An automobile travels to the right, with the center Aof the wheel moving at a constant speed of 48 mi/h. If the diameter of a wheel is 22in.,determine the velocities of points B, C, D, and E on the rim of the wheelassuming there is no slipping. [Hint: at any given instant the velocity of point C, which is in contact with the ground must be zero]. Please introduce axes and give your answers in vector form.

Answer:

C. A. B.

Explanation:

Credit to Lainey in the comments for the answer!

Answer:

1] C

2] A

3] B

Explanation:

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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"

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.

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

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:

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.

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.

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

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]

distance ; time

Explanation:

wavelength is in metres [m]

period is in seconds [s]

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]

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.

Answer:

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

Explanation:

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]

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.

Answer:

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

Explanation:

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

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

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

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:

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

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.

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}  }

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]  

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]

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.

Answer: ur mom

Explanation: