A 0.20-kg baseball is thrown with a speed of 20 m / s. If the speed of the ball at the start of the throw is zero, calculate the net work during the throw.

Answers

Answer 1

Explanation:

Work = change in energy

W = ½ mv²

W = ½ (0.20 kg) (20 m/s)²

W = 40 J


Related Questions

A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.

Answers

Answer:

Explanation:

force constant of spring  k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

where m is mass of the body attached with spring

Putting the values

[tex]\omega = \sqrt{\frac{71.2}{5} }[/tex]

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .

The mass of the crate can also be adjusted by clicking on the More Features tab and then using the slider bar in the right panel. How does the maximum angle for which the crate can remain at rest on the ramp depend on the mass of the crate

Answers

Answer:

The maximum angle does not depend on the mass

Explanation:

This is because In as much as the force acting on the normal which is the the maximum force of static friction increases as the mass increases, the component of the force of gravity parallel to the ramp increases at the same rate. Thus The maximum angle is independent of the mass.

A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a stopwatch?

Answers

Answer:

A stop watch measures time and the number of seconds making an osculation is the period T

Which is T = 2π√(L/g)........solving for L (the length of the pendulum),

Then the L = T²x g/( 4πr²) substitute and then u find length or height of the swing

Consider a vacuum-filled parallel plate capacitor (no dielectric material between the plates). What is the ratio of conduction current Jc to displacement current Jd at 10 MHz

Answers

Answer: 1798

Explanation:

Given that there is no dielectric material between the plates,

the permittivity of free space = 9 × 10^-12 f/m

The frequency F = 10 MHz

The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.

Please find the attached file for the solution

How many grams of water (H2O) have the same number of oxygen atoms as 7.0 mol of oxygen gas

Answers

Answer:

126 g of water, H2O.

Explanation:

First, we'll begin by calculating the number of atoms in 7 mole oxygen gas.

This is illustrated below:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This means that 1 mole of O2 also contains 6.02×10²³ atoms.

Now, if 1 mole of O2 contains 6.02×10²³ atoms, then 7 moles of O2 will contain = 7 × 6.02×10²³ = 4.214×10²⁴ atoms.

Finally, we shall determine the mass of H2O that will contain 4.214×10²⁴ atoms.

This is illustrated below:

6.02×10²³ atoms is present in 1 mole of any substance contains

1 mole of H2O = (2x1) + 16 = 18 g

6.02×10²³ atoms is present in 18 g of H2O.

Therefore, 4.214×10²⁴ atoms will be present in = (4.214×10²⁴ × 18)/6.02×10²³

= 126 g of water, H2O.

Therefore, 126 g of water, H2O will contain the same number of oxygen atoms as 7.0 mol of oxygen gas.

7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.

We want to know the mass of water (H₂O) that has the same number of oxygen atoms as 7.0 mol of oxygen gas (O₂).

First, we will calculate the number of oxygen atoms in 7.0 mol of O₂ considering the following relations.

1 mol of O₂ contains 6.02 × 10²³ molecules of O₂ (Avogadro's number).1 molecule of O₂ contains 2 atoms of O.

[tex]7.0 mol O_2 \times \frac{6.02 \times 10^{23}molecule O_2 }{1molO_2} \times \frac{2atomO}{1molecule O_2} = 8.4 \times 10^{24} atomO[/tex]

Now, we want to calculate the mass of H₂O that contains 8.4 × 10²⁴ atoms of O. We will consider the following relations.

1 molecule of H₂O contains 1 atom of O.1 mol of H₂O contains 6.02 × 10²³ molecules of H₂O (Avogadro's number).The molar mass of H₂O is 18.02 g/mol.

[tex]8.4 \times 10^{24} atomO \times \frac{1moleculeH_2O}{1atomO} \times \frac{1molH_2O}{6.02 \times 10^{23}moleculeH_2O } \times \frac{18.02gH_2O}{1molH_2O} = 250 gH_2O[/tex]

7.0 mol of oxygen gas contain the same number of oxygen atoms as 250 g of water.

