g=0.001\ kg\)

Conservation of volume

\(1\ Lt=0.001\ m^3\)

\(1\ mL=1\times10^{-6}\ m^3\)

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

\(\rho=\dfrac{m}{V}\)

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

\(\rho=\dfrac{0.275}{0.0011}\)

\(\rho=250\ kg/m^3\)

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

\(\rho=\dfrac{0.415}{0.001215}\)

\(\rho=341.56\ kg/m^3\)

\(\rho=342\ kg/m^3\)

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

\(\rho=\dfrac{0.563}{0.001425}\)

\(\rho=395.08\ kg/m^3\)

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

\(\rho=\dfrac{0.815}{0.001610}\)

\(\rho=506.21\ kg/m^3\)

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

\(\rho=\dfrac{0.954}{0.001742}\)

\(\rho=547.6\ kg/m^3\)

\(\rho=548\ kg/m^3\)

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

\(\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}\)

Hence, This is required solution.


Related Questions

A biologist studies a species of bacteria called FAST. Her research reveals that FAST bacteria reproduce prodigiously. If a bacteria culture contains 10 FAST bacteria at time t = 0, the number of FAST bacteria in the culture at later time t is N(t) = 10+6t+2t^2where t is expressed in seconds.

A) How many FAST bacteria occupy the culture at time t = 10 s? At time t = 11 s? From these answers, calculate how many net bacteria were born during the interval between t = 10 and t = 11 s. Assume no bacteria died.

B) Now use calculus to find the birth rate of bacteria (the rate at which new bacteria get born) at time t = 10 s. It may help you to sketch a rough graph of N vs. t.

C) Explain, conceptually, why your answers to part (a) and (b) are close but not exactly the same.

Answers

270 FAST bacteria occupy the culture at time t = 10 s and 318 FAST bacteria occupy the culture at time t = 11 s.

Define FAST bacteria.

A group of bacteria with the property of acid fastness is referred to as acid-fast bacteria, sometimes known as acid-fast bacilli or simply AFB. A bacterium's capacity to withstand acid decolorization during staining methods is known as acid fastness.

Mycobacteria's high mycolic acid concentration, which causes the staining pattern of poor absorption followed by significant retention, is what makes them acid-fast.

At t = 10 s

N(t) = 10+6t+2t^2

N(t) = 10 +6*10 + 2*10*10

      = 270

At t  = 11s

N(t) =  10+6*11+2*11*11

     =  318

Difference between two will be 318 - 270 =  48

Rate of bacterial growth will be (270-10/270) *100%

i.e. 96.2%

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Two blocks are attached by a string as shown. The blocks are released from rest and
allowed to move freely. The sloped ramp angle is θ = 30⁰. Find the coefficient of friction
on the blocks (assume they have the same coefficient), assuming the blocks have the
same mass, and given that the blocks travel a distance of 0.50 m in a time of 0.935 s.

Answers

The coefficient of friction between the two blocks is 0.194.

The problem involves finding the coefficient of friction between two blocks of equal mass attached by a string that moves over a sloped ramp with an angle of 30 degrees. The blocks start from rest and are allowed to move freely. The distance traveled by the blocks is 0.50 m, and the time taken to travel this distance is 0.935 s. To solve the problem, we need to consider the forces acting on the system of blocks. The forces acting on the blocks are the gravitational force, the tension in the string, and the frictional force. As the blocks are moving up the ramp, the force of gravity is pulling them down. The tension in the string is pulling the blocks up the ramp. The frictional force is opposing the motion of the blocks and is acting in the opposite direction to the tension in the string.

To determine the coefficient of friction, we can use the equations of motion to find the acceleration of the blocks. Once we have the acceleration, we can use Newton's Second Law to find the net force acting on the blocks. We can then use the force of friction to find the coefficient of friction. Using the equations of motion, we can find the acceleration of the blocks:

a = 2d/t^2

where d is the distance traveled by the blocks and t is the time taken to travel the distance. Plugging in the given values, we get:

a = 2(0.50 m)/(0.935 s)^2 = 1.15 m/s^2

Next, we can use Newton's Second Law to find the net force acting on the blocks:

ΣF = ma

where ΣF is the sum of the forces acting on the blocks. Plugging in the known forces, we get:

T - mg sin θ - μmg cos θ = ma

where T is the tension in the string, m is the mass of the blocks, g is the acceleration due to gravity, θ is the angle of the ramp, and μ is the coefficient of friction.

