A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.

• w = weight = 319 N

• n = normal force

• p = pushing force = 485 N

• f = friction = µ n, where µ is the coefficient of static friction

The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)

• vertical:

p sin(-35°) + n - w = 0

and solving for n,

- (485 N) sin(35°) + n - 319 N = 0

n ≈ 597 N

• horizontal:

p cos(-35°) - f = m a

where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get

p cos(35°) - µ n = (w / g) a

(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a

a ≈ (12.2 - 18.3 µ) m/s²

(a) If µ = 0.57, then the net acceleration on the box is

a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²

so that the time t required to move the box 4 m is

4 m = 1/2 a t ²

t ≈ √((8 m) / (1.75 m/s²))

t ≈ 2.14 s

(b) The box does not move.

If µ = 0.75, then

a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²

but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.

(a) The time taken to move the box 4 meters is 2.14 s.

(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

The given parameters;

The mass of the books is calculated as;

[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]

The normal force on the box is calculated as follows;

[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]

The frictional force when the coefficient of friction is 0.57;

[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]

The time taken to move the box 4 meters is calculated as;

[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]

(b) The frictional force when the coefficient of friction is 0.75;

[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]

The acceleration of the box is calculated as follows;

[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]

Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

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Answer:

Explanation:

step one:

given data

velocity v=34.2m/s

time t= 8.7s

Step two

Required is the distance the cheetah has covered on the condition

we know that speed= distance/time

make distance subject of formula we have

distance= velocity *time

distance= 34.2*8.7

distance = 297.54m

Therefore the displacement the cheetah covered at that velocity

is 297.54m

Answer:

Vₓ = -93.96 m/s

Vy = 34.2 m/s

Explanation:

The x-component of the velocity vector can be given as follows:

Vₓ = V Cos θ

where,

Vₓ = x-component of velocity vector = ?

V = Magnitude of Velocity Vector = 100 m/s

θ = Angle with positive x-axis = 160°

Therefore,

Vₓ = (100 m/s)Cos 160°

Vₓ = -93.96 m/s

The y-component of the velocity vector can be given as follows:

Vy = V Sin θ

where,

Vy = y-component of velocity vector = ?

V = Magnitude of Velocity Vector = 100 m/s

θ = Angle with positive x-axis = 160°

Therefore,

Vy = (100 m/s)Sin 160°

Vy = 34.2 m/s

Answer:

T' = 865.15 °C

Explanation:

In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.

FOR COPPER COLLAR:

dc' = dc(1 + ∝c*ΔT)

where,

dc' = final diameter of copper ring

dc = initial diameter of copper ring = 5.98 cm

∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature

Therefore,

dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT]   ------------- equation (1)

FOR STEEL SHAFT:

ds' = ds(1 + ∝s*ΔT)

where,

ds' = final diameter of steel shaft

ds = initial diameter of steel shaft = 6 cm

∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature

Therefore,

dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]   ------------- equation (2)

Comparing equation (1) and equation (2):

(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]

(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]

0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT

1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT

0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT

ΔT = 846.15 °C

but,

ΔT = T' - T = T' - 19°C = 846.15°C

T' = 846.15 °C + 19 °C

T' = 865.15 °C

Answer:

The images produced by a convex mirror are smaller than the object it reflects. The image produced by a concave lens is always virtual. The image in a convex mirror is always upright and is smaller than the object.

Explanation:

:)

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

Answer:

Explanation:

Step one:

given

Force F= 210N

mass m= 1600kg

velocity v=5500m/s

Step two

Required is the radius r

the expression for the force is

[tex]F_c = \frac{mv^2}{r}[/tex]

substitute

210=1600*5500^2/r

cross multiply we have

210r=48400000000

divide both side by 210

r=230476190.476m

r=230476.19km

Answer:

Explanation:

a) Maximum height id expressed as;

H = u²sin² theta/2g

H = 36.6²(sin42.2)²/2(9.8)

H = 1,339.56(0.6717)/19.6

H = 899.81/19.6

H = 45.91m

Hence the maximum height is 45.91m

b) The Time of flight is the total time in air expressed as;

T = 2usin theta/g

T = 2(36.6)sin42.2/9.8

T = 73.2sin42.2/9.8

T = 49.17/9.8

T = 5.017secs

Hence the total time in air is 5.017scs

c) Range = U²sin2(theta)/g

Range = 36.6²sin2(42.2)/9.8

Range = 1,339.5(0.9952)/9.8

Range = 1,333.11/9.8

Range = 136.03m

Hence the range is 136.03m

d) USing the rime of flight formula;

T =2Usintheta/g

1.5 = 2Usin42.2/9.8

2Usin42.2 =1.5*9.8

2Usin42.2 = 14.7

U = 14.7/2sin42.2

U = 14.7/1.3434

U = 10.94m/s

Hence the speed of the projectile is 10.94m/s

Answer: 4100 Mpc

Explanation:

Since H o = 70 km/s/Mpc

Redshift z = 5.82

Recessional velocity vr = 287,000 km/s

Then, the distance to the galaxy in light years will be:

= Recessional velocity / H o

= 287000 / 70

= 4100 Mpc

Answer:

It will be the same for both

Explanation:

from this question we have one similarity between these two spheres.