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A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. Assuming constant acceleration, how far has the cheetah run in this time

Answers

Answer:

The distance covered by the cheetah during the motion is 77.75 m

Explanation:

Given;

initial velocity of cheetah, u = 0

final velocity of cheetah, v = 25 m/s

time of acceleration, t = 6.22 s

Apply kinematic equation;

[tex]s = (\frac{v+u}{2} )t[/tex]

where;

s is the distance covered by the cheetah during the motion

[tex]s = (\frac{v+u}{2} )t\\\\s = (\frac{25+0}{2} )6.22\\\\s = 77.75 \ m[/tex]

Therefore, the distance covered by the cheetah during the motion is 77.75 m

A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?
A) What is rms speed?
B) What is the mean free path?
C) What is the thermal energy of gas?
D) What is the molar specific at a constant volume?

Answers

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

The rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.

Given that  

[tex]\bold{\dfrac {V_2}{V_1}= 0.25}[/tex]

In adiabatic process  

[tex]\bold {TV^\gamma^-^1}[/tex]= constant  

So  

[tex]\bold {\dfrac {T1}{T2}=(\dfrac {V1 }{V2})^ \gamma^-^1}}\\[/tex]  

So,

[tex]\bold {\dfrac {T_1} {T_2} = (\dfrac { 1}{0.25})^ 0^.^6^6= 2.5}[/tex]  

for pressure,

[tex]\bold {PV^\gamma = \ Constant }[/tex]  

So

[tex]\bold {\dfrac {P1}{P2}=(\dfrac {V1 }{V2})^ \gamma}}\\\\\bold {\dfrac {P1}{P2}=(\dfrac {1}{0.25})^ 0^.^6^6} = 9.98 } }[/tex]  

A. rms speed can be calculated as

 

[tex]\bold {\dfrac {Vrms 2}{Vrms1}= \sqrt {T2T1})}\\\\\bold {Vrms2 =\sqrt {2.5} = 1.6\ Vrms1 }[/tex]

B.  The mean free path can be calculated as

[tex]\bold {\dfrac {\lambda_1 }{\lambda_2} = \dfrac {V_1}{V_2} = 0.25 \\ }[/tex]

Mean free path is 0.25 times the first mean free path.

C.

[tex]\bold {Eth= \dfrac {3}{2}kT}}\\\\\bold {\dfrac {E_t2}{E_t1} = \dfrac {T_2}{T_1}}\\\\\bold {E_t2= 2.5\ E_t1}[/tex]  

 

D.  the molar specific at a constant volume can be calculated a using,

[tex]\bold {CV= \dfrac 3{2}R }[/tex]  

 [tex]\bold {Cv_f= Cv_i }[/tex]

So, molar specific heat constant will not change.

Therefore, the rms speed of the system is 1.6 while Mean free path is 0.25 times the first mean free path.

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What does this picture show?

Answers

Answer:

d poor accuracy,poor precision

The diagram represents how much accurate and precise the result data is. It is clear from the image that, the data points have poor accuracy and good precision.

What is accuracy ?

Accuracy of a result is the measure of the closeness of the experimental or calculated value to the true value or absolute value of a measurement.  For a reproducible experiment, the results for each trials can be differ or close.

The closeness between values of a set of experiments is called precision of the results. Not all accurate value be can be precise. Similarly precision of values does not need to meet accuracy.

Here, the central dot indicates the true value. The green points are a set of experimental values. They are not so close to the true value and hence it is less accurate. However, all the points and closer together. Hence, they are precise. Therefore, option B is correct.

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A grating has 320 lines/mm. How many orders of the visible wavelength 551 nm can it produce in addition to the m = 0 order

Answers

Answer:

6

Explanation:

Given that

dsinθ = mλ,

now, if sinθ = 1, then

m = d / λ, where

m = order of interference

d = distance between the slits

λ = wavelength of light

this is the formula we would use to solve the question

d = 1 / 320 lines/mm

d = 1 / 320*10^3

d = 3.125*10^-6 m

At λ = 551 nm, we have

m = 3.125*10^-6 / 551*10^-9

m = 5.67

5.67 ~ 6

thus, we can say that the orders of visible wavelength 551 nm, can produce is 6

For a conductor in electrostatic equilibrium, which of the following properties are true?
a) Any excess charge is uniformly spread throughout the volume of the conductor.
b) The electric field inside is zero.
c) The entire conductor is at the same potential.

Answers

Answer:

b) The electric field inside is zero.

Explanation:

This is simply because the charges in the field will be in motion, thus this charges kind of spread around to make the conductor field zero.

A plane moves 100 meters to the right in two seconds. What is its velocity?
0 m/s
O 30 m/s
50 m/s
O 100 m/s
200 m/s

Answers

Answer:

50 m/s

Explanation:

[tex]Distance = 100m\\Time = 2 secs\\\\Velocity = \frac{Distance}{Time} \\\\V = \frac{100m}{2s} \\\\= 50m/s[/tex]

The velocity of any point on a rigid body is _________ to the relative position vector extending from the IC to the point. Group of answer choices

Answers

Answer:

The velocity of any point on a rigid body is ___Always perpendicular______ to the relative position vector extending from the IC to the point

Explanation:

This is because The instantaneous center (IC) of zero velocity for the rigid body is at the point in contact with ground. The velocity direction at any point on the body is always perpendicular to the line connecting the point to the IC.

The velocity of any point on a rigid body is : Perpendicular   to the relative position vector extending from the IC to the point.

The IC ( instantaneous center ) is the point of the rigid body in contact with the ground when the rigid body is at zero velocity.

The velocity direction from any point of the rigid body will always be perpendicular to the ( IC ) since the IC is the reference point of the rigid body when the body is at rest ( zero velocity ).

Hence we can conclude that the velocity of any point on a rigid body will be  perpendicular.

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Although the options related to your question is missing an accurate answer is provided within the scope of your question

How far can a person run in 0.25hr if he or she runs at a average speed of 16km/he?

Answers

Answer:

4km

Explanation:

d = st

d=0.25*16

d=4km

Which describes the amplitude of a wave when it carries more ene
O It is higher.
O It is lower.
It is darker.
Oli is lighter​

Answers

Answer:

It is higher.

Explanation:

Hello,

In this case, the quantity of energy that a wave is able to transfer is directly related its amplitude exhibiting that the higher the carried energy, the higher its amplitude and on the flip side, the lower the energy the lower the amplitude, therefore, the answer is it is higher. You can verify this on the attached picture.

Regards.

Answer:

It is higher.

Explanation:

Simple Harmonic Motion: The quartz crystal in a digital watch has a frequency of 32.8 kHz. What is its period of oscillation

Answers

Answer:

Time Period of Oscillation = 3.04 x 10⁻⁵ s

Explanation:

The time period of oscillation and the the frequency of oscillation are two inter linked quantities. They are actually the reciprocals of each other. So, the time period of oscillation of quartz crystal is given by the following formula:

Time Period of Oscillation = 1/Frequency of Oscillation

where,

Frequency of Oscillation = 32.8 KHz

Frequency of Oscillation = 32800 Hz

Therefore,

Time Period of Oscillation = 1/32800 Hz

Time Period of Oscillation = 3.04 x 10⁻⁵ s

The rotating loop in an AC generator is a square 12.0 cm on a side. It is rotated at 70.0 Hz in a uniform field of 0.800 T. Calculate the following quantities as functions of time t, where t is in seconds. I need help with d and e.
(a) the flux through the loop
11.52cos(140πt) mT·m2
(b) the emf induced in the loop
5.067sin(140πt) V
(c) the current induced in the loop for a loop resistance of 2.50 Ω
2.03sin(140πt) A
(d) the power delivered to the loop
4.12sin2(140πt) W (wrong)
(e) the torque that must be exerted to rotate the loop
mN·m

Answers

Answer:

Explanation:

d )

power in an electrical loop = volt x current

= 5.067sin(140πt)  x 2.03sin(140πt)

= 10.286 sin²(140πt) .

10.286 ( 1 - cos 280πt ) / 2

= 5.143 ( 1 - cos 280πt )

e )

Torque on a current carrying loop in a magnetic field

= M B sinθ

M is magnetic moment of coil , B is magnetic field , θ is angle between area vector of coil and direction of  B .

Here magnetic moment of coil

= A i where A is area of coil and i is current .

= .12² x 2.03sin(140πt)

= .029 sin(140πt)

B = .8 T

MB = .8 x .029 sin(140πt)

= .0232  sin(140πt) x sin(140πt)

=.0232 sin²(140πt).

The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.44 m. She sets the pendulum swinging, and her collaborators carefully count 1.10×102 complete cycles of oscillation during 2.00×102 s. What is the result

Answers

Answer:

18808.7 m/s^2

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = [tex]2\pi \sqrt{\frac{L}{g} }[/tex]

where g is the acceleration due to gravity

substituting values, we have

0.55 = [tex]2\pi \sqrt{\frac{1.44}{g} }[/tex]

0.0875 = [tex]\sqrt{\frac{1.44}{g} }[/tex]

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = 18808.7 m/s^2

A family made the observation that more expensive brands of popcorn seem to produce more popped kernels than cheaper brands of popcorn. In order to test this hypothesis, the family conducted an experiment in which they followed the procedures below: The family bought 2 differing brands of popcorn to test: Wilbur Bockenreder Popcorn and PopWhisper. They measured 1 cup of kernels and popped 5 bowls of each type of popcorn. They ensured they popped the popcorn for the same amount of time, used the same wattage microwave, and did not have any other ingredients added to the kernels (butter, etc). They counted and recorded the number of unpopped kernels in each of the 10 bowls. As they analyzed the data collected, they were able to conclude that their initial hypothesis was supported. They formed the conclusion that the more expensive brand of popcorn popped more kernels than the cheaper brand of popcorn. From this experiment, let's answer some questions. Identify the independent variable.

Answers

Answer:

The different types of corn used.

Explanation:

Independent variable: In research methods, the term "independent variable" is determined as a variable that is being manipulated, changed, or altered in an experiment by the experimenter in order to see its effect on the dependent variable. The changes in the dependent variable in an experiment depends on the independent variable directly.

The independent variable in the popcorn experiment is brand of popcorn used.

INDEPENDENT VARIABLE:Independent variable in an experiment is the variable that the experimenter changes or manipulates in order to bring about a response.

According to this question, a family is conducting an experiment to test which brand of popcorn seem to produce more popped kernels. They used two brands of popcorn as follows: Wilbur Bockenreder Popcorn and PopWhisper.

However, the brand of popcorn was changed in this experiment, hence, the brand of the popcorn is the independent variable.

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How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 36.0°C greater than when they were laid? Their original length is 26.0 m.

Answers

Answer:

11 mm

Explanation:

Original length L' = 26 m

Increase in temperature dT = 36.0°C

The gap l =?

The coefficient of linear expansion for steel & = 12 × 10^−6 per °C

The gap will be gotten from the equation of linear expansion.

l = L' x & x dT

substituting, we have

L = 26 x 12 × 10^−6 x 36

L = 0.011 m = 11 mm

The x component of vector is 8.7 units, and its y component is -6.5 units. The magnitude of is closest to

Answers

Answer:

F = 10.86 units

Explanation:

The magnitude of a vector in terms of the magnitude of its rectangular components is given by the following formula:

F = √(Fₓ² + Fy²)

where,

F = Magnitude of the Vector = ?

Fₓ = magnitude of the x-component of vector = 8.7 units

Fy = magnitude of y component of vector = - 6.5 units

Therefore, using these values in the equation, we get:

F = √[(8.7 units)² + (- 6.5 units)²]

F = √(117.94 units²)

F = 10.86 units

Capacitor C1 is in series with capacitors C2 and C3 in parallel. Then three capacitor system is connected to battery with V0. Determine the charge stored by C1 when C1 = 20 μF, C2 = 10 μF, C3 = 30 μF, and V0 = 18 V.g

Answers

Answer:

  Q₁ = 2.4 10⁻⁴ C

Explanation:

We have a circuit with several capacitors, let's find the equivalent capacitor of the parallel

    [tex]C_{eq1}[/tex] = C₂ + C₃

    C_{eq1} = (10 +30) 10⁻⁶

    C_{eq1} = 40 10⁻⁶ F

There remains a series system between C₁ and C_{eq1}, let's find the equivalent capacitor

     1/C_{eq2} = 1 / C₁ + 1 / C_{eq1}

     1 /C_{eq2} = 1 / 20 10⁻⁶ + 1/40 10⁻⁶

     1 / C_{eq2} = 0.075 10⁶

     C_{eq2} = 13.33 10⁻⁶ F

let's use the relationship

        V = Q / C_{eq2}

        Q = V C_{eq2}

        Q = 18  13.33 10⁻⁶

        Q = 2.4 10⁻⁴ C

In a combination of capacitors in series the charge is constant, so the charge on C₁ is the same

        Q₁ = 2.4 10⁻⁴ C

A person whose near-point distance is 45.5 cm wears a pair of glasses that are 2.1 cm from her eyes. With the aid of these glasses, she can now focus on objects 25 cm away from her eyes.

Required:
a. Find the focal length of her glasses.
b. Find the refractive power of her glasses.