We can simplify this equation by substituting mg sin θ for the component of the weight of the blocks that is acting down the ramp and mg cos θ for the component of the weight that is acting perpendicular to the ramp:

T - mg sin θ - μmg cos θ = ma

T - mg(1/2) - μmg(√3/2) = ma

We can also use the equation for the tension in the string:

T = 2mg sin θ

Substituting this into the equation for net force, we get:

2mg sin θ - mg(1/2) - μmg(√3/2) = ma

Simplifying and solving for μ, we get:

μ = (2gsinθ - a)/(2gcosθ)

Substituting the given values, we get:

μ = (2(9.81 m/s^2)sin 30° - 1.15 m/s^2)/(2(9.81 m/s^2)cos 30°) = 0.194

Therefore, the coefficient of friction between the two blocks is 0.194.

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HELP THIS IS URGENT!!!!!

HELP THIS IS URGENT!!!!!

Answers

Multiply the number of Watts by the number of seconds in each case. Convert the hours into seconds by multiplying by 3600. Multiply the minutes by 60. This will give you the energy in Joules for each case.

The low-frequency speaker of a stereo set produces 10.0 W of acoustical power. If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity level 80.0 dB

Answers

Answer:

the required distance is 89.125 m

Explanation:

Given the data in the question;

we know that, sound intensity B in decibels of sound is;

β(dB) = 10log₁₀( \(I\) / \(I_0\) )

where intensity \(I\) = power / area carried by wave

\(I_0\) = 10⁻¹² W/m² { minimum threshold intensity }

Now,

intensity \(I\) = power / area carried by wave = P/A = P/4πr² { spherical  }

given that; β = 80.0 dB and P = 10 W

so

β(dB) = 10log₁₀( \(I\) / \(I_0\) )

we substitute

80 = 10log₁₀( P / 4πr²× \(I_0\))

80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )

8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )  

8 = 1 - log₁₀( 4πr²× 10⁻¹² )

8 - 1 = -log₁₀( 4πr²× 10⁻¹² )

7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )

7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]

7 = -[ -10.9 + log₁₀( r² ) ]

7 = 10.9 - log₁₀( r² )

-log₁₀( r² ) = 7 - 10.9

-log₁₀( r² )  = - 3.9

log₁₀( r² ) = 3.9

2log₁₀r = 3.9

log₁₀r  = 3.9 /2

log₁₀r = 1.95

r = 89.125 m

Therefore, the required distance is 89.125 m

Find the vector whose magnitude is 5 and which is in the direction of the vector 4i -3j +k

Answers

The vector with a magnitude of 5 and in the direction of the vector 4i - 3j + k is approximately (20/√26)i + (-15/√26)j + (5/√26)k.

To solve this problem

The given vector can be normalized before being multiplied by the desired magnitude. This is how to locate the vector:

The vector that has been provided should be normalized by dividing each of its components by its magnitude. The Pythagorean theorem can be used to determine the magnitude of the vector 4i - 3j + k:

Magnitude = √(4² + (-3)² + 1²) = √(16 + 9 + 1) = √26

Normalize the vector by dividing each component by the magnitude:

Normalized vector = (4/√26)i + (-3/√26)j + (1/√26)k

Multiply the normalized vector by the desired magnitude:

To obtain a vector with a magnitude of 5, multiply each component of the normalized vector by 5:

Desired vector = 5 * ((4/√26)i + (-3/√26)j + (1/√26)k)

Simplifying the expression gives:

Desired vector ≈ (20/√26)i + (-15/√26)j + (5/√26)k

So, the vector with a magnitude of 5 and in the direction of the vector 4i - 3j + k is approximately (20/√26)i + (-15/√26)j + (5/√26)k.

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1. What types of elements does an ionic bond occur between?

