If one has twice the length of the other, what is the ratio of the diameter of the longer wire to the diameter of the shorter wire

Answer:

A

Explanation:

cant be c because its asking for speed and speed doesnt have direction

Answer:302.5

Explanation:

Answer:

A) Average speed = 18.75 m/s

B) More time is spent at 15 m/s than at 25 m/s.

Explanation:

Let the first distance be d1 and the second distance be d2.

We are given;

d1 = 10 km = 10000 m

d2 = 10 km = 10000 m

Speed; v1 = 15 m/s

Speed; v2 = 25 m/s

Now, the formula for distance is; Distance = speed x time

Thus:

d1 = v1 x t1

t1 = d1/v1 = 10000/15 = 666.67 seconds

Also,

d2 = v2 x t2

t2 = d2/v2 = 10000/25 = 400 seconds

Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s

From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;

More time at 15 m/s than at 25 m/s.

Answer:

The distance moved is 44 metres.

The magnitude of displacement is 20 metres with northward direction.

Answer:

44 m.

North 20 m.

Explanation:

Distance moved = 26 + 12 + 6

=  44 m.

Magnitude of the displacement = 26 - 12 + 6

= 20m

Direction is Northward.

Complete Question

A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is: [tex]R = 0.0498 \ kPAkm^{-1}[/tex] Convert R to [tex]kJ*m^{-4}[/tex]

Answer:

The value  is       [tex]R  = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]

Explanation:

From the question we are told that

   The altitude is  [tex]R = 0.0498 \ kPAkm^{-1}[/tex]

Generally  

     [tex]1 k PA  =  1000 PA[/tex]

So  

   [tex]R = 0.0498 \frac{1000PA}{ km}[/tex]

Also  

   1 km  =  1000 m

So  

     [tex]R = 0.0498 \frac{1000PA}{ 1000m}[/tex]

=>   [tex]R = 0.0498 \frac{1 PA}{ 1 m}[/tex]

Now  PA  is Pascal which is mathematically represented as

       [tex]PA =  \frac{N}{m^2 }[/tex]

So

   [tex]R  = 0.0498 \frac{\frac{N}{m^2} }{m}[/tex]

    [tex]R  = 0.0498 \frac{N}{m^3}[/tex]

Looking the unit we are arrive at we see that it contains  J  which is mathematically represented as

      [tex]J =  N  *  m[/tex]

So  

  [tex]R  = 0.0498 \frac{ N \frac{m}{m} }{m^3}[/tex]

=>  [tex]R  = 0.0498 \frac{\frac{J}{m} }{m^3}[/tex]

=>  [tex]R  = 0.0498 \frac{J}{m^4}[/tex]

Generally  

      [tex]1 J \to 1.0*10^{-3} kJ[/tex]

      [tex]0.0498 J  \to x kJ[/tex]

=>      [tex]x =  \frac{0.0498 *  1.0*10^{-3}}{1}[/tex]

=>   [tex]0.0498 *10^{-3} kJ[/tex]

So

    [tex]R  = 0.0498 *10^{-3} \frac{kJ}{m^4}[/tex]

Answer:

b

Explanation:

a looks so silly

Answer:

The correct option is a

Explanation:

The alpha particle has the lowest penetrating power of the trio of alpha, beta and gamma particles and can be stopped by a sheet of paper and hence cannot penetrate a human skin. Beta particle has a higher penetrating power than alpha particle (some of it penetrates the human skin and some do not) while the gamma particle has the highest penetrating power (with all of it penetrating the human skin).

From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

[tex]\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}[/tex]  

therefore : Ra = [tex]\frac{sin120^o * 750}{sin 35^o}[/tex] = 1132 N

To determine the scalar component Rb

[tex]\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}[/tex]

therefore : Rb = [tex]\frac{sin 25^o * 750}{sin 35^o}[/tex]  = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos[tex]25^o[/tex] = 679.7 N

Answer:

  n = 1 + R / f

Explanation:

The equation of the constructor is optical is

          1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and image, respectively

The exercise tells us that it is a concave lens with focal length fo, in these lenses the focal length is negative. The relationship to calculate the focal length is

         1 / f = (n -n₀) (1 /R₁ - 1 /R₂)

where is n₀ the refractive index of the medium that surrounds the lens in this case it is air with n₀ = 1, you do not indicate the type of lens, but the most used lens is the concave plane, in this case R₂ = ∞, so which 1 / R₂ = 0, let's substitute