- they are both made from the same material,

The difference between both spheres is that:

- one of the spheres has its diameter to be twice as large as that of the other one.

We are to say the sphere with the greater bulk modulus.

If the material is the same thenthe Bulk modulus is also the same. It is not dependent on the material since it is a constant for that materia

Therefore the correct answer is:

It will be the same for both spheres.

Answer:

Gravity is dependent on the mass of two bodies and the distance between them. There is a strong gravitational attraction between Earth and the Moon because they’re relatively close to one another. There is a strong gravitational attraction between Earth and the Sun because the Sun is so massive

Answer:

Gravity is determined by the mass of two bodies and their separation. Because the Earth and the Moon are so close to one another, they have a tremendous gravitational attraction. Because the Sun is so large, there is a significant gravitational force between Earth and it.

Explanation:

Answer:

by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products. - This is the answer.

Explanation:

Answer:

its A(by counting each individual atom and making sure the number of each kind of atom is the same in the reactants and the products

Explanation:

Answer:

contains many young stars

Explanation:

Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.

However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.

The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.

They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.

Answer:

Contains many young stars

Explanation:

When calcium reacts with chlorine. Electrons move from the calcium atoms to the chlorine atoms. Option D is correct.

When one or more chemicals are transformed into one or more other compounds, a chemical reaction occurs.

In an ionic compound, calcium chloride is present. To build a complete outer shell of electrons, the calcium atom loses two electrons and each chlorine atom obtains one.

When calcium combines with chlorine, it forms calcium chloride. From the calcium atoms to the chlorine atoms, electrons travel.

Hence, option D is correct.

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Answer:

2) deflection must be towards the negative side of the voltage.

4) the correct statements are: b and c

Explanation:

2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by

           E = - N d [tex]\phi_{B}[/tex] / dt

where [tex]\phi_{B}[/tex] = B. A

The bold are vectors

Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.

4) From the Faraday equation we can see that the inductive electromotive force depends

* The magnitude of B that changes over time

* The area of ​​the loop that changes over time

* The angle between B and the area that changes over time

* A combination of the above

With this analysis we will review the different alternatives given

a) False. It takes a temporary change and an absolute value of B

b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases

c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller

d) False. A temporary change of area is needed, in addition to increasing the area the current increases

We can see that the correct statements are: b and c

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

Answer:

The resultant velocity has a magnitude of 38.95 m/s

Explanation:

Vector Addition

Given two vectors defined as:

[tex]\vec v_1=(x_1,y_1)[/tex]

[tex]\vec v_2=(x_2,y_2)[/tex]

The sum of the vectors is:

[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]

The magnitude of a vector can be calculated by

[tex]d=\sqrt{x^2+y^2}[/tex]

Where x and y are the rectangular components of the vector.

We have a plane flying due west at 34 m/s. Its velocity vector is:

[tex]\vec v_1=(-34,0)[/tex]

The wind blows at 19 m/s south, thus:

[tex]\vec v_2=(0,-19)[/tex]

The sum of both velocities gives the resultant velocity:

[tex]\vec v =(-34,-19)[/tex]

The magnitude of this velocity is:

[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]

[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]

d = 38.95 m/s

The resultant velocity has a magnitude of 38.95 m/s

Answer:

Explanation:

The car’s acceleration can be found by using the formula

[tex]a = \frac{v - u}{t} \\ [/tex]

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

From the question we have

[tex]a = \frac{25 - 15}{6} = \frac{10}{6} = \frac{5}{3} \\ = 1.666666...[/tex]

We have the final answer as

Hope this helps you

Answer:

C.  3.0 s

Explanation:

Well, I've taken this question before, so I went back and this was the answer I put, which was correct.

Hope this helps! <3 May I have brainliest if it helps you enough?

It will take 3 s for the stone to get to the surface of the water.

We'll begin by listing out what was given from the question. This includes:

Height (h) = 45 m

NOTE: Acceleration due to gravity (g) is 10 m/s²

From the data above, we can obtain the time taken for the stone to get to the surface of the water as follow:

45 = ½ × 10 × t²

45 = 5 × t²

[tex]t^{2} = \frac{45}{5} \\\\t^{2} = 9[/tex]

[tex]t = \sqrt{9}[/tex]

Therefore, it will take 3 s for the stone to get to the surface of the water.