Answers

Answer:

Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

Explanation:

when water in a brook or system of pipes flows from a wide region to a narrow region, the speed of water in the narrow region is

Answers

Answer:

more

hope this helps

plz mark brainliest

a small sphere of
mass 0.25 g that carries a charge of 9.0 × 10−10 C.
Two parallel vertical infinite charged sheets of
charge densities σ1= -30 × 10−6 C/m2 and σ2= ab ×
10−6 C/m2 respectively. The sphere is attached to
one end of a very thin silk string 5.0 cm long. The
other end of the string is attached to the 2nd sheet as
shown in the figure. At equilibrium, the string will
make an angle (ϴ) with the vertical. Calculate the
angle that the string makes with the vertical?

Answers

Answer:

θ = 39.7º

Explanation:

In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left

Y Axis

       \(T_{y}\) -W = 0

       \(T_{y}\) = W

X axis

         -\(F_{e1}\) - F_{e2} + Tₓ = 0

let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical

         sin θ = Tₓ / T

         cos θ = T_{y} / t

         Tₓ = T sin θ

         T_{y} = T cos θ

let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.

         F = ∫ E. dA = \(q_{int}\) / ε₀

  in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate

        F = 2E A = q_{int} / ε₀

let's use the concept of surface charge density

        σ = q_{int} / A

we substitute

        2E A = σ A /ε₀

          E = σ / 2ε₀

this is the field created by each plate. The electric force is

        \(F_{e}\) = q E

for plate 1 with σ₁ = -30 10⁻⁶ C / m²

         F_{e1}  = q σ₁ /2ε₀

for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²

          F_{e2} = q σ₂ /2ε₀

we substitute and write the system of equations

           T cos θ = mg

          - q σ₁ / 2ε₀  - q σ₂ /2ε₀  + T sinθ = 0

we introduce t in the second equations

          - q /2 ε₀  (σ₁ + σ₂) + (mg / cos θ) sin θ = 0

          mg tan θ = q /2ε₀   (σ₁ + σ₂)

          θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)

data indicates the mass of 0.25 g = 0.25 10⁻³ kg

give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²

let's calculate

         θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2  8.85 10⁻¹² 0.25 10⁻³ 9.8))

         θ = tan⁻¹ 8.3 10⁻¹)

         θ = 39.7º

HELPPPPPP

i need to turn this assignment in soon!

HELPPPPPP i need to turn this assignment in soon!

Answers

Answer:

A B C

Explanation:

I think that is right. Sorry if it's not

Hope this helps :)

It takes a continent approximately 250,000,000 years to move half-way around Earth, a distance of about 19,200 km or 1,920,000,000 cm. The speed of the continent in what per what can be found by dividing 1,920,000,000 cm by 250,000,000 years.

Answers

The speed of the continent is found to be 7.68 centimeters per hour when 1920000000cm is divided by 250000000years.

The distance travelled by the continent is 1920000000cm.

The time taken by the continent to complete its path is 250000000.

The speed S of the body,

S = D/T

Where,

D is the distance travelled by the body,

T is the time taken by the body to cover that distance.

The speed S of the continent is,

S = D/T

S = 1920000000cm/250000000year

S = 7.68 cm/year

Hence, the speed of the continent is 7.68 centimeters per year.

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1. Which object would have more inertia — a 1,000 kg car or a 150 kg golf cart?

Answers

Answer:a car with 1,000kg

Explanation: A car with a mass of 1,000kg has more inertia than 150 kg

Out of the two objects a 1000-kilogram car and a 150-kilogram golf cart, the 1000-kilogram cart will have more inertia as the inertia of any object is directly proportional to its mass.

What is Newton's first law?

According to Newton's first law, until pushed to alter its condition by the intervention of an external force, every object will continue to be at rest or in uniform motion along a single direction.


The larger the mass of the object, the greater would be the inertia of the object.

Thus, a 1000-kilogram automobile will have higher inertia than a 150-kilogram golf cart since any object's inertia is directly proportionate to its mass.

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How is an object’s position in the solar system affect its motion and temperature?

Answers

Answer:

it will be base on the climate change.

Explanation: so let just say tomorrow it will be cold so the atmosphere will circle on the coolest atmosphere away from the sun.

