Jerry is conducting an experiment to test friction. His setup is shown below. If he wants to test the friction between different types of materials, what should he change in the setup above? A. Replace the mass with a different size mass B. Replace the string with a different length string C. Replace the spring scale with a different kind of scale D. Replace the wooden block with a different kind of block

Answer:

a) to the right

b) up

c) down

d) there will be no force on the proton

e) there will be no force on the proton

Explanation:

The complete question is

If the velocity of a proton is straight up(thumb pointing up) then RHR2 (right hand rule) shows that the force points to the left. What would the direction of the force be if the velocity were a)down, b)to the right, c)to the left, d)into the page, and e)out of the page?

The right hand rule for a positive charge states that...

Hold the thumb of the right hand at right angle to the rest of the fingers, and the rest of the fingers parallel to one another, and pointing away from the body. If the thumb shows the velocity of a positive charge in a magnetic field, and the fingers all point in the direction of the magnetic field, then the palm will push in the direction of the force on the positive charge (proton).

From this, we can deduce from the original statement about this proton that the direction of the field is into the screen of this computer. with that field direction held constant, we can work out that

a) if the thumb point down, the force will be to the right

b) if the thumb points to the right, the force will be upwards

c)  if the thumb points to the left, the force is downwards

d)  if the thumb points into the page, then there will be no force on the proton since the proton must travel perpendicularly to the magnetic field for a force to be induced on it

e) explanation is the same as foe option d.

Answer:

6?

Explanation:

Answer:

did you get the answer bc I think its -3

Explanation:

Hey There!! ~

The answer to this is: Complete these sentences.

If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is 15J.

If the kinetic energy of the ball decreases, then the potential energy will Increase.

Hope It Helped!~

[tex]ItsNobody[/tex]~

Answer:

A ) [tex]5.197 \times 10 ^{-9} inches[/tex]

b) There are approximately 100,000,000 atoms on the 1.32 cm lines

Explanation:

To ensure the accuracy of our answer, we will have to make sure we work in the appropriate units.

A

To convert    [tex]1.32 \times 10^{-10} m[/tex] to inches, we can use this factor.

1 m = 39.3701 inches

Therefore [tex]1.32 \times 10^{-10} m[/tex] =  

B

To find the number of atoms on a 1.32 cm line, we will first of all need to convert 1.32 cm to meters.

1.32 cm = 0.0132 metres (divided 1.32 by 100 to convert to metres)

Assuming all the atoms are arranged side by side, with their edges touching on the 0.0132m line, the number of atoms present will be

[tex]0.0132/1.32\times 10^{-10}[/tex] = 100,000,000 atoms

Answer:

smooth

go far

Answer:

a

 The  charge is  [tex]Q =  2.0*10^8 \  C[/tex]

b

The number of electrons  [tex]N  =  1.25 *10^{27} \  electrons[/tex]

Explanation:

From the question we are told that

    The  maximum rate of charge flow is  [tex]\frac{dq}{dt} = I =  20000 \  C/s[/tex]

Here  I means current

    The time interval is [tex]t=100 \mu s  = 100*10^{-6}[/tex]

Generally the amount of charge is mathematically represented as

        [tex]Q =  \frac{I}{t}[/tex]

=>      [tex]Q =  \frac{20000}{ 100*10^{-6}}[/tex]

=>    [tex]Q =  2.0*10^8 \  C[/tex]

The number of electrons that flow during that time is evaluated as

    [tex]N  =  \frac{Q}{e}[/tex]

Here e is the charge on a single electron with value [tex]e=  1.60*10^{-19} \  C[/tex]

So

         [tex]N  =  \frac{2.0*10^8}{ 1.60*10^{-19}}[/tex]

         [tex]N  =  1.25 *10^{27} \  electrons[/tex]

Answer:

Clockwise direction

Explanation:

In a case of a wire carrying a current, the right hand rule is used.

The thumb in the direction of current while the finger curl around in the direction of the magnetic field.

The right hand rule applies to a current in a straight line wire.

If the direction a wire carrying a current perpendicularly into this page, the direction of the magnetic field will be in a clockwise direction .

Given :

Distance of point P from the origin is , d = 9.83 mm .

To Find :

The electric field produced by this charge at point P , if charge is placed at origin .

Solution :

Let , charge on origin is q .

Electric field at point P is given by :

[tex]E=\dfrac{kq}{r^2}[/tex]    ..... ( 1 )

Here , k is constant , [tex]k=9\times 10^{9}\ N \ m^2/C^2[/tex] .

Putting value of k and r in above equation , we get :

[tex]E=\dfrac{q\times 9\times 10^9}{(9.83\times 10^{-3})^2}\ N/C\\\\E=(9.31\times 10^{13})q\ N/C[/tex]

Hence , this is the required solution .

All we need to do is divide the momentum and weight to find the velocity:

v = m / w

Solve using the values we are given.

v = 952/68 = 14 m/s

Best of Luck!

Answer:

330 km

55*6= 330 km. An easy formula for this is multiplying time with speed.

Explanation:

330 kilometros

55 * 6 = 330 km. Una fórmula fácil para esto es multiplicar el tiempo por la velocidad.

