mester Exam 1 11 of 35
A car has an oil drip. As the car moves, it drips oil at a regular rate, leaving a trail of spots on the road. Which diagram shows the spots
of car that is continuously slowing down?

Answers

Answer 1
please show picture of diagrams

Related Questions

i just got hit on the head by a basketball.
story of my life.

Answers

Season Leaders

Go to Stats

Points Per Game

1.

Talen Horton-Tucker

26.0

2.

Nikola Jokic

26.0

3.

Zion Williamson

26.0

4.

Giannis Antetokounmpo

24.5

5.

Patty Mills

24.0

Rebounds Per Game

1.

Rudy Gobert

14.0

2.

Harry Giles III

13.5

3.

Clint Capela

13.0

4.

Nikola Vucevic

13.0

5.

Giannis Antetokounmpo

12.0

Assists Per Game

1.

Ja Morant

9.0

2.

Bam Adebayo

8.0

3.

T.J. McConnell

7.5

4.

DeMar DeRozan

7.0

5.

Spencer Dinwiddie

7.0

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Wavelength is a measurement of ___________ , while period is a measurement of ___________ . *
Speed; time
Time; speed
Distance; time
Time; distance

Distance; speed
Speed; distance

Answers

distance ; time

Explanation:

wavelength is in metres [m]

period is in seconds [s]

A street bridge is 5.5m long if the linear expansion of steel is 0.00001 oc How much will it expand when temperatures is by 10oc? Give answer in Cm

Answers

Answer:

[tex]l_o=550.055\ cm[/tex]

Explanation:

Given that,

Length of a street bridge, l = 5.5 m

The coefficient of bridge, [tex]\alpha =0.00001 ^0 C[/tex]

We need to find how much will it expand when temperatures is by 10°C.

The change in length per unit original length is given by :

[tex]\dfrac{\Delta l}{l}=\alpha \Delta T\\\\\Delta l = l\alpha \Delta T\\\\=5.5\times 0.00001 \times 10\\\\\Delta l=0.00055\\\\(l_o-l)=0.00055\\\\l_o=0.00055+5.5\\\\=5.50055\ m\\\\l_o=550.055\ cm[/tex]

Hence, the length will expanded 550.055 cm.

if an atom was a scale, in which the nucleus is the size of an apple the electron.....

Answers

Answer:

the nucleus is the size of an apple, approximately 5 cm of radius e, the atom has a radius of R = 5 cm 104 = 50000 cm = 50 km

Explanation:

In the Rutherford experiments it was proved that the atomic nucleus has the volume 10-4 the volume of the atom.

If we make a scale design in which the nucleus is the size of an apple, approximately 5 cm of radius e, the atom has a radius of R = 5 cm 104 = 50000 cm = 50 km

This shows that almost the entire volume of the atom is empty.

Briefly describe the Rutherford atomic model

Answers

Answer:

The Rutherford model was made by Ernest Rutherford, to describe a atom. That is a brief explanation

Explanation:

A spiral spring of 8cm extended to 9.2cm when a load of 1.6N is applied. what is the force constant of the spring, provided the elastic is not exceeded.​

Answers

Explanation:

By Hooke's Law, Fe = kx.

Since Fe = 1.6N and x = 9.2cm - 8cm = 1.2cm,

k = Fe/x = 1.6N/1.2cm = 1.33N/cm.


vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum
might be:

A. Zero
B.5
C.9
0 15
E.4
and how ? ​

Answers

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

[tex]R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45[/tex]

The minimum sum of the resultant vector,

[tex]R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10[/tex]

So, the magnitude of this sum either 45 or -10.

A little girl pushes a 5.0 kg toy baby stroller at constant speed 7.0 m across the floor. She pushes on the handle with a force of 40 N at an angle of 30o with the horizontal. All parts are 4 points each.

Answers

Complete Question

1 a A little girl pushes a 5.0 kg toy baby stroller at constant speed 7.0 m across the floor. She pushes on the handle with a force of 40 N at an angle of 30o with the horizontal.  How much work is done by the girl on the wagon?

