The movement of a car on a road is represented in this figure. Between t = 0 and t = 0.6 hrs, what is the displacement made by the car?

1). 4.0 km.
2). 0.0 km.
3). -4.0 km. 4
). 8.0 km.

Answer:

2.7s

Explanation:

The solution of time required is shown below:-

In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time

Now, the

63% that is equal to 0.63 which is full equilibrium charge

Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s

So, the correct answer is 2.7s

Answer:

V = 451.47 volts

Explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0

So,

[tex]d=\dfrac{1}{2}(v-u)t[/tex]

[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]

Now applying the conservation of energy i.e.

[tex]\dfrac{1}{2}mv^2=qV[/tex]

V is voltage

[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]

So, the voltage is 451.47 V.

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

Answer:

  B. matching your speed

Explanation:

To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*

_____

* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.

Answer:

It's C "signaling your intent"

Explanation:

The key thing to look at is they are asking the rest of the first step and that;s C

Answer:

Δx = 9 x 10⁻³ m = 9 mm

Explanation:

The formula for fringe spacing in Young's Double Slit Experiment is given as follows:

Δx = λL/d

where,

Δx = fringe spacing = ?

λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m

L = Distance between slits and screen = 2 m

d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m

Therefore,

Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)

Δx = 9 x 10⁻³ m = 9 mm

Answer:

Explanation:

Area of the loop = a b

current = I

magnetic moment of the loop M  = area x current

= ab I

Torque on the loop = MB sinθ

here θ = 90

Torque = MB

= abIB

In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .

Answer:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

The friction force between the tires of the truck and the road is 22500 N.

It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.

Distance traveled before coming to rest, s = 10m

The final velocity of the truck will be zero, v = 0

When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.

The frictional force is given by:

f = -ma

the acceleration of the truck = -a

The negative sign indicates that the acceleration is opposite to the motion.

Applying the third equation of motion we get:

v² = u² -2as

0 = 15² - 2×a×10

225 = 20a

a = 11.25 m/s²

So the magnitude of frictional force is:

f = ma = 2000 × 11.25 N

f = 22500 N

Learn more about friction force:

https://brainly.com/question/1714663?referrer=searchResults

Answer:

Power = Force × Distance/time

Power = Force × Velocity

Power = 2,300.0 N × 28.0 m/s²

Power = 64400 Nm/s

Explanation:

First show the formula of Power

Re-arrange formula and used to work out Power

Pretty simple stuff!

Hope this Helps!!

Answer:

The  potential energy is [tex]PE = 2.031 *10^{-7} \ J[/tex]

Explanation:

From the question we are told that

    The inner radius is  [tex]r_i = 5 \ mm = 0.005 \ m[/tex]

      The outer radius is  [tex]r_o = 11 \ mm = 0.011 \ m[/tex]

     The  common length is  [tex]l = 160 \ m[/tex]

      The  potential  difference is   [tex]V = 6 \ V[/tex]

Generally the capacitance of the cylindrical capacitor is mathematically represented as

       [tex]C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l[/tex]

Where  [tex]\epsilon _o[/tex] is the permitivity of free space with the values [tex]\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

and  k  is the dielectric constant  of the dielectric material here the  dielectric material is free space so  k  =   1

     Substituting values

             [tex]C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160[/tex]

             [tex]C = 1.129 *10^{-8} \ F[/tex]

The potential energy stored is mathematically represented as

       [tex]PE = \frac{1}{2} * C * V ^2[/tex]

substituting values

      [tex]PE = 0.5 * 1.129 *10^{-8} * (6)^2[/tex]

      [tex]PE = 2.031 *10^{-7} \ J[/tex]

Answer:

5.33333... seconds

Explanation:

800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....

Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

We are given;

Charge;Q = 3.40 nC = 3.4 × 10^(-9) C

Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

Answer:

c. 66 V

Explanation:

p =IV

I =P/V

P1/V1=P2/V2

60/110=36/V2

0.55 = 36/V2

V2 =36/0.55 = 65.5V

V2 = 66V

Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm

Answer:

d = 1.0128×10⁻⁵m

Explanation:

given:

length L = 4.0m

maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m

wavelength λ = 633nm = 633×10⁻⁹m

note:

dsinθ = mλ (constructive interference)

where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength

for small angle

sinθ ≈ tanθ

[tex]d (\frac{y}{L}) =[/tex] mλ

[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]

[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]

d = 1.0128×10⁻⁵m

Answer:

The force is  [tex]F = 172 \ N[/tex]

Explanation:

From the question we are told that

    The  mass of the block is  [tex]m_b = 27.0 \ kg[/tex]

     The  coefficient of  static friction is  [tex]\mu_s = 0.65[/tex]

     The coefficient of kinetic friction is  [tex]\mu_k = 0.50[/tex]

The  normal force acting on the block is  

      [tex]N = m * g[/tex]

substituting values

     [tex]N = 27 * 9.8[/tex]

     [tex]N = 294.6 \ N[/tex]

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

        [tex]F_f = \mu_s * N[/tex]

substituting values

        [tex]F_f = 0.65 * 264.6[/tex]

        [tex]F_f = 172 \ N[/tex]

Now for this  block to move the force require is  equal to [tex]F_f[/tex] i.e

       [tex]F= F_f[/tex]

=>    [tex]F = 172 \ N[/tex]

Answer:

Current that flows through any one of the resistors has a value of 12 amperes.

Explanation:

When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:

[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]

[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]

Where:

[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.

[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.

[tex]i[/tex] - Current, measured in amperes.

If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:

[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]

Current that flows through any one of the resistors has a value of 12 amperes.

The current flows via any of the resistors should have a value of 12 amperes.

At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.

Here the circuit should be

vbatt = i.(R1 + R2+ R3)

i = Vbatt/R1 + R2 + R3

here

Vbatt means the voltage of the battery

R1,R2, and R3 means the resistance of the first, second and third resistors

I means the current

So, in the case when

R1 = R2 = R3 = R

So,

i = Vbatt/3.R

Learn more about current here: https://brainly.com/question/14956680

Answer:

The time taken is [tex]t = 0.0225 \ s[/tex]

Explanation:

From the question we are told that

    The length of the wire is [tex]l = 4500 \ km = 4500000 \ m[/tex]

      The  refractive index is  [tex]n_f = 1.5[/tex]

The velocity of the signal is mathematically represented as

       [tex]v = \frac{c}{n_f }[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

 substituting values  

         [tex]v = \frac{3.0 *10^{8}}{1.5}[/tex]

         [tex]v = 2.0*10^{8} \ m/s[/tex]

The time taken is mathematically evaluated as

      [tex]t = \frac{d}{v}[/tex]

substituting values  

      [tex]t = \frac{4500000}{2.0 *10^{8}}[/tex]

      [tex]t = 0.0225 \ s[/tex]

Explanation:

Newton's third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs. Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel out.

A wedge of glass of refractive index 1.64 has a silver coating on the bottom, as shown in the image attached below.

Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.

Answer:

the smallest distance x  = 2.74 × 10⁻³ m or 2.74 mm

Explanation:

From the given information:

The net phase change is zero because both the light ray reflecting from the air-glass surface and silver plate undergo a phase change of [tex]\dfrac{\lambda}{2}[/tex] , as such the condition for the  constructive interference is:

nΔy = mλ

where;

n = refractive index

Δy = path length (inside the glass)

So, from the diagram;

[tex]\dfrac{y}{x}=\dfrac{10^{-5} \ m}{0.2 \ m}[/tex]

[tex]\dfrac{y}{x} = 5 \times 10^{-5}[/tex]

[tex]y = 5 \times 10^{-5} x[/tex]

Now;

Δy can now be = 2 ( 5 × 10⁻⁵ [tex]x[/tex])

Δy =1 ×  10⁻⁴[tex]x[/tex]

From nΔy = mλ

n( 1 ×  10⁻⁴[tex]x[/tex] ) = mλ

[tex]x = \dfrac{m \lambda}{n \times 1 \times 10^{-4} }[/tex]

when the thickness is minimum then m = 1

Thus;

[tex]x = \dfrac{1 \times 450 \times 10^{-9} \ m}{1.64 \times 1 \times 10^{-4} }[/tex]

x =  0.00274 m

x = 2.74 × 10⁻³ m or 2.74 mm

Answer: B. The surface of the coating is rough, so light that shines on it gets scattered in many directions.

