Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.

Answer:

E = 134.85 N/C

Explanation:

We are given;

Electric field location; b = <-0.2, -0.4, 0> m

Charge: q = 3 nc = 3 × 10^(-9) C

From the location given, we have total distance;

r = √((-0.2²) + (-0.4)² + (0²))

r = √0.2

Formula for Electric field is;

E = kq/r²

where;

k is a constant with a value of 8.99 x 10^(9) N.m²/C²

q is charge on the proton particle with a costant value = 1.6 × 10^(-19) C

r is the distance

Plugging in the relevant values, we have;

E = (8.99 x 10^(9) × 3 × 10^(-9))/(√0.2)²

E = 134.85 N/C

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Answer:

26 lbft

Explanation:

Given that

Force exerted by the woman, F = 5 lb

Length of the ramp, d = 6 ft

angle of inclination, θ = 30° above the horizontal

Work done on the box, W = ?

This is very much a straightforward question..

Work done, W = F * d, where the force takes the factor of the Angie if inclination. So that,

W = Fcosθ * d

On substituting, we have

W = 5 * cos 30 * 6

W = 5 * 0.866 * 6

W = 30 * 0.866

W = 25.98 lbft

Therefore, the work done on the box is 25.98 or approximately 26 lbft

Answer:

Using sold paper is better than to burn it because burning paper may cause pollution.

Explanation:

Hope it will help ^_^

Answer:

5m/s²

Explanation:

v²-v¹ =50 - 25 = 25m/s = 5m/s²

t 5 5

Answer:

Explanation:

[tex]Distance = 490 \:km\\Time = 4.2 \:hours\\\\Average\:speed = \frac{Distance}{Time} \\\\A.V = \frac{490}{4.2} \\\\A.V = 116.67km/h[/tex]

Answer:

provide new testble ideas

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  [tex]E =248.2 \ N/C[/tex]

Explanation:

From the question we are told that

   The  diameter is  [tex]d = 12 \ cm = 0.12 \ m[/tex]

    The charge  is  [tex]Q = 4.4 nC = 4.4 *10^{-9} \ C[/tex]

    The  distance from the center  is  [tex]k = 0.1 \ mm = 1*10^{-4} \ m[/tex]

Generally the radius is mathematically represented as

        [tex]r = \frac{d}{2}[/tex]

=>     [tex]r = \frac{0.12}{2}[/tex]

=>       [tex]r = 0.06 \ m[/tex]

Generally electric field is mathematically represented as  

       [tex]E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ][/tex]

substituting values  

      [tex]E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ][/tex]

     [tex]E =248.2 \ N/C[/tex]

Answer:

B.  The brain moves inside the skull with a force equal and opposite to that of the impact.

Explanation:

Just took the test and got it correct.

Applying Newton's laws the brain is in danger of injury when an impact is made to our head because ; ( B ) The brain moves inside the skull with a force equal and opposite to that of the impact

According to Newton's law an object will continue to be at rest except a force ( impact ) acts on it. also the third law of Newton states that when two objects interact ( The brain and the impact to the head ) they apply forces of equal magnitude to each other but in opposite directions from one another.

The Impact to the head will have a direct effect on the brain because the brain will move in opposite direction to the direction of the impact and with the same magnitude which will lead to an injury to the brain because the brain is suppose to be at rest at all times and not in motion.

Hence we can conclude that Applying Newton's laws the brain is in danger of injury when an impact is made to our head because The brain moves inside the skull with a force equal and opposite to that of the impact.

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Answer:

yes it would.

Explanation:

Answer:

1319.14 lb

Explanation:

From the question above,

Density of the Blazblue extract = mass of Blazblue extract/volume of Blazblue extract.

D = m/v.............. Equation 1

make m the subject of the equation

m = D×v.................... Equation 2

Given: D = 0.770 grams per millilitre, v = 29053 liters = 29053000 millilitres

Substitute these values into equation 2

m = 0.770×29053000

m = 22370810 grams

If,

1 gram = 0.00220462 pounds,

Then,

22370810 grams = (22370810×0.00220462) = 41319.14 lb

Answer:

I think the answer is D.

Complete Question

A grating has 767 lines per centimetre.Find the angles of the first three principal maxima above the central fringe when this grating is illuminated with 672 nm light.

Answer:

The  first is  

      [tex]\theta_1 = 2.92^o[/tex]

 The second is  

       [tex]\theta _2 = 5.93^o[/tex]

 The third is  

        [tex]\theta _3 = 8.92^o[/tex]

Explanation:

From the question we are told that

        The  number of lines per cm is  [tex]I = 767 \ lines/cm = 76700 \ lines / m[/tex]

        The  wavelength is  [tex]\lambda = 672nm = 672 *10^{-9} \ m[/tex]

Generally the condition for constructive  interference is  

            [tex]dsin (\theta ) = n\lambda[/tex]    

Here d is the distance of separation of the slit which is mathematically represented as

             [tex]d = \frac{1}{l}[/tex]

             [tex]d = \frac{1}{76700}[/tex]

             [tex]d = 1.30*10^{-5} \ m[/tex]

So          

           [tex]\theta_1 = sin^{-1} [\frac{ n \lambda }{ d} ][/tex]

For the  first angle  n =  1  

  So  

        [tex]\theta_1 = sin^{-1} [\frac{ 1 * 672 *10^{-9} }{ 1.30 *10^{-5}} ][/tex]

          [tex]\theta_1 = 2.92^o[/tex]

For  the second angle  n  =  2

      So  

            [tex]\theta_2 = sin^{-1} [\frac{ 2 * 672 *10^{-9} }{ 1.30 *10^{-5}} ][/tex]

            [tex]\theta _2 = 5.93^o[/tex]

For  the second angle  n  =  3

      So  

           [tex]\theta_3 = sin^{-1} [\frac{ 3 * 672 *10^{-9} }{ 1.30 *10^{-5}} ][/tex]

          [tex]\theta _3 = 8.92^o[/tex]

84km/hr

Explanation:

To find one light yr

= 3E8m/s x 24x 365/1000

= 996980E7kms

So

Radius of orbit for 24000light years

=996980E7kms x 24000

Total orbit distance is = 2πr

= 2 *π x 996980E7kms x 24000

Total time will be

= 194E6 x 24x365 hrs

Finally speed = distance/ time

= 2 *π x 996980E7kms x 24000/194E6 x 24x365 hrs

= 84km/hr

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

Initial speed, u = 0

Final speed, v = 25 m/s

Distance, d = 200 m

We need to find the net force. Firstly, let's find acceleration using equation of motion.

[tex]v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2[/tex]

Net force, F = ma

[tex]F=1600\times 1.56\\\\F=2496\ N[/tex]

So, the net force is 2496 N.

The amount of net force that will be required to accelerate a 1600kg SUV from rest to a speed of 25 m/s in a distance of 200m is 2500N

HOW TO CALCULATE NET FORCE:

Where:

Since a = 1.563m/s²

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Answer:

The potential energy of a 2 kg mass at a height of 40 meters is 784 J

Explanation:

Potential energy is that energy that a body possesses due to the height at which it is located and whose unit of measurement of the International System of Units is the joule (J).

The potential energy of a body is the result of multiplying its mass by its height and by gravity:

Ep=m*g*h

Potential energy Ep, is measured in joules (J), mass m is measured in kilograms (kg), gravity, g, in meters / second-squared ([tex]\frac{m}{s^{2} }[/tex]), and height, h , in meters (m).

In this case:

Replacing:

Ep= 2 kg* 9.8 [tex]\frac{m}{s^{2} }[/tex] * 40 m

Solving:

Ep= 784 J

The potential energy of a 2 kg mass at a height of 40 meters is 784 J

Answer:

Let's take the frequency receive by police officer to be

f¹/f²= u-v/u+v

So 1255/1205= 340-v/340+v

V= 6.9m/s positive value implies that the car is moving toward the police

2 to find frequency we use

f/1205= 340+25/340-25

f= 1396Hz

Answer:

The  value is  [tex]g =  16.104 \  m/s[/tex]

Explanation:

From the question we are told that

 The  length is  [tex]l = 53.0 \ cm = 0.53 \ m[/tex]

 The  number of cycle is  [tex]n = 105[/tex]

  The time taken is  [tex]t = 125 \ s[/tex]

The period is mathematically represented as

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

Also the period is mathematically represented as

      [tex]T    =  \frac{ t}{n}[/tex]

      [tex]T  =  \frac{125}{110}[/tex]

      [tex]T  =  1.14 \ s/cycle[/tex]

So

      [tex]1.14  =  2 * 3.142  \sqrt{\frac{ 0.53}{g} }[/tex]

=>   [tex]g =  \frac{ 4 *  3.142^2 *  0.53}{1.14^2}[/tex]

=>     [tex]g =  16.104 \  m/s[/tex]

Answer:

Explanation:

 |F-->1| =  E-->1 x charge on proton

= 300 x 1.6 x 10⁻¹⁹

Magnitude of electric force on proton

|F-->1| = 480 x 10⁻¹⁹ N .

