Two identical point charges q=71.0 pCq=71.0 pC are separated in vacuum by a distance of 2d=29.0 cm.2d=29.0 cm. Calculate the total electric flux ΦΦ through the infinite surface placed at a distance dd from each charge, perpendicular to the line on which the point charges are located.

Answer:

I believe the answer is 254.8 J, or rounded 255 J.

Explanation:

The formula for potential energy is:

PE=m(h)g

This means the mass (m) times height (h) times gravity (m). Gravity is 9.8 m/s (meters per second). Putting all of the numbers into it would equal:

PE=2(13)9.8

This equals 254.8 exactly, or if the assignment calls for you to round, 255.

Answer:

[tex]q=4\times 10^{-16}\ C[/tex]

Explanation:

It is given that,

The charge on an object is 2500 e.

We need to find how many coulombs in the object. The charge remains quantized. It says that :

q = ne

[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]

So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].

Answer:

The new wavelength is 112.5 nm.

Explanation:

It is given that,

When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10⁻¹⁹ J. We need to find the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

The energy of incident electron is given by :

[tex]E=\dfrac{hc}{\lambda}[/tex]

New energy is 2 E and new wavelength is [tex]\lambda'[/tex]. So,

[tex]\dfrac{E}{2E}=\dfrac{\lambda'}{\lambda}\\\\\dfrac{1}{2}=\dfrac{\lambda'}{\lambda}\\\\\lambda'=\dfrac{\lambda}{2}\\\\\lambda'=\dfrac{225}{2}\\\\\lambda=112.5\ nm[/tex]

So, the new wavelength is 112.5 nm.

Answer:

Element C will be best for a nuclear fission reaction

Explanation:

Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process. For the reaction to be continuous in a chain reaction, the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

Answer:

a) m₁ = m₂  F₁ₓ = F₂ₓ

b) m₁ << m₂   F₂ₓ =0

Explanation:

This interesting exercise is unclear your statement, so that in a center of mass system has an acceleration of zero it is necessary that the sum of the forces on each axis is zero, to see this we write Newton's second law

     ∑ F = m a

for acceleration to be zero implies that the net force is zero.

we must write the expression for the center of mass

        [tex]x_{cm}[/tex] = 1 / M (m₁ x₁ + m₂ x₂)

now let's use the derivatives

      [tex]a_{cm}[/tex] = d² x_{cm}/dt² = 1 / M (m₁ a₁ + m₂a₂)

where M is the total mass M = m₁ + m₂

     so that the acceleration of the center of mass is zero

               0 = 1 / M (m₁ a₁ + m₂a₂)

               m₁ a₁ = - m₂ a₂

In the case that we have components on the x axis, the modulus of the two forces are equal and their direction is opposite, therefore

   F₁ₓ = -F₂ₓ

b)r when the two masses are equal , in the case of a mass greater than the other m₁ << m₂

      acm = d2 xcm / dt2 = 1 / M (m1 a1 + m2a2)

so that the acceleration of the center of mass is zero

               0 = 1 / M (m1 a1 + m2a2)

               m1 a1 = - m 2 a2

with the initial condition, we can despise m₁, therefore

                0 = m₂a₂

 if we use Newton's second law

              F₂ = 0

I tell you that in this case with a very high mass difference the force on the largest mass must be almost zero

Answer:

B. t = 0.250s

Explanation:

A. An image with the sketch of the bat emitting a sound, which reflects on a surface and return to the bat is attached below.

B. In order to calculate the time that the pulse emitted by the bat, return to the bat, you first calculate the time that pulse takes to arrive to the object.

You use the following formula:

[tex]x=vt[/tex]      (1)

x: distance to the object = 43m

t: time = ?

v: speed of sound beat = 343 m/s  

You solve the equation (1) for t:

[tex]t=\frac{x}{v}=\frac{43m}{343m/s}=0.125s[/tex]

The time on which the bat hears the echo is twice the value of t, that is:

[tex]t'=2(0.125s)=0.250s[/tex]

The time on which bat heart the echo of its sound, from the moment on which bat emitted it, is 0.250s

Answer:

Explanation:

We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth

[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]

[tex]L=\frac{.5}{.24}[/tex]

L= 2.05

The length will appear to be 2.05 m . and width will appear to be .5 m  to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.  

