Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recent years allows us to perform the double slit experiment rather easily in class. Your professor shines a green laser (564 nm) on a double slit with a separation of 0.108 mm. The diffraction pattern shines on the classroom wall 4.0 m away. Calculate the fringe separation between the fourth order and central fringe.

Answer:

69°

Explanation:

Projectiles that land at the same elevation they're launched from will have the same range if the launch angles are complementary (add up to 90°).

The complement of 21° is 69°.

The second ball hit at a 69° angle.

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Answer:

v = 2.41 m/s

Explanation:

Here net force on the Skater + snowball is zero

so here we can use momentum conservation for skater + snowball

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

now we have

[tex]66.2(2.47) + (0.136)(2.47) = 66.2 (v) + 0.136(31.4)[/tex]

[tex]163.85 = 66.2 v + 4.27[/tex]

[tex]163.85 - 4.27 = 66.2 v[/tex]

[tex]v = \frac{159.6}{66.2}[/tex]

[tex]v = 2.41 m/s[/tex]

so the final speed of skater will be 2.41 m/s

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

Answer:

6.0625 x 10^13

Explanation:

Charge of one electron, e = 1.6 x 10^-19 C

Total charge, Q = 9.7 μ C = 9.7 x 10^-6 C

The number of electrons is given by

n = Total charge / charge of one electron

n = Q / e

n = (9.7 x 10^-6) / (1.6 x 10^-19)

n = 6.0625 x 10^13

Answer:

100 cm³

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

n and R are constant, so:

P₁V₁/T₁ = P₂V₂/T₂

If we say point 1 is at 40m depth and point 2 is at the surface:

P₂ = 1.013×10⁵ Pa

T₂ = 20°C + 273.15 = 293.15 K

P₁ = ρgh + P₂

P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa

P₁ = 4.933×10⁵ Pa

T₁ = 4.0°C + 273.15 = 277.15 K

V₁ = 20 cm³

Plugging in:

(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)

V₂ = 103 cm³

Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.

Answer:

49220 J

Explanation:

The work done by the tension force on the elevator is given by:

W = F d

where

F is the magnitude of the tension

d is the vertical distance through which the elevator has moved

Here we have

[tex]F=2.14\cdot 10^4 N[/tex]

d = 2.30 m

Substituting,

[tex]W=(2.14\cdot 10^4 N)(2.30 m)=49220 J[/tex]

The work done by the tension force on the elevator as it rises through a vertical distance of 2.30 m while exerting a constant upward tension of 2.14 x 10^4 N, is calculated using the work formula (W = Fd). The calculated work done is 4.92 x 10^4 Joules.

The work done by the tension force on the elevator can be calculated using the formula for Work (W), which is force (F) multiplied by distance (d). Since tension is the force applied here and it's constant as the elevator moves vertically upwards, we can apply the formula as follows: W = Fd.

In this case, F (tension) is 2.14 x 10^4 N, and d (distance) is 2.30 m. Substituting these values into the formula will give us:

W = Fd = (2.14 x 10^4 N) x 2.30 m = 4.92 x 10^4 Joules

So, the work done by the tension force on the elevator as it rises 2.30 m is 4.92 x 10^4 Joules.

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Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

[tex]F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N[/tex]

Answer:

18816π ≈ 59112 Joules

Explanation:

Draw a picture (like the one attached).  The cone has a radius R and height H.  The spout has height a.  The tank is filled with water to height h.

Cut a thin slice from the volume of the water.  This slice is a cylindrical disc with a radius r, a thickness dy, and a height y.

Using similar triangles, we can say:

r / y = R / H

The work required to lift this slice up to the spout is:

dW = dm g (H + a − y)

where dm is the mass of the slice and g is the acceleration due to gravity.