Inside lithium nucleus , there are three protons so charge on it

= 3 x 1.6 x 10⁻¹⁹ = 4.8 x 10⁻¹⁹ C .

Magnitude of electric force on lithium nucleus

|F-->Li|  =   E-->1 x charge on lithium nucleus

= 300 x  4.8 x 10⁻¹⁹

= 1440 x 10⁻¹⁹ N .

Electron has same charge as that possessed by proton so charge on electron = 1.6 x 10⁻¹⁹ C

Magnitude of force on electron

|F-->1| =  E-->1 x charge on electron

= 300 x 1.6 x 10⁻¹⁹

=  480 x 10⁻¹⁹ N .

Answer:

5 N

Explanation:

From the question,

The magnitude of the force that would be required to just loosen the nut when the force is applied perpendicularly at the end of the handle is

Fy = Fsinθ................. Equation 1

Where Fy = force acting perpendicular at the end of the handle, F = Force applied to the handle, θ = angle of inclination of the force to the end of the handle.

Given: F = 10 N, θ = 30°

Substitute these values into equation 1

Fy = 10(sn30°)

Fy = 10(0.5)

Fy = 5 N.

Answer this question

Explanation:

Answer:  C. the motion of a spacecraft under gravitational influence.

Explanation:

A is Metallurgy, B is Biology, C is astro-physics, I am not sure what D is, but it's   safe to say it's not physics, E, micro-biology, and the study of radiation. C is the only one involving physics.

Explanation:

When light travels from air into water, due to the change in refractive index of the two medium, refraction of light occurs.

Velocity of light in a medium is given by :

[tex]v=\dfrac{c}{n}[/tex]

c is speed of light

n is refractive index

The frequency remains the same in every medium. It means that the its velocity and wavelength change, but its frequency does not change.

Answer:

Explanation:

Given the the path of an object at a 45° with initial velocity of 80 feet per second modeled by the equation [tex]h(x) = (-\dfrac{32}{80^2} )x^2 + x[/tex] where;

x is the horizontal distance traveled and h(x) is the height in feet, if x is given as 100 ft, the height of the object will be gotten by simply substituting x = 100 into the modeled function as shown;

[tex]h(x) = (-\dfrac{32}{80^2} )x^2 + x\\\\h(100) = (-\dfrac{32}{80^2} )100^2 + 100\\\\h(100) = (-0.005 )100^2 + 100\\\\h(100) = (-0.005 )10,000 + 100\\\\h(100) = -50 + 100\\\\h(100) =50 feet[/tex]

Hence height of the object when it has traveled 100 feet away horizontally is 50 feet.

Answer:

1) Only the perpendicular component of wind is responsible for the rotation, because gravity points downward.

2) static friction is greater than kinetic friction and the applied force acts oppositely

3)   Inertia decreases with decreasing radius of the orbit.

4) constant is the mass of the car

Explanation:

1) In order to know which statement is correct, let's analyze the movement of the mill.

The rotation of the blades due to a force perpendicular to each one, which creates a torque that results in its rotation.

The true force does not affect the movement because it is applied at the center of mass of the mill that coincides with its axis of rotation.

Of the statements, the correct one is:

The perpendicular to the wind is responsible for the movement

2) static friction is the result of microscopic bonds between two surfaces when there is no relative motion between them

kinetic friction is the result of the unions between the two surfaces when there is a relative movement between them.

From these statements it follows that friction is contrary to motion, static friction must be greater than kinetic friction.

Of the statements the correct one is:

static friction is greater than kinetic friction and the applied force acts oppositely

3) The inertia of a particle is

         I = m R²

where m is the mass of the satellite and R is the distance from the center of the Earth, decreasing with each orbit

The correct answer is:

  Inertia decreases with decreasing radius of the orbit.