Answer:330 km

Explanation:55*6= 330 km. An easy formula for this is multiplying time with speed.

Answer:

2.52 s

Explanation:

We can use equation of motion to solve this.

[tex]s \: = ut + \frac{1}{2} a {t}^{2} [/tex]

Assuming initial velocity to be 0 and no air resistance,

[tex] s \: = \frac{1}{2} a {t}^{2} \\ 31.16 = \frac{1}{2} (9.81) {t}^{2} \\ t = 2.52s[/tex]

The time of flight is 3.54 s.

There are three equation of motions that can be used when motion of the object is under constant acceleration and on a straight path.

They are listed below as:

[tex]v = u + at\\\\s = ut + \dfrac{1}{2} at^2\\\\v^2 = u^2 + 2as[/tex]

where the symbols have following meanings:

u = initial velocity of the considered object

v = final velocity of the object

a = acceleration of the object

s = distance traveled by the object in 't' time.

It is given that A woman leaning over the railing of a tall bridge accidentally drops her cell phone. If the bridge is 31.16 m high, then  how long will it take the phone to hit the river below the bridge.

We can use equation of motion to solve this;

[tex]s = ut + \dfrac{1}{2} at^2[/tex]

where:

s = 31.16 m is the vertical displacement of the phone

u = 0 is the initial vertical velocity

Assuming initial velocity to be 0 and no air resistance,

[tex]s = 1/2 (9.8)(t)^2\\31.16 = 1/2 (9.8)(t)^2\\t = 2.52 s[/tex]

Hence, The time of flight is 3.54 s.

Learn more about free fall here:

brainly.com/question/1748290

#SPJ2

Answer:

a. 1.80 %

b. 0.60 %

c. 1.20 %

d. 0.60 %

e. 0.60 %

Explanation:

In order to calculate percent uncertainties in each case, we forst need to calculate the average value:

Average Diameter = (16.4 mm + 16.8 mm + 16.9 mm + 16.6 mm +16.8 mm)/5

Average Diameter = 16.7 mm

Now, the formula for percent uncertainty is:

% Uncertainty = (Uncertainty/Average) * 100 %

where,

Uncertainty = |Value - Average|

For Each Case:

a. 16.4 mm:

Uncertainty = |16.4 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.3 mm/16.7 mm) * 100%

% Uncertainty = 1.80 %

b. 16.8 mm:

Uncertainty = |16.8 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

c. 16.4 mm:

Uncertainty = |16.9 mm - 16.7 mm| = 0.2 mm

Therefore,

% Uncertainty = (0.2 mm/16.7 mm) * 100%

% Uncertainty = 1.20 %

d. 16.6 mm:

Uncertainty = |16.6 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

e. 16.8 mm:

Uncertainty = |16.8 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

Answer:

[tex]\huge\boxed{\sf Frequency = 10\ Hz}[/tex]

[tex]\huge\boxed{\sf Time\ Period = 0.1\ secs}[/tex]

Explanation:

Given:

No. of times = 600 times

Time = 60 seconds

Required:

Frequency = f = ?

Time Period = T = ?

Solution:

Frequency = 600 / 60

Frequency = 10 Hz

Time Period = 1 / 10

Time Period = 0.1 secs

Answer:

7.5 meters

Explanation:

Answer:

7.5 kilometers

Explanation:

The distance can be found by multiplying the speed by the time.

[tex]d=s*t[/tex]

The car's speed is 1.5 kilometers per minute. It is traveling for 5.0, or 5  minutes. Therefore,

[tex]s= 1.5 km/min\\t=5 min[/tex]

Substitute the values into the formula.

[tex]d=1.5 km/min * 5 min[/tex]

Multiply. When you multiply, the minutes, or "min" will cancel each other out.

[tex]d= 1.5 km * 5[/tex]

[tex]d=7.5 km[/tex]

The distance the car travels is 7.5 kilometers.

Answer:

It will take 15 seconds

Answer:

provides a first hand account of an event or time period and are considered to be authoritative

Answer:

main source

Explanation:

Answer:

Jun 29, 2016 · Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year. (a) What distance does it move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?

Explanation:

Answer:50  miles per hour 50/1hr

Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1

Answer:

50 miles/hr

Explanation:

[tex]Distance = 100\: miles\\Time = 2\:hours\\\\Speed = \frac{Distance}{Time} \\\\Speed = \frac{100\:miles}{2\:hours}\\\\ Speed = 50\:miles/hour[/tex]

Answer:

Speed, [tex]v=1.6\times 10^8\ m/s[/tex]

Explanation:

Given that,

Distance covered by the electron, d = 32 cm = 0.32 m

Time, t = 2 ns

We need to find the speed of an electron. Speed is equal to distance covered divided by time. So,

[tex]v=\dfrac{0.32}{2\times 10^{-9}}\\\\v=1.6\times 10^8\ m/s[/tex]

So, the speed of the electron is [tex]1.6\times 10^8\ m/s[/tex].

Answer:

D

Explanation:

The Angular Position: The angular position is also called the angular displacement, and it is the angle through which a point revolves around the centre line that has been rotated in a particular manner about a given axis.