Mass is density times volume, so:

dW = dV ρ g (H + a − y)

Substituting the volume of the cylindrical disc:

dW = dy π r² ρ g (H + a − y)

From our similar triangles equation, we know that r = R/H y, so:

dW = dy π (R/H y)² ρ g (H + a − y)

Rearranging:

dW = π (R/H)² ρ g y² (H + a − y) dy

dW = π (R/H)² ρ g ((H + a)y² − y³) dy

The work to lift all the slices between y=0 and y=h is:

W = ∫ dW

W = π (R/H)² ρ g ∫₀ʰ ((H + a)y² − y³) dy

Integrating:

W = π (R/H)² ρ g (⅓(H + a)y³ − ¼y⁴) |₀ʰ

W = π (R/H)² ρ g (⅓(H + a)h³ − ¼h⁴)

Given:

R = 2 m

H = 5 m

a = 1 m

h = 2 m

g = 9.8 m/s²

ρ = 1000 kg/m³

W = π (2/5)² (1000) (9.8) (⅓(5 + 1)(2)³ − ¼(2)⁴)

W = 18816π

W ≈ 59112 Joules

It takes approximately 59.1 kJ of work.

Final answer:

To calculate the work needed to pump water from a conical tank, integrate the volume of water being moved over the distance it must be lifted, considering the tank's dimensions, the spout's height, and the effects of gravity.

Explanation:

The question involves finding the work needed to pump water from a conical tank with a spout at the top. To do this, we need to consider the volume of water to be moved, the height it must be lifted, and the forces involved, including gravity. The water's density (1000 kg/m3) and gravitational acceleration (9.8 m/s2) are key to calculating this work.

The work needed to lift water involves the product of the force required to lift the water, the distance it must be moved, and the volume of water. Given the tank's dimensions and the additional height from the spout, one would typically integrate along the height of the water to compute the total work, considering the changing cross-sectional area of the water column as it is being pumped out.

To find this, one could use the formula for the volume of a cone to model the water's volume at any height and then integrate this volume over the distance it must be moved, accounting for the force of gravity. This problem combines principles of physics and calculus to find a solution.

1. [tex]1.0\cdot 10^{-6}m[/tex]

First of all, let's convert the energy of the absorbed photon into Joules:

[tex]E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J[/tex]

The energy of the photon can be rewritten as:

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

where

h is the Planck constant

c is the speed of light

[tex]\lambda[/tex] is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m[/tex]

2. 1.24 eV

In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

[tex]\lambda=1.66\cdot 10^{-6}m[/tex]

So the energy of the emitted photon is given by the formula used previously:

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

and using

[tex]\lambda=1.66\cdot 10^{-6}m[/tex]

we find

[tex]E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J[/tex]

converting into electronvolts,

[tex]E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV[/tex]

EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.

Answer:

The emf and internal resistance of the battery are 9 volt and 2.04 Ω

Explanation:

Given that,

First resistor[tex]R = 100\ \Omega[/tex]

Power[tex]P = 0.794\ W[/tex]

Second resistor[tex]R=200\ \Omega[/tex]

Power[tex]P=0.401\ W[/tex]

We need to calculate the emf and internal resistance

Using formula of emf

[tex]P=\dfrac{e^2}{R+r}[/tex]

For first resistor

[tex]0.794=\dfrac{e^2}{100+r}[/tex]....(I)

For second resistor

[tex]0.401=\dfrac{e^2}{200+r}[/tex].....(II)

From equation (I) and (II)

[tex]\dfrac{795}{401}=\dfrac{200+r}{100+r}[/tex]

Using component of dividend rule

[tex]\dfrac{393}{401}=\dfrac{100}{100+r}[/tex]

[tex]r =2.04\ \Omega[/tex]

Now, Put the value of internal resistance in equation (I)

[tex]0.794=\dfrac{e^2}{100+2.04}[/tex]

[tex]e^2=0.794(100+2.04)[/tex]

[tex]e=\sqrt{0.794(100+2.04)}[/tex]

[tex]e=9\ volt[/tex]

Hence, The emf and internal resistance of the battery are 9 volt and 2.04 Ω

The emf and internal resistance of a battery can be calculated using Ohm's Law (V=IR) and the power formula (P = I²R). By evaluating the current and emf for two different resistors, we can arrive at two equations that, when solved simultaneously, give the emf and internal resistance.