4) From the figure it can be seen that the motorist subjected to a constant force, for which we can apply Newton's second law

          F = m a

The force equals the applied weight.

The quantity that is held constant is the mass of the car

since the applied force and therefore the acceleration can be changed by changing the applied weight.

Answer:

1. The perpendicular components of gravity and the force of wind are responsible for the rotation.  

2. Static friction is greater than kinetic friction, but they both act opposite the applied force.

3. Rotational inertia decreases proportional to the decrease in the radius of rotation.

4. mass

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2[/tex]

The formula that is used to find the rms value of the electric field is as follows :

[tex]I=\epsilon_o cE^2_{rms}[/tex]

c is speed of light and [tex]\epsilon_o[/tex] is permittivity of free space

So,

[tex]E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m[/tex]

Hence, this is the required solution.

Answer:

[tex]\Huge \boxed{\mathrm{2298.57 \ seconds}}[/tex]

[tex]\rule[225]{225}{2}[/tex]

Explanation:

[tex]\displaystyle \sf Speed = \frac{Distance \ covered }{Time \ taken}[/tex]

[tex]\displaystyle \sf s = \frac{d }{t}[/tex]

The speed is 0.7 m/s.

The distance covered is 1609 m.

[tex]\displaystyle \sf 0.7 = \frac{1609 }{t}[/tex]

Multiplying both sides by t.

Then dividing both sides by 0.7.

[tex]\displaystyle \sf t = \frac{1609 }{0.7}[/tex]

[tex]\sf t= 2298.57[/tex]

It would take 2298.57 seconds.

[tex]\rule[225]{225}{2}[/tex]

Answer:

[tex]\huge\boxed{\sf t = 2298.57\ secs}[/tex]

Explanation:

Given:

Speed = v = 0.7 m/s

Distance = S = 1 mile = 1609 metre

Required:

Time = t = ?

Formula :

v = S/t

Solution:

v = S/t

t = S/v

t = 1609 / 0.7

t = 2298.57 secs

Answer:

Volume flow rate = [tex]1.41 \times 10^{-3}m^3/s[/tex]

Explanation:

The volume flow rate through a channel can be gotten by multiplying its area and its velocity.

The channel under consideration is a circular channel. Hence, the cross-sectional area can be calculated by using the relation for calculating the area of a circle

Cross-sectional area = [tex]\pi \times d^{2}/4[/tex]

[tex]A = \pi \times (3\times 10^-2)^2 /4 = 7.07 \times 10 ^-4 m^2[/tex]

volume flow rate= [tex]A = 7.07 \times 10 ^-4 m^2 X 2.0 = 1.41 \times 10^{-3}m^3/s[/tex]

Volume flow rate = [tex]1.41 \times 10^{-3}m^3/s[/tex]

Answer:

A) 10.243 s

B) 514.64 m

C) 153.645 m

Explanation:

From projectile motion;

y(t) = h - ½gt²

Where;

h is the height of cliff

t is the time taken to fall from top of cliff down to half the height of the cliff

y(t) is height of the rock as function of time

We are told the rock falls half way of the cliff height.

Thus;

y = h/2

So;

h/2 = h - ½gt²

This gives;

h - h/2 = ½gt²

h/2 = ½gt²

h = gt² - - - - (1)

Also,the total free fall time from top of the cliff to ground would be;

T = t + 3

Thus, distance at this point is zero.

So;

0 = h - ½gT²

h = ½gT²

Putting t + 3 for T, we have;

h = (1/2)g(t + 3)² - - - 2

Inspecting eq 1 and eq 2,we have;

gt² = (1/2)g(t + 3)²

g will cancel out to give;

2t² = t² + 6t + 9

t² - 6t - 9 = 0

The roots are -1.243 or 7.243.

We will pick the positive value as time can't be negative.

Thus, T = 7.243 + 3 = 10.243 s

h = gt² = 9.81 × 7.243²

h = 514.64 m

Formula for horizontal distance is given by;

x = uT

Where u is initial speed given as;

u = 15 m/s

Thus;

x = 15 × 10.243

x = 153.645 m

Answer:

2.3*10^8m/s

Explanation:

Using

V = c/n

Where c= speed of light

n = refractive index of water

By substituting

We have

V= 3*10^8m/s/1.33

= 2.3*10^8m/s