         1 / f = (n-1) / R₁

         n - 1 = R₁ / f

let's calculate

         n = 1- R₁ / f

remember that the radius of curvature is negative, so the equation is

         n = 1 + R / f

Answer:

a. the minimum amplitude of vibration for the NO molecule A [tex]\simeq[/tex] 4.9378 pm

b. the minimum amplitude of vibration for the HCl molecule A [tex]\simeq[/tex] 10.9336 pm

Explanation:

Given that:

The effective spring constant describing the potential energy of the HBr molecule is 410 N/m

The effective spring constant describing the potential energy of the NO molecule is 1530 N/m

To calculate the minimum amplitude of vibration for the NO molecule, we use the formula:

[tex]\dfrac{1}{2}kA^2= \dfrac{1}{2}hf[/tex]

[tex]kA^2= hf[/tex]

[tex]A^2= \dfrac{hf}{k}[/tex]

[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]

[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 5.63 \times 10^{13} \ s^{-1})}{1530 \ N/m}}[/tex]

[tex]A = \sqrt{\dfrac{3.730438 \times 10^{-20} \ m}{1530 }[/tex]

[tex]A = \sqrt{2.43819477 \times 10^{-23}\ m[/tex]

[tex]A =4.93780799 \times 10^{-12} \ m[/tex]

A [tex]\simeq[/tex] 4.9378 pm

The effective spring constant describing the potential energy of the HCl molecule is 480 N/m

To calculate the minimum amplitude using the  same formula above, we have:

[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]

[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 8.66 \times 10^{13} \ s^{-1})}{480 \ N/m}}[/tex]

[tex]A = \sqrt{\dfrac{5.738116\times 10^{-20} \ m}{480 }[/tex]

[tex]A = \sqrt{1.19544083\times 10^{-22}\ m[/tex]

[tex]A = 1.09336217 \times 10^{-11} \ m[/tex]

[tex]A = 10.9336217 \times 10^{-12} \ m[/tex]

A [tex]\simeq[/tex] 10.9336 pm

Answer:

Acceleration, a = -3 m/s²

Explanation:

Given that,

Initial speed, u = 15 m/s

Final speed, v = 0

Time, t = 5 s

We need to find the acceleration of the train. It is equal to the change in velocity divided by time taken. So,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-15\ m/s}{5\ s}\\\\a=-3\ m/s^2[/tex]

So, the acceleration of the train is 3 m/s² and it is deaccelerating.

Answer:

(a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]

(b).  The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

[tex]P_{1}=\rho g h_{1}[/tex]

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

[tex]P_{1}=\rho gh_{1}[/tex].....(I)

Pressure for second pipe,

[tex]P_{2}=\rho gh_{2}[/tex].....(II)

From equation (I) and (II)

[tex]P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]

Put the value of P₁ and P₂

[tex]\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]

[tex]gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)[/tex]

[tex]2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2[/tex]....(III)

We know that,

The continuity equation

[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]

[tex]v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})[/tex]

Put the value of v₂ in equation (III)

[tex]2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2[/tex]

[tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2[/tex]

Here, [tex]\dfrac{A_{1}}{A_{2}}=\gamma[/tex]

So, [tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)[/tex]

[tex]v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]

Hence, (a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]

(b).  The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]

Answer:

A. Using

Fl= ( v+vl/v+vz)fz

= (340+19/340+31) x 2020

= 1954.7Hz

Then to find the frequency of sound when reflected from the truck such that the driver becomes the listener

we use

F"= ( v+vz/v+vl) fz

= 340+31/340+19 x 2020

2087.5Hz

B to find the wavelength of sound we use

Wavelength= V+vl/ F"

= 340+31/2087.5= 0.18m

Answer:

When inertia increases, it's because the mass increased, which increases the normal force, which ultimately increases friction.

Answer:

Here are some examples of orderliness In nature​

1. The proposal that the order of nature showed evidence of having its own human-like "intelligence" goes back to the origins of Greek natural philosophy and science, and its attention to the orderliness of nature, often with special reference to the revolving of the heavens.

2. For Stillman, the orderliness of Astaire-like dance is an actual cure for destructive emotions.

3. Tells about the orderliness of the crowds, and of the dispatch with which the trains were being filled and emptied.

Your question has been heard loud and clear.

An alpha particle , can move in any direction randomly. But with a magnetic field , we can deflect the alpha particle in any direction we want.