The question is asking for the calculation of the EMF (electromotive force) and internal resistance of the battery. First, we need to understand the relationship between power, voltage, current, and resistance, which is described by Ohm's Law and the power formulas. For the 100 Ω resistor, we can use the formula for power P = I²R to find the current, which comes out to be √P/R = √0.794/100 = 0.0281 A. Then, we can use Ohm's Law V = I(R+r) to find the voltage of the battery (emf), plugging in the values we get V = 0.0281(100+r).

When the resistor is changed to 200 Ω, we repeat the process, giving us the equation V = 0.0281(200+r). Solving these two equations will give us the EMF and internal resistance of the battery.

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After the collision, the three coupled cars will move with a common velocity of 1.73 m/s. The kinetic energy lost in the collision is 124,017 J.

After the collision, the three coupled cars will move with a common velocity. To find this velocity, we can use the law of conservation of momentum. The total momentum before the collision is the sum of the individual momenta of the cars. After the collision, the total momentum will be the same. Therefore, the mass of the three coupled cars multiplied by their common velocity will be equal to the sum of the individual momenta before the collision. To find the common velocity, we can set up the equation: (3.05 ✕ 10^4 kg + 3.05 ✕ 10^4 kg + 3.05 ✕ 10^4 kg) * v = (3.05 ✕ 10^4 kg * 2.60 m/s) + (3.05 ✕ 10^4 kg * 1.20 m/s) + (3.05 ✕ 10^4 kg * 1.20 m/s). Solving for v, the common velocity of the three coupled cars, gives us a value of 1.73 m/s.

The kinetic energy lost in the collision can be calculated by comparing the initial and final kinetic energies. The initial kinetic energy is given by (1/2) * (3.05 ✕ 10^4 kg * (2.60 m/s)^2 + 3.05 ✕ 10^4 kg * (1.20 m/s)^2 + 3.05 ✕ 10^4 kg * (1.20 m/s)^2) = 237,660 J. The final kinetic energy is (1/2) * (9.15 ✕ 10^4 kg * (1.73 m/s)^2), which is equal to 113,643 J. Therefore, the kinetic energy lost in the collision is 237,660 J - 113,643 J = 124,017 J.

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By using the law of conservation of momentum, we can calculate the final speed of the three linked railroad cars. By calculating the initial and final kinetic energy, we can determine how much energy is lost during the collision.

The question involves concepts of Physics, specifically Conservation of Momentum and Kinetic Energy in the context of colliding objects. In a collision, if external forces can be neglected, the total momentum of the system is conserved. If we let m represent the mass of a car and v its velocity, we have a single car moving (m1v1 = 3.05 x 10^4 kg x 2.60 m/s) and two joined cars moving (2m2v2 = 2 x 3.05 x 10^4 kg x 1.20 m/s). The final velocity of the three aligned and linked cars can be calculated using the law of conservation of momentum: m1v1 + 2m2v2 = 3m3v3.

After solving for v3, we can find the initial and final kinetic energy, and hence calculate the difference in kinetic energy. The kinetic energy before the collision is given by (1/2)m1v1² + (1/2)m2(2v2)². The kinetic energy after the collision is given by (1/2)m3v3², where m3 = m1 + 2m2. The difference in kinetic energy before and after the collision will give us the kinetic energy lost during the collision.