1b  A farmhand pushes 20 ㎏ bale of hay 4m across the floor of the barn if she exerts a horizontal force of 60 N on the hay, how much work is done? (5 pts)

All parts are 4 points each

Answer:

1a

  [tex]W = 242.5 \ J[/tex]

1b

    [tex]W = 240 \ J[/tex]

Explanation:

Considering question a

From the question we are told that

       The mass of the toy baby stroller is  [tex]m = 5.0 \ kg[/tex]

        The distance covered is    [tex]d = 7.0\ m[/tex]

          The force the girl applies on the handle  is  [tex]F = 40 \ N[/tex]

          The angle at which this force is applied is  [tex]\theta = 30^o[/tex]

Generally the workdone is mathematically represented as

          [tex]W = F_x * d[/tex]

Here  [tex]F_x[/tex] is the force along the horizontal axis , this is mathematically represented as

           [tex]F_x = F cos (\theta )[/tex]

=>         [tex]F_x = 40 * cos(30 )[/tex]

=>         [tex]F_x = 34.64 \ N[/tex]

So

              [tex]W = 34.64 * 7[/tex]

=>           [tex]W = 242.5 \ J[/tex]

Considering question b

From the question we are told that

       The mass of the toy baby stroller is  [tex]m = 20 \ kg[/tex]

        The distance covered is    [tex]d = 4 \ m[/tex]

          The force the girl applies on the handle  is  [tex]F = 60 \ N[/tex]      

Generally the workdone is mathematically represented as

          [tex]W = F * d[/tex]

=>     [tex]W = 60 * 4[/tex]

=>     [tex]W = 240 \ J[/tex]

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 59.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.852 m/s2. Calculate her mass.

Answers

Answer:

The value is  [tex]m = 69.24 \ kg[/tex]

Explanation:

From the question we are told that

   The value of the external force is  [tex]F = 59.0 \ N[/tex]

    The magnitude of the astronaut's acceleration is  [tex]a = 0.852 \ m/s[/tex]

Generally Newton's Second Law of Motion from the mass of the astronauts is mathematically represented as

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

=>         [tex]m = \frac{59 }{0.852 }[/tex]

=>         [tex]m = 69.24 \ kg[/tex]  

A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
that you must drop into the bucket. Where should the bucket be when you drop the mass (so it is moving
purely vertically when it lands in the bucket) if your goal is to:
(a) Maximize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.
(b) Minimize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.

Answers

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = [tex]\sqrt{2y_{o}/g}[/tex]

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = [tex]\sqrt{k/m}[/tex]

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - [tex]\sqrt{k/3}[/tex] [tex]\sqrt{2y_{o}/g}[/tex]

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

types of magnetic resonance image​

Answers

Answer: ur mom

Explanation:

How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires
0.05 Amp to operate it.

Answers

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

t is the time

Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs  = 1,000,000/3600

1,000,000secs = 277.78 hours

Hence it will take 277.78 hours for the charge to flow through the diode

Using a scale of 1 cm to represent 10 N, find the size and direction of the resultant of forces of 50 N
and 30 N acting at the angle of 45 O to each other.

Answers

Answer:

[tex]74.31\ \text{N}[/tex]

[tex]16.59^{\circ}[/tex]

Explanation:

P = 50 N

Q = 30 N

[tex]\theta[/tex] = Angle between the vectors = [tex]45^{\circ}[/tex]

Resultant is given by

[tex]R=\sqrt{P^2+Q^2+2PQ\cos\theta}\\\Rightarrow R=\sqrt{50^2+30^2+2\times 50\times 30\times \cos45^{\circ}}\\\Rightarrow R=74.31\ \text{N}[/tex]

Angle of resultant

[tex]\phi=\tan^{-1}\dfrac{Q\sin\theta}{P+Q\cos\theta}\\\Rightarrow \phi=\tan^{-1}\dfrac{30\times \sin45^{\circ}}{50+30\cos45^{\circ}}\\\Rightarrow \phi=16.59^{\circ}[/tex]

Magnitude of the resultant is [tex]74.31\ \text{N}[/tex]

Direction of the resultant is [tex]16.59^{\circ}[/tex]

A girl rides her bike at 15 m/s for 20 s. How far does she travel in that time?