So , the magnetic field must be placed to the west of the alpha particle , so that the particle gets deflected and moves towards the north direction.

Thank you.

Like AL2006 said reasons for what

Answer:

The value is  [tex]q_o = Q[/tex]

Explanation:

From the question we are told that

   The net charge is [tex]Q = + 3Q[/tex]

     The  charge place on the  inside of the cavity is  [tex]q = -2Q[/tex]

Since we are told from the question that small charge placed inside the cavity is isolated from the conductor

  Then it implies that the electric flux is  Zero

Which mean that the charge place within the conductor +  the charge  on the inner region of the conductor  =  0  

i.e

     [tex]q + q_i = 0[/tex]

=>   [tex]-2Q + q_i = 0[/tex]

=>   [tex]q_i = 2Q[/tex]

Now the net charge on the conductor is mathematically represented as

   [tex]Q = q_o + q_i[/tex]

Here  [tex]q_o[/tex] is the charge on the outer surface

So

    [tex]3Q = q_o + 2Q[/tex]

=>   [tex]q_o = Q[/tex]

Answer:

Explanation:

We shall apply law of conservation of momentum to solve the problem .

Helium-3 collides with nitrogen-14 at rest . After the collision the newly formed deuterium atom and oxygen-15  atom moves .

momentum before the collision

= 3.016 x 6.346 x 10⁶ + 14.003 x 0 = 19.14 x 10⁶ unit

momentum after collision

2.014 x 1.531 x 10⁷ + 15.003 V

3.083 x 10⁷ +  15.003 V units

Applying the law of conservation of momentum ,

19.14 x 10⁶ = 3.083 x 10⁷ +  15.003 V

1.914 x 10⁷ = 3.083 x 10⁷ +  15.003 V

15.003 V = - 1.169 x 10⁷

V = .077917 x 10⁷

= 7.79 x 10⁵  m /s

= .0779 x 10⁷ m /s

mass of helium atom = 3.016 u = 3.016 x 1.67 x 10⁻²⁷ kg

velocity = 6.346 x 10⁶ m /s

kinetic energy = 1 /2 x  3.016 x 1.67 x 10⁻²⁷ x (6.346 x 10⁶ )²

= 101.42 x 10⁻¹⁵ J  

kinetic energy of nitrogen atoms = 0

Total energy before collision =  101.42 x 10⁻¹⁵ J  

Similarly kinetic energy after collision

= 1 /2 x [ 2.014 x 1.531² + 15.003 x .0779² ] x 1.67 x 10⁻²⁷ x 10¹⁴

= .835 x [ 4.72  + .09 ] x 10⁻¹³ J

=  4.016 x 10⁻¹³ J

= 401.6 x 10⁻¹⁵  J  

value of kinetic energy is increased .

Answer:

False

Explanation:

Faraday's law gives the relationship between the induced emf and the rate of change of magnetic flux i.e.

[tex]\epsilon=\dfrac{-d\phi}{dt}[/tex]

The given statement "A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change" is false. The reason is that if the rate of change of magnetic flux is greater, then its will induce more emf. It would mean it does not say about emf.

Hence, it is false.

Answer:

13.51 nm

Explanation:

To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians

y/L=tan θ ≈ θ

and ∆θ ≈∆y/L

Where ∆y= wavelength distance= 2.92 mm =0.00292m

L=screen distance= 2.40 m

=0.00292m/2.40m

=0.001217 rad

The grating spacing is d = (90000 lines/m)^−1

=1.11 × 10−5 m.

the small-angle

approx. Using difraction formula with m = 1 gives:

mλ = d sin θ ≈ dθ →

∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad

=0.000000001351m

= 13.51 nm

Answer:

a)   v = - 4.4168 m / s  , b)    V = 0

Explanation:

The average speed is the variation of the displacement in time used.

          v = Δx /Δt = (x₂-x₁) / (t₂-t₁)

let's apply this equation to our case

Let's reduce the magnitudes of the system Yes

         Δx = 5150 km (1000 m / 1km) = 5.15 106 m

         Δt = 13.5 day (86,400 s / 1 day) = 1,166 10 6 s

a) return trip

the vector is negative because it points long towards the center of the system

         v = - 5.15 106 / 1.166 106

         v = - 4.4168 m / s

the negative sign indicates that he is coming back, to the lair

b) In a complete trip the distance is zero, because it is a vector, consequently the mean fickleness is also zero

                     V(j_ = 0

Answer:

a. True

Explanation:

Distance is described with only magnitude. It is defined as the total path covered by an object, in other words it  is the length of a path followed by a particle.