Answer:

the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Explanation:

The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:

[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]

p: momentum

m: mass

[tex]\Delta v[/tex]:  change in the velocity

The sign of the change in the velocity determines the direction of rate of change. Then you have:

[tex]\Delta v=v_2-v_1[/tex]

v2: final velocity = 35m/s

v1: initial velocity = 40m/s

[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]

Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Answer:

185 N

Explanation:

Sum of forces in the x direction:

Fₓ = -(80 N cos 75°) + (120 N cos 60°)

Fₓ = 39.3 N

Sum of forces in the y direction:

Fᵧ = (80 N sin 75°) + (120 N sin 60°)

Fᵧ = 181.2 N

The magnitude of the net force is:

F = √(Fₓ² + Fᵧ²)

F = √((39.3 N)² + (181.2 N)²)

F = 185 N

We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that  the magnitude of the resultant force is

R=200N

From the question we are told

Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?

A) 47.5 N

B) 185 N

C) 198 N

D) 200 N

Generally the equation for the Resultant force is mathematically given as

For x axis resolution

[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]

For y axis resolution

[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]

Therefore

[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]

For more information on this visit

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Answer:

8kg

Explanation:

For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :

M ×3 = 12 ×2

M = 24/3 = 8kg

Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.

Explanation:

impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)

Answer:

3750Ns

Explanation:

Impulse is defined as Force × time

Force = mass × acceleration,

Hence impulse is;

mass × acceleration × time.

From Newton's second law

Force × time = mass × ∆velocity

750× 5 = 3750Ns

∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.

Answer:

# _units = 1000

Explanation:

This exercise we can use a direct proportion rule.

If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?

    # _units = V₁₀ / V₁

The volume of a body of radius 1 is

       V₁ = 4/3 π r₁³

        V₁ = 4/3π

the volume of a body of radius r = 10

        V₁₀ = 4/3 π r₂³

        V10 = 4/3 π 10³

the number of times this content is

         #_units = 4/3 π 1000 / (4/3 π 1)

        # _units = 1000

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

v₀ₓ = 63.5 m/s

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

v₀y = 54.2 m/s

Answer:

Explanation:

Given That:

radius of spherical core r₁ = 4cm

radius of tubular r₂ = 0.5cm

length of tubular l = 8cm

Volume of spherical V₁

[tex]=\frac{4}{3} \pi r_1^3[/tex]

[tex]=\frac{4}{3} \pi(4)^3\\\\=\frac{4}{3} \pi 64\\\\=268.1cm^3[/tex]

Volume of tabular V₂

[tex]=\pi r ^2_2h[/tex]

[tex]=\pi(0.5)^2\times 8\\\\ =\pi 90.250\times8\\\\ =\pi 2\\\\=6.283cm^3[/tex]

F ∝ V

[tex]F_1 \propto V_1[/tex] and [tex]F_2 \propto V_2[/tex]

As  V₁ is greater than V₂

⇒ F₁ is greater than F₂

F is force

V is volume

This is the required answer

Answer:

[tex]\delta = 0.385\,m[/tex] (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]

Where:

[tex]P[/tex] - Load experimented by the bar, measured in newtons.

[tex]L[/tex] - Length of the bar, measured in meters.

[tex]A[/tex] - Cross section area of the bar, measured in square meters.

[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])

[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]

Where:

[tex]D_{o}[/tex] - Outer diameter, measured in meters.

[tex]D_{i}[/tex] - Inner diameter, measured in meters.

[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]

[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]

The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)

[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]

[tex]\delta = -3.852\times 10^{-4}\,m[/tex]

[tex]\delta = -0.385\,mm[/tex]

[tex]\delta = 0.385\,m[/tex] (Compression)

Given that,

First charge = Q₁

Second charge = Q₂

The potential experienced by Q2 due to Q1 increases

We know that,

The electrostatic force between two charges is defined as

[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]

Where,

k = electrostatic constant

[tex]Q_{1}[/tex]= first charge

[tex]Q_{2}[/tex]= second charge

r = distance

According to given data,

The potential experienced by Q₂ due to Q₁ increases.

We know that,

The potential is defined from coulomb's law

[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]

[tex]V\propto\dfrac{1}{r}[/tex]

If r decrease then V will be increases.

If V decrease then r will be increases.

Since, V is increases then r will decreases that is moving Q₁ closer  to Q₂.

Hence, Moving Q₁ closer to Q₂.

(c) is correct option.

Answer:

36s

Explanation:

Let the objects be A and B.

Let the initial velocity of A be U and the initial velocity of B be 3U

The height sustain by A will be;

The final velocity would be zero

V2 = U2-2gH

Hence

0^2= U2 -2gH

H = U^2/2g

Similarly for object B, the height sustain is;

V2 = (3U)^2-2gH

Hence

0^2= 3U^2 -2gH

U2-2gH

Hence

0^2= U2 -2gH

H = 3U^2/2g

By comparism. The object with higher velocity sustains more height and so should fall longer than object A.