An object's position in the solar system affects both its motion and temperature. Objects closer to the Sun experience a stronger gravitational force, causing faster or slower orbits. Other planets and moons also influence an object's orbit and speed. Objects closer to the Sun receive more solar radiation and are hotter, while those farther away receive less radiation and are colder. Additionally, the composition of an object's surface can affect its temperature.

What is the solar system?

The solar system refers to the group of celestial bodies that orbit around the Sun, including planets, dwarf planets, moons, asteroids, comets, and other small bodies. It is located in the Milky Way galaxy and spans a distance of about 4.6 billion years in age. The eight planets in the solar system, in order from the Sun, are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. The solar system is held together by the gravitational force of the Sun and is a subject of scientific study to understand the formation and evolution of our own planet and the other bodies in the solar system.

Here in the Question,

An object's position in the solar system can affect both its motion and temperature. Here's how:

Motion: The motion of an object in the solar system is affected by its distance from the Sun and the gravitational forces of other celestial bodies. An object closer to the Sun will experience a stronger gravitational force than an object farther away. This can cause the object's orbit to be faster or slower depending on its position. Additionally, the gravitational pull of other planets and moons can affect the object's orbit and speed.

Temperature: An object's temperature in the solar system is affected by its distance from the Sun and the composition of its surface. Objects closer to the Sun receive more solar radiation and thus have a higher temperature. Objects farther away receive less radiation and are colder. Additionally, the composition of the object's surface can affect its temperature. For example, objects with high albedo (reflectivity) will reflect more solar radiation and have a lower temperature than objects with low albedo.

Therefore, an object's position in the solar system has a significant impact on its motion and temperature. By studying the positions of celestial bodies and their effects on each other, scientists can gain a better understanding of the dynamics of the solar system and the properties of objects within it.

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A long, cylindrical solenoid with 100 turns per centimeter has a radius of 1.5 cm. (a) Neglecting end effects, what is the self-inductance per unit length of the solenoid? (b) If the current through the solenoid changes at the rate 5.0 A/s, what is the emf induced per unit length?

Answers

The solenoid's induced emf is 0.011 V/m per unit length. Be aware that the negative sign denotes an opposition to the change in current in the direction of the induced emf.

What is the amount of induced emf per unit length if the current through the solenoid changes at a rate of 5.0 A/s?

Faraday's law of induction provides the emf induced per unit length in a solenoid as follows:

ε = -N(dΦ/dt)

A solenoid's magnetic flux is determined by:Φ = μ₀n²AI

where I is the solenoid's current (passing through it).

When we adjust for time in both sides of this equation, we obtain:

dI/dt = 0n2A(dd)

When we add this to Faraday's law, we obtain:

= -0n2A(dI/dt)N

Inputting the values provided yields:

= - (4 10 7 T m/A) (100 000 turns/m)² (7.07×10^−4 m²)(5.0 A/s)

ε = - 0.011 V/m

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imagine that the blue light and orange light from the source were blocked. what color would how be present in the spectrum of light observed

Answers

Everything but blue & orange would now be present in the spectrum of light observed.

Spectrum refers to a range of different wavelengths of electromagnetic radiation. Electromagnetic radiation is a form of energy that travels through space and includes different types such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Each type of electromagnetic radiation has a different wavelength and frequency, and together they make up the electromagnetic spectrum.

The concept of spectrum is used in a variety of fields, including physics, astronomy, and telecommunications. The spectrum of electromagnetic radiation is essential for many technologies, such as radios and televisions, cell phones, and medical imaging devices, as they all rely on the transmission and reception of specific wavelengths of electromagnetic radiation.

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Complete Question: -

Imagine that the blue light and orange light from the source were blocked. What color(s) would now be present in the spectrum of light observed?

how many legs a cow has​

Answers

the cow has 4 legs :)

PLS HELP URGENT A moon orbits a planet as shown above.
The moon has a mass 0.66 x 1022 kg and the gravitational field strength at distance R is 0.0012 N/kg. What is the magnitude of the gravitational force exerted by the planet onto the moon while it is in orbit of radius R?
Type your answer in scientific notation with two decimal places. Example: 1.23*10^4

Answers

The "magnitude of the gravitational force exerted by the planet onto the moon" (whew!) is just the moon's weight.

