Light from a fluorescent lamp is observed through a cloud of cool nitrogen gas. Again, two students are having a discussion about the kind of spectra that they would see.
Student 1: We would see absorption line spectra and the missing lines would correspond to the light from the fluorescent lamp.
Student 2: I disagree. We would see an emission line spectrum corresponding to nitrogen. This would happen because the nitrogen gas would absorb some energy from the fluorescent lamp and would reemit this energy which would result in an emission line spectrum.
Which student, if any, do you agree with and why?

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)

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:

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:

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:

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:

 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:

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

Answer:

Seismic waves

Explanation:

seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.

Answer:

Siesmic waves

Explanation:

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:

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]

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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:

Work is defined as a force doing a movement, for example, if with a force F we move an object a distance D, the work done is:

W = F*D

(note, the force is causing the movement, the product here is a dot product, this means that if the force and the displacement are perpendiculars, the product is zero)

So the examples where there are not work being done may be:

We do not have any movement:

For example, you can go to a wall in your house and push it really hard.

There is a force, but the wall will not move, so we have D = 0

and W = F*D = F*0 = 0

Because we have no motion.

Another case is where the force and the direction of motion are perpendiculars.

If we have for example a car, moving at a constant speed, and you push it sidewise (perpendicular to the direction of movement) we have a force being applied and movement, but those are in different directions (so the force does not cause the movement) so we dont have work being done.

Answer:

a) [tex]s_{T} = 30\,m[/tex], b) [tex]t = 5\,min[/tex], c) [tex]\Delta t = 6.667\,s[/tex], d) [tex]\Delta s_{R} = 33.333\,m[/tex], e) [tex]t' = 11.667\,s[/tex], f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

[tex]s_{T} = v_{T}\cdot t[/tex]

Where:

[tex]t[/tex] - Time, measured in seconds.

[tex]v_{T}[/tex] - Velocity of the turtle, measured in meters per second.

[tex]s_{T}[/tex] - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

[tex]s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)[/tex]

[tex]s_{T} = 30\,m[/tex]

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

[tex]t = \frac{s_{T}}{v_{T}}[/tex]

[tex]t = \frac{150\,m}{0.5\,\frac{m}{s} }[/tex]

[tex]t = 300\,s[/tex]

[tex]t = 5\,min[/tex]

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

[tex]v_{R} = v_{o,R} + a_{R}\cdot \Delta t[/tex]

Where:

[tex]v_{R}[/tex] - Final velocity of the rabbit, measured in meters per second.

[tex]v_{o,R}[/tex] - Initial velocity of the rabbit, measured in meters per second.

[tex]a_{R}[/tex] - Acceleration of the rabbit, measured in [tex]\frac{m}{s^{2}}[/tex].

[tex]\Delta t[/tex] - Running time, measured in second.

[tex]\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}[/tex]

[tex]\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }[/tex]

[tex]\Delta t = 6.667\,s[/tex]

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

[tex]v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}[/tex]

[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]

[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]

Where [tex]\Delta s_{R}[/tex] is the travelled distance of the rabbit from rest to maximum speed.

[tex]\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\Delta s_{R} = 33.333\,m[/tex]

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

[tex]t' = \frac{\Delta s_{R}}{v_{R}}[/tex]

[tex]t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }[/tex]

[tex]t' = 11.667\,s[/tex]

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

[tex]t_{T} = 300\,s[/tex]

Rabbit:

[tex]t_{R} = 60\,s + 6.667\,s + 11.667\,s[/tex]

[tex]t_{R} = 78.334\,s[/tex]

The rabbit won the race as [tex]t_{R} < t_{T}[/tex].

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.

Answer:

b

Explanation:

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:

B) The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.

Explanation:

For every negative charge of an electron, there is an equal positively charged proton in the nucleus of the atom. This is why the overall charge on an atom is zero.

Answer:

B

Explanation:

This is described by Gauss a scientist.

The positive charge is found in the proton in the nucleus.

The neutron has no charge.

The positive charge radiates in all directions and a counter negative charge ensues.

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

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:

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:

The electric flux at the infinite surface is ZERO

Explanation:

From the question we are told that  

    The point charge are identical and the value is  [tex]q = 71.0 pC = 71 * 10^{-12} \ C[/tex]

    The distance of separation is  [tex]D = 29.0 \ cm = 0.29 \ m[/tex]

    The distance of both from the infinite surface is  d

Generally the electric force exerted by each of the  charge on the infinite surface is

       [tex]\phi = \frac{q}{\epsilon_o}[/tex]

Now given from the question that they are identical, it then means that the electric flux of the first charge on the infinite surface will be nullified by the electric flux of the second charge hence the electric flux at that infinite surface due to this two identical charges is ZERO

Answer:

The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons

Explanation:

Binding energy can be given by the formula:

Binding energy = Binding energy per nucleon * Number of nucleons

This means that if the binding energy per nucleon = x MeV

Where x is a positive real number

The nucleus with 60 nucleons will have Binding energy = 60x MeV

The nucleus with 200 nucleons will have binding energy = 200x MeV

For a +ve x, 200x > 60x

Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.

Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.

It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.

Thus the response above is correct.

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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].

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

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:

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:

10 m/s²

Explanation:

Acceleration is the slope of the velocity vs time graph.

At t = 1.0 s, the slope of the line is:

a = Δv/Δt

a = (20 m/s) / (2.0 s)

a = 10 m/s²

Answer:

10m/s2

Explanation:

Acceleration is defined as change in velocity over the time period

Now at 1second the velocity is 10m/s.

Hence the acceleration is V-U/t;

Where V is velocity at 10s and U is velocity when the object is at rest.

So we have;

Acceleration = 10-0/ 1 = 10m/s2

Answer: Option A

Explanation:

The potential energy decreases in the case when the charges are opposite and they attract each other.

In this case there is no external energy required in order to put the charges together.

This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.

Example: a negative charge and a positive charge.