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Answer:

Part a)

f = 1112.3 Hz

Part b)

L = 94.2 dB

Explanation:

Part a)

As we know by Doppler's Effect of sound that when source of sound and observer moves relative to each other then the frequency of sound observed by the observer is different from actual frequency

Here when source and observer is moving towards each other then the observed frequency is given by

[tex]f = f_o\frac{v + v_0}{v- v_s}[/tex]

now plug in all values in it

[tex]f = 965 (\frac{343 + 15.7}{343 - 31.8})[/tex]

[tex]f = 1112.3 Hz[/tex]

Part b)

Power of the siren is given as

[tex]P = 45.2 W[/tex]

Now the intensity of sound received by driver of car is given as

[tex]I = \frac{P}{4\pi r^2}[/tex]

[tex]I = \frac{45.2}{4\pi (36.8)^2}[/tex]

[tex]I = 2.66 \times 10^{-3} W/m^2[/tex]

now the decibel level of sound is determined by

[tex]L = 10 Log \frac{I}{I_o}[/tex]

now we have

[tex]L = 10 Log\frac{2.66 \times 10^{-3}}{10^{-12}}[/tex]

[tex]L = 94.2 dB[/tex]

Hello There!

When an object moves in uniform circular motion, the direction of its acceleration is directed toward the center of its circular path.

Uniform circular motion is a certain object or thing that that is traveling in a circular path. The reason why "uniform" is there because a part of it has to stay constant.

Answer:

C) is directed toward the center of its circular path.

Explanation:

Uniform Circular Motion happens when its trajectory is a circumference and its velocity modulus remains constant over time.

In our daily life we often observe the movement made by fans, car wheels and also by the blender. These are all examples of devices that use MCU.

Within the study of uniform circular motion there is the presence of centripetal acceleration, ie when there is speed variation there is acceleration.

Centripetal acceleration is always directed to the center of the circumference. It does not change the modulus of velocity and its mathematical representation is given by the equation:

                                 acp = v2 / R or acp = ω2 R

Answer:

it will move by d = 8.00 m in x direction before it will turn back

Explanation:

Here the initial velocity of particle is along +x direction

it is given as

[tex]v_i = 4.90 m/s[/tex]

now its acceleration is given as

[tex]a_x = -1.50 m/s^2[/tex]

[tex]a_y = 3.00 m/s^2[/tex]

now when it turns back then the velocity in x direction will become zero

so we will say

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 4.90^2 = 2(-1.50) d[/tex]

[tex] d = 8.00 m[/tex]

so it will move by d = 8.00 m in x direction before it will turn back

Final answer:

The particle moves 8.01 m in the x direction before it turns around, determined by using kinematic equations with the given initial velocity and constant acceleration.

Explanation:

To determine how far the particle moves in the x direction before turning around, we need to find the point at which its velocity in the x direction is zero. Since the particle starts with an initial velocity of 4.90 m/s in the x direction and has a constant acceleration of ax = -1.50 m/s², we can use the kinematic equation v = u + at to find the time when the velocity becomes zero:

v = 0 m/s (the velocity at the turnaround point)
u = 4.90 m/s (initial velocity)
a = -1.50 m/s² (constant acceleration)

Solving for the time (t) gives:

0 = 4.90 + (-1.50)t
t = 4.90 / 1.50
t = 3.27 s

Now using the equation s = ut + (1/2)at² to find the displacement in the x direction:

s = 4.90 m/s ⋅ t + (1/2)(-1.50 m/s²) ⋅ t²
s = 4.90 ⋅ 3.27 + (0.5) ⋅ (-1.50) ⋅ (3.27)²
s = 8.01 m

The particle moves 8.01 m in the x direction before turning around.

Due Sun 06/09/2019 11:59 pm (you're already more than a week late) Skip Navigation Questions, correct Q 1 [1/1], correct Q 2 [1/1], correct Q 3 [1/1], untried Q 4 (0/1), untried Q 5 (0/1), untried Q 6 (0/1), untried Q 7 (0/1), untried Q 8 (0/1), untried Q 9 (0/1), untried Q 10 (0/1), Grade: 3/10, Print Version, Start of Questions:

Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

Start of Answer:

-- When they started, they were 42 miles apart.  When they met, 2 hours later, they were no miles apart.  The distance between them shrank at the rate of 21 miles per hour.  Since they rode directly toward each other, the sum of their individual speeds must have been 21 miles per hour.