The Angular Velocity: The angular velocity of an object is the measure of the speed, or say, how fast an object rotates at one point with respect to a point.

The Angular Acceleration: This is very simple as it's a derivative of angular velocity. They both share the same relationship velocity and acceleration share. Angular acceleration is the rate of change of angular velocity.

The only options that are equal at all points on the rod are;

1. Angular position.

2. Angular velocity.

3. Angular acceleration.

Let us look at each of the options;

1) Angular Position: This is also referred to as angular displacement. In centripetal motion, it is defined as the angle through which a a body has been rotated from the initial position or reference position. This must always be equal at all points on the rod.

2) Angular Velocity: This is defined as the speed with which an object is moving in a circular motion. This must always be equal at all points on the rod.

3) Angular Acceleration: This is simply defined as the rate of change of angular velocity with time. Now, since angular velocity must be equal at all points on the rod, then angular acceleration must also be equal at all points on the rod.

4) Tangential acceleration; This is defined as the rate of change of tangential velocity with time. Tangential velocity is the linear component of the speed and it is not equal at all points on the rod and as such the tangential acceleration will also not be equal at all points on the rod.

Read more at; https://brainly.com/question/13014974

Answer:

If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear.

Answer:

Explanation:

From the question we are told that

  The  height is  [tex]h  =  0.518 \  m[/tex]

    The  sphere density is  [tex]\rho =  22.2g/cm^3 =  \frac{22.2 }{1000}  *  1*10^{6} =  22200 kg/m^3[/tex]

    The  radius is  [tex]r =  2.12 \  mm =0.00212 \ m[/tex]

Generally the power required  is mathematically represented as

   [tex]P  =  \frac{16 *  \pi *  \rho *  r *  g *  h^2 * c }{3}[/tex]

substituting values  

    [tex]P  =  \frac{16 * 3.142 *  22200 *  0.00212 *  9.8 *  0.518^2 *  3.0*10^{8}}{ 3}[/tex]

   [tex]P  = 6.22*10^{11} \  W[/tex]

Answer:

The field of view is reduced.

Explanation:

Given that,

The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.

We know that,

Field of view :

When the view is observed at a point in a defined field then these field called field of view.

The normal angle of  field of view is 90°.

The formula of field of view is define as,  

[tex]field\ of\ view = \dfrac{field\ number}{magnification}[/tex]

We can say that,

The field of view is inversely proportional to the magnification.

When magnification is low then field of view will be large.

When magnification is higher then field of view will be small .

According to question,  

When the magnification adjust from 4 to 10 to 43, the field of view is reduced.    

Hence, The field of view is reduced.

Answer:

h = 1.07 m

Explanation:

It is given that,

The average sped of the kingfisher, v = 4.6 m/s

Time, t = 1.4 s

Final speed, v = 0 (as it hits the water)

We need to find the height from which the bird dove. Let the height is h. According to the conservation of energy.

[tex]mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(4.6)^2}{2\times 9.8}\\\\h=1.07\ m[/tex]

So, the bird dove from a height of 1.07 m.

Complete Question

A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?

Answer:

The  values is  [tex]s =  7.49 \  m[/tex]

Explanation:

From the question we are told that

   The  distance it travels eastward is  [tex]x =  5.0 \ m[/tex]

   The distance it travel towards the southeast  is [tex]l  =  3.0\  m[/tex]

   The distance it travel towards the south is  [tex]z =  1 \  m[/tex]

Let x-axis  be east

      y-axis  south

       z-axis into the ground

The angle made between east and south is  [tex]\theta  =  45^o[/tex]

The displacement toward x-axis is

       [tex]x =  5 +  3cos(45)[/tex]

       [tex]x =  7.12[/tex]

 The  displacement toward the y-axis is  

     [tex]y  =  3 *  sin (45)[/tex]

      [tex]y =  2.123[/tex]

Now the overall displacement of the rat is mathematically evaluated as

        [tex]s =  \sqrt{7.12^2 +  2.12^2 +  1^2}[/tex]

      [tex]s =  7.49 \  m[/tex]

Answer:

50.2m/s

Explanation:

Using first equation of motion we have

V= u+gt

V= final speed

U= initial speed

So v = 1.2m/s + 9.8(5s)

50.2m/s

Given :

A kayak on a lake is moving at a constant speed of 4.1 m/s in a direction 41.6° north of west.

To Find :

After 10 min :

(a) how far west the kayak has traveled km.

(b) how far north the kayak has traveled km.

Solution :

a) Time taken in seconds :

[tex]t=10\times 60 \ s = 600\ s[/tex] .

Distance travelled in 600 s is :

[tex]D=s\times t\\\\D=4.1\times 600\ m\\\\D=2460\ m[/tex]

Distance in km is :

[tex]D=\dfrac{2460}{1000}=2.46\ km[/tex]

b) Now , distance from the north is given by :

[tex]d=D\times sin\ 41.6^o\\\\d=2.46 \times sin\ 41.6^o\\\\d=1.63\ km[/tex]

Hence , this is the required solution .