Answers

Answer:

a

   [tex]f = 0.4848 \ m[/tex]

b

  [tex]p = 2.063 D[/tex]

Explanation:

From the question we are told that

   The  near point distance is  [tex]k = 45.5 \ cm[/tex]

    The  distance of the glasses from the eye is  [tex]y = 2.1 \ cm[/tex]

    The  distance of an object she can focus with the glass is  [tex]i= 25 \ cm[/tex]

Generally the image distance is mathematically evaluated  as

     [tex]v = -(45.5 - 2.1)[/tex]

     [tex]v = -43.4 \ cm[/tex]

Generally the object distance is mathematically represented as

     [tex]u = (25 -2.1)[/tex]

     [tex]u = 22.9 \ cm[/tex]

The  negative sign tells us that the image was formed behind the eye  

Generally the lens formula  is mathematically represented as  

        [tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]

=>   [tex]\frac{1}{f} = \frac{1}{22.9} + \frac{1}{ - 43.4}[/tex]

=>  [tex]f = 48.48 \ cm[/tex]

 converting to meters  

   [tex]f = 0.4848 \ m[/tex]

Thus the refractive power is mathematically represented as

    [tex]p = \frac{1}{f}[/tex]

=>  [tex]p = \frac{1}{0.4848 }[/tex]

=>   [tex]p = 2.063 D[/tex]

   

       

The temperature of a plastic cube is monitored while the cube is pushed 8.6 m across a floor at constant speed by a horizontal force of 19 N. The monitoring reveals that the thermal energy of the cube increases by 120 J. What is the increase in the thermal energy of the floor along which the cube slides

Answers

Answer:

Answer:

43.4J

Explanation:

We know that

Work done = total heat energy

But work done is force x distance

=> F = 19 x8.6 = 163.4 J

So the total heat. Will be Heat of cube + heat of floor = 163.4J

So that heat of floor will now be

floor = 163.4 J - 120 J = 43.4 Joules

Explanation:

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on each? A. 4.0x10-6 C 4.0x10-6 C B. 7.4x10-6 C 0.6x10-6 C C. 6.6x10-6 C 1.4x10-6 C D. 5.0x10-6 C 3.0x10-6 C

Answers

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂   -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

Answers

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

I would appreciate it!

Answer:

[tex]R\approx12.43 \,\, \Omega[/tex]

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

[tex]V=I\,\,R\\R = \frac{V}{I} \\R=\frac{87}{7} \\R\approx12.43 \,\, \Omega[/tex]

If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour

Answers

Answer:

Mass of flake = 67.716 kg

Explanation:

Given:

Length of lawn = 21 ft

Width of lawn = 20 ft

Note:

Each snow flake mass = 1.90 mg

Find:

Weight of total flake in a minute

Computation:

Area of lawn = 21 × 22

Area of lawn = 440 ft²

Amount of flake per minute = 440 × 1350

Amount of flake per minute = 594000 flake/ minutes

Mass of flake = 594000 × 1.90 mg × 60 minutes

Mass of flake = 67,716,000 mg

Mass of flake = 67.716 kg

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

Explanation:

The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow

putting the values

π x .2945² x 12.1

= 3.3  m³ /s

To know the pipe's diameter at the refinery we shall apply the following formula

a₁ v₁ = a₂ v₂

a₁ v₁ and  a₂ v₂ are volume rate of flow of liquid which will be constant .

3.3 = a₂ x 6.29

a₂ = .5246 m³

π x r² = .5246

r = .4087 m

= 40.87 cm

diameter

= 81.74 cm

When your scalpel gets dull or is broken, it should be disposed of by
a. Placing in the broken glass/sharps container
b. Thrown in the trash
c. Kept in your dissection kit

Answers

I would assume that the answer is A

1) A cyclist moves a distance of 2000 meters during a time of 20 minutes. How fast is the cyclist?

2) A car moves for 20 minutes on a road with constant speed. If he traveled 60 kilometers, what is his speed?

Answers

Answer:

1)  speed of a cyclist = 100 m/min.

2) speed of a car moving = 3 km/min.

Explanation:

1)  speed of a cyclist = distance over time

                                  = 2000 m / 20 min

                                  = 100 m/min.

2) speed of car moving = distance over time

                                       = 60 km / 20 min

                                        = 3 km/min.

Where are the most reactive elements located on the periodic table.

Answers

Answer:

The most reactive elements are at the bottom left corner of the periodic table.  Those are the elements that have the most active/most reactive.  For example, lithium, sodium, and potassium all react with water!

Explanation:

Don't know what to explain :v

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