Weight = m•g ...... We know both of these

(0.66 x 10^22 kg) • (0.0012 N/kg) =

7.92 x 10^18 Newtons


A football player kicks a ball with a mass of .55 kg. The average acceleration of the
football was 14.9 m/s2. How much force did the kicker supply to the football?

A.8.2 m/s2
B.)8.2 N
C.)35.24 N
D.)35.24 m/s2

Answers

Answer:

8.2N

Explanation:

I took the test and got it right :)

Two friends grab different sides of a textbook and pull with forces of 6.7 N to the east and 4.4 N to the west, respectively. What would be the magnitude and direction of the force a third friend would need to exert on the textbook in order to balance the other two forces?

Answers

Explanation:

This equation for acceleration can be used to calculate the acceleration of an object that is acted on by a net force. For example, Xander and his scooter have a total mass of 50 kilograms. Assume that the net force acting on Xander and the scooter is 25 Newtons. What is his acceleration? Substitute the relevant values into the equation for acceleration:

Answer:

a=Fm=25 N50 kg=0.5 Nkg

The Newton is the SI unit for force. It is defined as the force needed to cause a 1-kilogram mass to accelerate at 1 m/s2. Therefore, force can also be expressed in the unit kg • m/s2. This way of expressing force can be substituted for Newtons in Xander’s acceleration so the answer is expressed in the SI unit for acceleration, which is m/s2:

a=0.5 Nkg=0.5 kg⋅m/s2kg=0.5 m/s2

A semi circle has a radius of 2000km and a thickness of 3000 meters. How many cubic centimeters does it have

Answers

The volume of the semi circle in cubic centimeters  is 3.75 * 10^22 cm^3

What is the cubic centimeters ?

We know that the first step that we would have to take is to be able to get the area of the semi circle and when we do that we are going to have from the values that we have in the question;

A = πr^2

A = area

r = radius

Where r = 2000km or 2 * 10^8 cm

A = 3.142 * (2 * 10^8)^2

A = 1.25 * 10^17 cm^2

Then we would have;

Volume = area * thickness

Thickness = 3000 meters or 3 * 10^5 cm

Volume = 1.25 * 10^17 cm^2 * 3 * 10^5 cm

= 3.75 * 10^22 cm^3

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A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown.

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825

Answers

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Angular Speed of the pulley

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\(\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\\)

\(\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\\)

Substitute the given parameters and solve for the angular speed;

\(\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s\)

Linear speed of the block

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

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A car traveled a distance of 30 km in 20 minutes (1/3 hours). What was the
speed of the car?
A. 90 km/hr
OB. 60 km/hr
O C. 30 km/hr
D. 10 km/hr

Answers

The answer is, (A). 90 I’m/hr
The answer will be A.

A ball X of mals 1 kg travelling at 2 m/s has a
head-on collision with an identical ball Y at rest.
X stops and Y moves off. What is Y's velocity?

Answers

Answer:

A

Explanation:

Answer:

2m/s

Explanation:

If X stops Y takes momentum and it has the same velocity

4. When you are holding a book, energy is stored between the book and the Earth.
This type of energy is called
potential energy.
A. Elastic potential energy
B. Chemical potential energy
C. Gravitational potential energy
D. Kinetic energy

Answers

Answer:

gravitational potential energy

What part of the sweating process promotes cooling?

wiping sweat from the surface of the skin
condensation of liquid sweat on the skin
liquid (sweat) sitting on the skin’s surface
evaporation of liquid sweat from the skin

Answers

Answer: Condensation of liquid sweat on the skin

Explanation:

How much is 813 in grams
I Math

Answers

To convert 813 into grams if it is already in grams, then the value remains as 813 grams. If it is in pounds, 813 pounds would be approximately 368,646.696 grams. If it is in kilograms, 813 kilograms would be equal to 813,000 grams.

To convert 813 into grams, we need to know the unit of measurement you're referring to. Grams are typically used to measure the weight of objects, but without additional context, it's difficult to provide an accurate answer.

If you are referring to 813 grams, then the conversion is straightforward, as it is already in grams.