Now object A would take;

From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;

V=10×12=120m/s let g be 10m/S2

Similarly for object B,

The final velocity for B when it's falling it should be 3×that of A

Meaning

3V= gt

t =3V/g = 3× 120/10 = 36s

Answer:

6.69m/s,529N

Explanation:

Now we have 3 forces acting on the boy at its least point, the two tensions on the string and its weight. The tensions are acting upwards why its weight is acting downwards.

Hence the net force causes the child to swing in a circular fashion without skidding off the swing.

This net force is the centripetal force.

The weight of the child is mass × g

35×9.8=343N

The net force = 436+436-343 = 529N

The centripetal force is defined mathematically as;

F = mass × velocity square/ length of string.

Hence 529 = 35V^2/2.96

V^2 = 529×2.96/35 =44.7383

V=√44.7383 =6.69m/s

B. The force exerted by the seat on the child is the net force which keeps the boy from falling.

529N

Answer:

b

Explanation:

Answer:

a)     d = (6.00 t i ^ + 0.500 t²) m , b)   v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d)  v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

       v = x / t

        x = v t

we substitute

        x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

         y = v₀ t + ½ a t²

         y = ½ at t²

we substitute

        y = ½ 1.00 t²

        y = 0.500 t²

in vector position is

       d = x i ^ + y j ^

       d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

       vₓ = v₀

       vₓ = 6.00 m / s

y Axis  

there is an acceleration and the initial speed is zero

         [tex]v_{y}[/tex] = v₀ + a t

         v_{y} = a t

         v_{y} = 1.00 t

the velocity vector is

         v = vₓ i ^ + v_{y} j ^

         v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

        d = (6.00 4 i ^ + 0.50 4 2 j⁾

        d = (24.00 i ^ + 8.00 j^ ) m

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

        v = (6 i ^ + 1 5 j ^)

         v = (6.00 i ^ + 5 j^ ) m/s

The correct answer is A. It is subjective

Explanation:

In 1943, the recognized psychologist Abraham Maslow proposed a theory to understand and classify human needs. The work of Maslow included five different categories to classify all basic needs, psychological needs, and self-esteem needs; additionally, in this, Maslow proposed individuals need to satisfy the needs of previous levels to satisfy more complex needs. For example, the first level includes physiological needs such as hunger and these are necessary to get to more complex needs such as the need for safety or self-satisfaction.

This hierarchy is still used all around the world to understand human needs; however, it was been widely criticized because the classification itself is related to Maslow's perspective as this was mainly based on Maslow's ideas about needs, which makes the hierarchy subjective. Also, due to its subjectivity,  the hierarchy may apply only in some individuals or societies.

Answer: it takes 3 seconds (b)

Explanation:

Answer: B. 3

Explanation:

Each black point on the map represents one second. There are three black points with vectors representing Z's movement before Y begins to move.

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:  

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed = 14 m/s

[tex]V_{0}[/tex]: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]  

b) The maximum height is:

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]

c) The time can be found using the following equation:

[tex] V_{f} = V_{0} - gt [/tex]

[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]

d) The flight time is given by:

[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]

I hope it helps you!    

Answer:

The velocity is  [tex]v = 8.85 m/s[/tex]

Explanation:

From the question we are told that

    The mass of the skier is [tex]m_s = 60 \ kg[/tex]

      The initial speed is [tex]u = 4.0 \ m/s[/tex]

       The height is  [tex]h = 10 \ m[/tex]

According to the law of energy conservation

     [tex]PE_t + KE_t = KE_b + PE_b[/tex]

Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as

       [tex]PE_t = mg h[/tex]

substituting values

       [tex]PE_t = 60 * 4*9.8[/tex]

      [tex]PE_t = 2352 \ J[/tex]

And  [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill

And  [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as

         [tex]KE_b = 0.5 * m * v^2[/tex]

  substituting  values

         [tex]KE_b = 0.5 * 60 * v^2[/tex]

=>     [tex]KE_b = 30 v^2[/tex]

Where v is the velocity at the bottom

   And [tex]PE_b[/tex] is the potential  energy at the bottom which equal to zero due to the fact that height  is zero at the bottom of the hill

So  

        [tex]30 v^2 = 2352[/tex]

=>      [tex]v^2 = \frac{2352}{30}[/tex]

=>       [tex]v = \sqrt{ \frac{2352}{30}}[/tex]

        [tex]v = 8.85 m/s[/tex]

Answer:

The Skier's velocity at the bottom of the hill will be 18m/s

Explanation:

This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.

That is

[tex]mgh = 0.5 mv^{2}[/tex]

solving for velocity, we will have

[tex]v= \sqrt{2gh}[/tex]

hence her velocity will be

[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]

This is the velocity she gains from the slope.

Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s

The Skier's velocity at the bottom of the hill will be 18m/s

Point C would the greatest

ok so im not that good when it comes to physics but i think its C)