-- One biker was two times the speed of the other.  Slower biker: 1 time.  Faster biker: 2 times.  Sum of their speeds:  3 times = 21 miles per hour.  Each 'time' = 7 miles per hour.

-- Slower cycler = 1 time = 7 mi/hr

Faster cycler = 2 times = 14 mi/hr

Quid pro quo sexual harassment involves offering work benefits in exchange for sexual favors or threatening consequences if the person refuses. True.

Answer:

Minimum stopping distance = 164.69 ft

Explanation:

Speed of car = 35.6 mi/hr = 15.82 m/s

Stopping distance = 40.8 ft = 12.44 m

We have equation of motion

             v² = u² + 2as

             0²=15.82²+ 2 x a x 12.44

              a = -10.06 m/s²

Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.

          Speed of car = 71.5 mi/h = 31.78 m/s

We have

           v² = u² + 2as

          0² = 31.78² -  2 x 10.06 x s

           s = 50.2 m = 164.69 ft

Minimum stopping distance = 164.69 ft

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, [tex]\theta_c=52^o[/tex]

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

[tex]n_1sin\theta_c=n_2sin\theta_2[/tex]

Here, [tex]\theta_2=90[/tex]

[tex]sin\theta_c=\dfrac{n_2}{n_1}[/tex]

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

[tex]n_1=\dfrac{n_2}{sin\theta_c}[/tex]

[tex]n_1=\dfrac{1}{sin(52)}[/tex]

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

Answer:

28.23 years

Explanation:

It takes 28.23 years to take one electron to travel the full length of the cable.

3.156 107 = number of seconds in a year.

230000 / (2.58 x 10^-4) = 8.91 x 10^8 second.

Answer:A}

Explanation:Because it happened to me lol

Answer:

The average induced emf is 2.5 V.

(A) is correct option.

Explanation:

Given that,

Number of turns = 10

Length of side = 1 m

change in magnetic field = 4-2=2 T

Time t = 8 sec

We need to calculate the average induced emf

Using formula of induced emf

[tex]\epsilon=NA\times\dfrac{\Delta B}{\Delta t}[/tex]

Where,

N = number of turns

A = area of cross section

B = magnetic field

t = time

Put the value into the formula

[tex]\epsilon=10\times1\times\dfrac{2}{8}[/tex]

[tex]\epsilon = 2.5\ V[/tex]

Hence, The average induced emf is 2.5 V.

This question involves the concepts of induced emf and magnetic field.

The average induced emf will be "2.5 volts".

The induced emf in the coil can be given by the following formula:

[tex]EMF= \frac{NA\Delta B}{\Delta t}[/tex]

where,

Therefore,

[tex]EMF=\frac{(10)(1\ m^2)(2\ T)}{8\ s}\\\\[/tex]

EMF = 2.5 volts

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Try this option (shortly):

According to the 'law of consevation of pulse' the passenger in the smaller car receives greater energy then the one in the larger car.

Almost all the received energy is spent to injure the passenger in the smaller car. :(

Final answer:

Passengers in a smaller mass car are at greater risk in a head-on collision due to higher acceleration and force resulting from the collision. Cars with similar masses experience a more even distribution of momentum changes, potentially reducing impact severity. The duration of impact also affects the decelerating force felt by occupants, illustrating the importance of seatbelts and crumple zones.

Explanation:

A passenger in the smaller mass car is more likely to be injured in a head-on collision with a larger mass car because the smaller car will experience a larger acceleration due to the force of impact, according to Newton's second law (force equals mass times acceleration). This greater acceleration translates to a stronger force exerted on the occupants of the smaller car. Additionally, according to the principle of conservation of momentum, if both cars are moving at the same speed and collide, the car with less mass will undergo a more significant change in velocity.