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Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  [tex]X = 38.3 \ m[/tex]

Explanation:

From the question we are told that

The acceleration along the x axis is  [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  [tex]v_{0}x = 7.10 m/s[/tex]

Generally from the equation for acceleration along x axis we have that

     [tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]

=>   [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]

=>   [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

At  t =0  s   and  [tex]v_{0}x = 7.10 m/s[/tex]

=>   [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]

=>   [tex]K_1 = 7.10[/tex]      

So  

      [tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

=>  [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]

=>  [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]

At  t =0  s   and   x = -14.0 m

  [tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]

=>   [tex]K_2 = -14[/tex]

So

     [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]

At  t = 10.0 s

      [tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]

=>   [tex]X = 38.3 \ m[/tex]

Answer: 13062.5 N

What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h? i got the answer to be 13062.

Answer:

The answer is 13062.5 Newtons (N).

Explanation:

The mass of the car is 1100 kg.

The acceleration is 95 km/h.

Using this information, we can use Newton's 2nd Law, F=MA.

1100 kg * 95km/h = 104500 N

Because the answer wants average force, we need to divide the answer by 8 seconds, giving us 13062.5 N.

104500 N / 8 seconds = 13062.5 N (avg force).

Answer:

1000x Newton

Explanation:

Step one

given data

acceleration= 1000 m/s²

The question did not specify the mass of the mass of the space ship.

So, let's assume the mass is x kg

Step two:

Required is the force F in Newton

From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.

F=mass x Acceleration

F=ma

Substituting our given data we have

F=1000x Newton

Answer: Option C) -mv / M is the correct answer

Explanation:

given that;

A rifle of mass M is initially at rest but free to recoil

velocity = v

using conversation of moment

pi = pf

i.e initial moment = final moment

But initial moment was zero (0) since everything  was rest but free to recoil

so

pi = pf

0 = mv + MV

V is the recoil velocity of the rifle

MV = -mv

V = -mv / M

Therefore Option C) -mv / M is the correct answer

Answer:

The time taken by the ant is 1204.82 seconds or 20.08 minutes

Explanation:

Given:

Speed of the ant (s) = 0.083 m/s

Distance traveled by the ant (D) = 100 m

We know that distance traveled by a body is equal to the product of the speed of the body and the time taken for its travel.

Let the time taken by the ant be 't' seconds.

Now, as per the formula:

Distance = Speed × Time

⇒ 100 m = 0.083 m/s × t

⇒ [tex]t = (100\ m)/(0.083\ m/s)[/tex]

[tex]\therefore t=1204.82\ s[/tex]

Now, we know that,

60 seconds = 1 minute

So, 1 second = [tex]\frac{1}{60}\ min[/tex]

Therefore, [tex]1204.82\ s = \frac{1204.82}{60}=20.08\ min[/tex]

Hence, the time taken by the ant is 1204.82 seconds or 20.08 minutes

Answer:

3.125 meters.

Explanation:

(3.0*10^8)/(96*10^6)

= 3.125 meters.

Hope this helped!

I have assumed a weight of 120 N on Earth.

Answer:

The object weighs 20 N on the moon

Explanation:

Weight

The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.

The formula to calculate the weight is:

W = m.g

If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:

[tex]g_m=1/6 g_e[/tex]

Dividing by ge:

[tex]g_m/g_e = 1/6[/tex]

An object of weight We=120 N on planet Earth has a mass of:

[tex]m = 120 / g_e[/tex]

Multiplying by gm:

[tex]m.g_m=120 g_m/g_e[/tex]

Substituting the ratio of accelerations of gravity:

[tex]m.g_m=120 * 1/6[/tex]

Since m.gm is the weight on the Moon Wm:

[tex]W_m=20~N[/tex]

The object weighs 20 N on the moon

Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, the weight of the object on the moon would be one-sixth of its weight on Earth.

Knowing an object's weight on Earth and applying the ratio of the acceleration caused by gravity on the moon to that on Earth will allow us to determine an object's weight on the moon.

Assume that the object's weight on Earth is W (in newtons).

On Earth, the acceleration due to gravity = 9.8 [tex]m/s^2[/tex].

On the moon, = (1/6) * 9.8 [tex]m/s^2[/tex] = 1.6333 [tex]m/s^2[/tex]

Weight = Mass * Acceleration due to gravity

Weight on the moon = W * (1/6)

Thus, the weight of the object on the moon would be one-sixth of its weight on Earth.

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