Displacement is described with both magnitude and direction. It is distance traveled in a specified direction or  change in position in some time interval.

Therefore, the correct option is " a. True"

Answer:

1.2J

26 N

(-28, 14) m/s

Explanation:

energy

U = kQq / d = 8.99*10^9 * (2.5*10^-6C)² / 0.047m

U = 0.0562/0.047

U = 1.20 J

to two significant figures

tension

T = kQq / d²

T = U / d

T = 1.2 / 0.047

T = 25.53 N = 26 N to 2 sf

Momentum is conserved, and the initial momentum is zero:

0 = 0.0020 * V2 + 0.0040 * V4

so

V2 = -2 * V4

Energy is also conserved:

½ * 0.0020 * (-2V4)² + ½ * 0.0040 * (V4)² = 1.2 J

-½ * 0.0080 * V4² + ½ * 0.0040 * V4² = 1.2 J

-0.0040V4² + 0.002V4² = 1.2 J

0.0060V4² = 1.2 J

V4² = 1.2/0.0060

V4² = 200

V4 = √200

V4 = 14 m/s

and since V2 = -2 * V4

V2 = -28 m/s

(V2, V4) = (-28, 14)

The energy of this system is 1.2J

The  tension in the string is 26 N

The speed of each sphere when they are far apart is (-28, 14) m/s

The energy should be

U = kQq / d

= 8.99*10^9 * (2.5*10^-6C)² / 0.047m

U = 0.0562/0.047

U = 1.20 J

The tension should be

T = kQq / d²

T = U / d

T = 1.2 / 0.047

T = 25.53 N

= 26 N

The speed should be

Since Momentum should be conserved, and the initial momentum is zero:

So,

0 = 0.0020 * V2 + 0.0040 * V4

Now

V2 = -2 * V4

Due to this, Energy is also conserved:

½ * 0.0020 * (-2V4)² + ½ * 0.0040 * (V4)² = 1.2 J

-½ * 0.0080 * V4² + ½ * 0.0040 * V4² = 1.2 J

-0.0040V4² + 0.002V4² = 1.2 J

0.0060V4² = 1.2 J

V4² = 1.2/0.0060

V4² = 200

V4 = √200

V4 = 14 m/s

and now V2 = -2 * V4

V2 = -28 m/s

(V2, V4) = (-28, 14)

Learn more about energy here: https://brainly.com/question/15182235

Answer:

Question 1: D because the height the ball bounces depends on all the other factors in the experiment.

Question 2: B because the the temperature of the water is not affected by the other variables.

Question 3: A because the more that they can control in the experiment, the more accurate the results will be.

Hopefully this helps :)

1:D 2:B 3:A

just took the test

Answer:

resultant force = 14 N  ( East direction)

Explanation:

A = [tex]\sqrt{(4*7)^2 + (4*7)^2}[/tex]

A = 39.6 N

B = 4 * 7

B = 28 N

C = 2 * 7

C = 14 N

∑ y forces = Ay - B = (4*7) - 28 = 0

∑ x forces = Ax - C = (4*7) - 14 = 14 N

so the resultant force = 14 N  ( East direction)

Answer:

resultant = 14 N to the right

Explanation:

A→+B→+C→

adding all forces acting on x

and

adding all forces acting on y

A = sqrt(4*7)^2 + (4*7)^2 = 39.6

B = 4 * 7 = 28

C = 2 * 7 = 14

forces acting on x = (4*7) - 28 = 0

forces acting on y = (4*7) - 14 = 14 N

so the resultant = 14 N to the right

Vector  explanation is apex and I got 100 on the test and it’s the definition

Answer:

that's too much to read

Answer:

Density is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). Remember, grams is a mass and cubic centimeters is a volume (the same volume as 1 milliliter).

Explanation:

Hey, there!

Density is defined as mass per unit volume. so, when we keep is as a formula we get like,

[tex]density = \frac{mass}{volume} [/tex]

So, you can state that density is calculated by dividing mass by its unit volume.

Hope it helps...

Answer:

Given x = bt-c²

We know that t= time (s)

x= distance (m)

So

bxt= meters

m/s x s= m

And then c= m/s²

And b= m/s

Answer:

3.5s

Explanation:

Using equation of motion number 2

S= ut + 1/2 gt²

Substituting

10= 0.5+1.62t²

t= 3.5seconds