If you are referring to another unit, such as pounds or kilograms, we can convert it to grams:

If you meant 813 pounds, one pound is approximately equal to 453.592 grams. Therefore, 813 pounds would be approximately 368,646.696 grams.

If you meant 813 kilograms, one kilogram is equal to 1,000 grams. Therefore, 813 kilograms would be equal to 813,000 grams.

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3. A car with a mass of 1600 kg has a kinetic energy of 125 000 J. How fast is it moving?​

Answers

The car is moving at approximately 12.5 meters per second.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * m * \(v^2\)

where

KE = kinetic energy,

m =Mass of the object, and

v = velocity.

In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .

Substituting the  values , we have:

125,000 J = 1/2 * 1600 kg *\(v^2\)

Now, we can solve for v by rearranging the equation:

\(v^2\) = (2 * 125,000 J) / 1600 kg

\(v^2\) = 156.25 \(m^2/s^2\)

Taking the square root, we find:

v = √156.25\(m^2/s^2\)

v ≈ 12.5 m/s

Therefore, the car is moving at approximately 12.5 meters per second.

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You apply 10 N to a machine and the machine applies 10 N to another object. What is the mechanical advantage ? Rocket Science

Answers

Answer:

The mechanical advantage is 1.

Explanation:

Given;

input force, F₁ = 10 N

output force, F₀ = 10 N

mechanical advantage is called force ratio and it is given as;

M.A = output force / input force

M.A = 10 N / 10 N

M.A = 1

This is an ideal machine scenario.

Therefore, the mechanical advantage is 1.

One ball is dropped at rest from a height of h = 65 m. At the same time, another ball is thrown upward from the ground with initial velocity v2 = 25 m/s. It takes Δt time for them to meet on the way. Defining the positive direction going upward, what is the acceleration of the two balls, a in m/s2?

Answers

If one ball is dropped at rest from a height of h = 65 m.  The acceleration of the two balls is: -9.8 m/s^2 for both balls.

How to find the acceleration?

Let's start by finding the time it takes for the two balls to meet. We know that the ball thrown upward starts from rest, so its initial velocity is 0 m/s. We can use the following kinematic equation:

y = v_i*t + (1/2)at^2

where y is the displacement (in this case, it is the distance between the two balls), v_i is the initial velocity, a is the acceleration, and t is the time. We can set y equal to the initial height of the dropped ball, which is h = 65 m. For the ball thrown upward, the initial position is y = 0.

For the dropped ball:

y = h = 65 m

v_i = 0 m/s

For the ball thrown upward:

y = 0

v_i = 0 m/s

Using the given information, we can solve for t:

h = (1/2)at^2

65 m = (1/2)*(-9.8 m/s^2)t^2

t = sqrt(65 m / (1/2(-9.8 m/s^2))) ≈ 3.64 s

So it takes about 3.64 seconds for the two balls to meet.

Now, we can find the acceleration of the two balls. For the dropped ball, the acceleration is simply the acceleration due to gravity, which is -9.8 m/s^2. For the ball thrown upward, the acceleration is also the acceleration due to gravity, but with a negative sign since it is moving in the opposite direction of gravity. Therefore, the acceleration of the two balls is:

a = -9.8 m/s^2 for both balls

This means that both balls experience the same acceleration due to gravity, regardless of their initial velocities.

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external criticism and internal criticism​

Answers

External criticism:External criticism, which is also known as lower criticism, is a tool used by historians and exegetes to determine the validity of a document, particularly a document with some sort of historical significance.

Internal Criticism:

Internal criticism is applied to check the credibility of the document whether the contents given in it are believable or not.

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External criticism is a process by which historians determine whether a source is authentic by checking the validity of the source. Internal criticism looks at the reliability of an authenticated source after it has been subjected to external criticism.

how is solar impulse is different than conventional airplane

Answers

Answer:

Solar Impulse is a Swiss long-range experimental solar-powered aircraft project and Conventional Aircraft are the atmosphere-only aircraft that have been around since the Wright Flyer's first takeoff. Though modern craft are much more advanced and capable than that early model, Conventional Aircraft have been superseded in modern times by the AeroSpace Fighter, which can operate in both atmosphere and space.

Explanation:

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