Answers

We are given:

Initial Velocity(u) = 15 m/s

Time interval(t) = 20s

Solving for the distance covered:

Since the girl keeps riding her bike at 15 m/s, her speed is constant and hence, acceleration of the bike is 0 m/s²

acceleration(a) = 0 m/s²

We know that:

s = ut + 1/2(at²)                [second equation of motion]

s = (15)(20) + 1/2(0)(400)  [plugging the values]

s = 150 + 0

s = 300 m

Hence, the girl covered 300 m in 20 seconds

What’s the answer to this

Answers

The Cartesian product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set {{1.09m, 12.8s}}

2. Two identical spheres having charges Q and -
2Q experience a force F at a certain distance
If the spheres are kept in contact and then
placed at same initial distance, the force
between them will be​

Answers

force between the two charges (+q1 and +q2),if they are at a distance 'a' is

F1=1/4pieEo q1q2/d^2................ (1)

when the metal spheres are in contact the charge flow from one sphere to another till both the sides acquires the same charge. here q1 and q2 are of same sign,hence after contact each sphere will have a charge

[tex] \binom{q1 + q2 }{2} [/tex]

now,the force between them,

f2=1/4pieEo

(q1+q2/2)^2/d^2

from eq (1)and eq (2)

f2=f1 (q1+q2)^2/4q1q2

2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from
west to east, while along the second track, train B moves with a speed of 60 m/s from east to west calculate speed of
Bw.rt. A​

Answers

Answer:

[tex]Relative\ Velocity = 105m/s[/tex]

Explanation:

Given

[tex]V_A = 45m/s[/tex]

[tex]V_B = 60m/s[/tex]

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a [tex]sum\ of\ velocities\ of[/tex] both trains:

This is shown below:

[tex]Relative\ Velocity = V_A + V_B[/tex]

[tex]Relative\ Velocity = 45m/s + 60m/s[/tex]

[tex]Relative\ Velocity = 105m/s[/tex]

(a) Calculate the linear acceleration of a car, the 0.220-m radius tires of which have an angular acceleration of 11.0 rad/s2. Assume no slippage.

Answers

Answer:

The value is  [tex]a_t = 2.42 \ m/s^2[/tex]

Explanation:

From the question we are told that

  The radius of the tires is  [tex]r = 0.22 \ m[/tex]

   The  angular acceleration is  [tex]\alpha = 11.0 \ rad/s^2[/tex]

Generally the linear acceleration is mathematically represented as

     [tex]a_t = r * \alpha[/tex]

=>  [tex]a_t = 0.22 * 11[/tex]

=>  [tex]a_t = 2.42 \ m/s^2[/tex]

b) A satellite with mass m orbits the Earth at a radius r. A second satellite also with mass m orbits the
Earth at twice the radius.
How does the force of Earth's gravity acting on the two satellites
compare? PLEASE HURRY

Answers

Answer:

So, given the eqn Fg=G(m1+m2/r^2) where G is the gravitational constant, m is the mass of the satellite and m2 is the mass of the earth and r is the distance from earth to the satellite, the force of earths gravity should be quartered.

Cause (2r)^2 gets turned into (4r^2) where 4r^2 is compared to r^2

Explanation:

Can you help with this question please thanks

Answers

Answer:

no .the blue runner began 16 m ahead of the red runner

calculate the force necessary to keep a mass of 2 kg moving on a circular path of radius 0.2 m with a period of 0.5 second. what is the direction of this force

Answers

Brideghdsrgjjncddionc

A car moving initially at 20 m/s accelerates up to 60 m/s during the
course of 5 seconds. The average acceleration of the car is m/s2

Answers

Gain in velocity = 60 - 30 = 30 m/s.

Time taken for this gain = 5 seconds

Average acceleration = 30 / 5 = 6 m/s/s.

So, 6 metres per second squared.
472 viewsView 2 Upvoters
Related Questions (More Answers Below)

Help please I would appreciate it

Answers

Answer:

i think the red runner travels greater distance ie.40 m

Explanation:

although ,the blue one travels 21 m but in the backward direction .so the correct ans is "the red runner travels 40 m"

What is the lithosphere?
A. the outer layer of the Earth's crust
B. the inner core
C. the middle portion of the mantle
D. the outer core

Answers

A. The outer layer of the earth

Answer:

a. outer layer

Explanation:

lithosphere is right underneath the continental and ocean crust. it is approximately 100 km in deep and it is a brittle layer. It is broken into tectonic plates.

the inner core is located at the very center and its full of iron and nickel (so its not B)

on top of that is the outer core which is liquid (not D)

the middle portion of the mantle is the asthenosphere and mesosphere. they are right beneath the lithosphere. (not C)

so the best answer is A

А
Each sentence describes potential and kinetic energy
at various locations on a slide. Complete each
statement by selecting the position of the object on the
slide.
An object at position has all kinetic energy
An object at position has all potential energy
An object at position has about half potential energy
and half kinetic energy.
B
с

Answers

Answer:

C. A. B.

Explanation:

Credit to Lainey in the comments for the answer!