Explanation: On Edge!!!!!!!!!!!!!!!!!!!!

Answer:

A) 1.76U×10⁻³N

B) 2.716×10⁻³N

C) 264.5⁰

Explanation:

See detailed workings for (a), (b), (c) attached.

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

Answer:

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

Answer:

a

Explanation:

The correct answer would be that the apparent depth of the object is less than the real depth.

The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.

Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.

The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.

Correct option: a

Answer:

The  time interval is  [tex]t = 3 \ s[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = 4.0 \ rad/s^2[/tex]

     The  time taken is  [tex]t = 4.0 \ s[/tex]

      The angular displacement is  [tex]\theta = 80 \ radians[/tex]

The angular displacement can be represented by the second equation of motion as shown below

          [tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]

where  [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval

So substituting values

        [tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]

=>    [tex]w_i = 12 \ rad/s[/tex]

Now considering this motion starting from the start point (that is rest ) we have

       [tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]

Where  [tex]w__{0}}[/tex] is the angular velocity at rest which is zero  and  [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s

        [tex]12 = 0 + 4 t[/tex]

=>       [tex]t = 3 \ s[/tex]

Following are the response to the given question:

Given:

[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]

[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]

To find:

[tex]\to \omega=?\\\\\to t=?\\\\[/tex]

Solution:

Using formula:

[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]  

It would be the angle for rotation at the start of the 4-second interval.

This duration can be estimated by leveraging the fact that the wheel begins from rest.  

[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]

Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".

Learn more:

brainly.com/question/7464119

Answer:

C) Each successive stage lasts for approximately the same amount of time.

Explanation:

Nuclear burning is a series of nuclear processes through which a star gets its energy. The energy within a star is due to nuclear fusion of lighter elements (hydrogen) into more massive element (helium), with a release of a large amount of energy due to the conversion of some of the mass into energy. Each stage leads to a loss of some of the mass which is converted into energy (option A is valid).

The fusion of four hydrogen atoms into one helium atom means that there is a creation of element with a higher atomic weight (option D is valid), and the energy output of each stage exceeds its energy input, meaning that each stage will require a higher temperature than its previous stages (option B is valid).

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

Answer:

   R = r_protón / 2

Explanation:

The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement

      F = m a

Since the magnetic force is perpendicular, the acceleration is centripetal.

       a = v² / R

the magnetic force is

       F = q v x B = q v B sin θ

the field and the speed are perpendicular so the sin 90 = 1

we substitute

          qv B = m v² / R

          R = q v B / m v²

in the exercise they indicate

the charge  q = 2 e

the mass     m = 4 m_protón

        R = 2e v B / 4m_protón v²

we refer the result to the movement of the proton

         R = (e v B / m_proton) 1/2

the data in parentheses correspond to the radius of the proton's orbit

         R = r_protón / 2

Answer:

 x = 0.006 m

Explanation:

The potential at one point is given by

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

remember that the potential is to scale, let's apply to our case

          V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

          0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

         3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

         3 / x = 500

          x = 3/500

          x = 0.006 m

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

 [tex]a_{max} = 0.00246 \ m/s^2[/tex]

b

   [tex]k =722.2 \ N/m[/tex]

Explanation:

From the question we are told that

     The  amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]

     The period is [tex]T = 17 \ s[/tex]

    The test weight is  [tex]W = 13 \ N[/tex]

Generally the radial acceleration is mathematically represented as

        [tex]a = w^2 r[/tex]

at maximum angular acceleration

       [tex]r = A[/tex]

So  

       [tex]a_{max} = w^2 A[/tex]

Now [tex]w[/tex] is the angular velocity which is mathematically represented as

      [tex]w = \frac{2 * \pi }{T}[/tex]

Therefore

       [tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]

substituting values

       [tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]

       [tex]a_{max} = 0.00246 \ m/s^2[/tex]

Generally this test weight is mathematically represented as

     [tex]W = k * A[/tex]

Where k is the spring constant

Therefore

        [tex]k = \frac{W}{A}[/tex]

substituting values        

      [tex]k = \frac{13}{0.018}[/tex]

     [tex]k =722.2 \ N/m[/tex]