Answer:

1] C

2] A

3] B

Explanation:

what are the very small particles that make up matter​

Answers

Atoms
Hope this helps!!

Answer:

The very small particles that make up matter are I) Atoms

Matter - Anything that have mass and occupies space is called matter . it is made up of atoms and molecules

Atoms - The smallest part of matter is called atom.

Molecule - Group of atoms combine together to form a molecule.

More to know -

Atom is made up of even smaller particles called neutron, proton and electron.

Electron moves around nucleus ( nucleus is made up of neutron and proton)

Different types of atom combine and form molecule ( Nitrogen dioxide No2 has 1 atom of nitrogen and 2 atoms of oxygen)

What kind of energy is produce when sun reaches solar panel?

Answers

Answer:

Radient to ElEcTrIcAAl

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

The energy produced when the sun reaches solar panel is nuclear fusion

A has a frequency of 300 Hz and a wavelength of 1.10 m. What is the velocity of the wave?

Answers

Hello!!

For calculate the Velocity of the wave let's applicate the formula:

[tex]\boxed{V=f*\lambda}[/tex]

[tex]\textbf{Being:}[/tex]

[tex]\sqrt{}[/tex] V = Velocity = ?

[tex]\sqrt{}[/tex] f = Frequency = 300 Hz

[tex]\sqrt{}[/tex] [tex]\lambda[/tex] = Wavelength = 1,1 m

⇒ [tex]\text{Then let's \textbf{replace it according} we information:}[/tex]

[tex]V = 300 \ Hz * 1,1 \ m[/tex]

⇒ [tex]\text{Let's resolve it: }[/tex]

[tex]V = 330 \ m / s[/tex]

[tex]\textbf{Result:}\\\text{The velocity is \textbf{330 meters per second}}[/tex]

Answer:

For calculate the Velocity of the wave let's applicate the formula:

V = Velocity = ?

f = Frequency = 300 Hz

 = Wavelength = 1,1 m

⇒  

⇒  

Explanation:

What is the change in potential energy in moving a 10kg box from the floor to a table 1 m high?

Answers

Explanation:

potential energy formula is 1/2 mv2(this 2 is square).

m means mass and v means velocity.

now, you can calculate

The change in potential energy in moving a 10kg box from the floor to a table 1 m high is 98 Joules.

To find the potential energy, the given values are,

Mass of the box = 10Kg

Height h = 1 m

What is Potential energy?

The potential energy can be defined as the energy that is stored and that can be determined through various parts in a particular system.

Also, the energy will gets stored when the object is not moving.

For example: Spring. Even when the spring is not stretched or contracted it stores energy.

The formula for Potential energy PE= mgh joules.

m is the Mass,

g is the acceleration due to gravity = 9.8 m/s²,

h is the height.

Substituting all the given values,

Potential Energy= 10 (9.8)(1)

PE= 98 J

Thus the potential energy of the box= 98 J.

To learn more about the potential energy,

https://brainly.com/question/13548111

#SPJ2

Which term best describes the motion of the rope particles in relation to the motion of the rope wave shown in the photograph

Answers

Answer:

A: Perpendicular

Explanation:

The question is incomplete as it lacks the image of the rope wave motion.

However, as found on "estudyassistant", the options are;

A) Perpendicular

B) Circular

C) Longitudinal

D) Parallel

From all that, we can say that;

The rope's are moving simultaneously in the same pattern without touching each other.

This is therefore a mechanical wave being created with the motion having oscillations that are perpendicular to the direction of energy transfer of the ropes.

This is a definition of transverse waves because the rope particle motion is perpendicular to the wave motion.

Answer:

A: Perpendicular

Explanation:

Read above explanation.

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