When cars with more similar masses collide, the change in momentum is distributed more evenly between the two, potentially reducing the severity of the impact on both vehicles. This is due to the momentum of the two-car system remaining constant if external forces are negligible (as stated by the law of conservation of momentum). The impulse, or change in momentum, is also important to consider. If cars crumple together instead of rebounding, there's a smaller change in momentum, leading to a smaller force on the occupants, which is why crumple zones are critical safety features in cars.

The scenario with a seatbelt demonstrates the importance of the time over which a force acts. A seatbelt helps extend the time of a collision's impact, reducing the average decelerating force experienced by the occupant, compared to a sudden stop against the dashboard.

Answer:

The linear density of the wire is 0.314 g/m.

Explanation:

It is given that,

Acceleration, [tex]a=5.52\ m/s^2[/tex]

Mass of the ball, m = 112 gm

Speed of the transverse wave, v = 44.4 m/s

The speed of the transverse wave is given by :

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]

Where

T = tension in the wire

[tex]\mu[/tex] = mass per unit length

[tex]\mu=\dfrac{T}{v^2}[/tex]

[tex]\mu=\dfrac{ma}{v^2}[/tex]

[tex]\mu=\dfrac{112\ g\times 5.52\ m/s^2}{(44.4\ m/s)^2}[/tex]

[tex]\mu=0.3136\ g/m[/tex]

or

[tex]\mu=0.314\ g/m[/tex]

So, the linear density of the wire is 0.314 g/m. Hence, this is the required solution.

The correct answer is b) Between 5 V and 10 V.

Given data:

- Work done (W) = 10 J

- Charge (q) = 2 C

The electric potential (V) can be calculated using the formula:

[tex]\[ V = \frac{W}{q} \][/tex]

Substitute the values:

[tex]\[ V = \frac{10\, \text{J}}{2\, \text{C}} = 5\, \text{V} \][/tex]

So, the electric potential relative to its starting position is 5V

Electric potential (V) is the amount of electric potential energy per unit charge at a point in an electric field. It is measured in volts (V). The formula to calculate electric potential is [tex]\( V = \frac{W}{q} \)[/tex], where W is the work done (electric potential energy) and q is the charge.

In this case, 10 J of work is done to push 2 C of charge into the electric field. Dividing the work done by the charge gives us the electric potential, which is 5 V.

So, the correct answer is b) Between 5 V and 10 V, as the calculated electric potential is 5 V, falling within that range.

Complete Question:
If 10 J of work (electric potential energy) is used in pushing 2 C of charge into an electric field, its electric potential relative to its starting position is

a) Less than 5 V

b) Between 5 V and 10 V

c) More than 20 V

d) 20 V

Answer:

Total momentum of the system is 378 kg-m/s

Explanation:

It is given that,

Mass of first bumper car, m₁ = 222 kg

Velocity of first bumper car, v₁ = 3.10 m/s (in right)

Mass of other bumper car, m₂ = 165 kg

Velocity of second bumper car, v₂ = -1.88 m/s (in left)

Momentum of the system is given by the product of its mass and velocity. So, the total momentum of this system is given by :

[tex]p=m_1v_1+m_2v_2[/tex]

[tex]p=222\ kg\times 3.10\ m/s+165\ kg\times (-1.88\ m/s)[/tex]

p = 378 kg-m/s

Hence, the total momentum of the system is 378 kg-m/s

Answer:

because the troposphere has more oxygen than the stratosphere so when he felt like he was floating even though he was free falling it was due to the absence of air.

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

[tex]F = ma[/tex]

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

[tex]-10000 =2000\times a[/tex]

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

[tex]a = -\dfrac{10000}{2000}[/tex]

[tex]a=-5\ m/s^2[/tex]

For the distance,

Using third equation of motion

[tex]v^2-u^2=2as[/tex]

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

[tex]0-30^2=2\times(-5)\times s[/tex]

[tex]s = 90\ m[/tex]

Hence, The car traveled the distance before stopping is 90 m.

Answer:

The frequency is 0.34 Hz and the wave speed is 1.05 m/s.

Explanation:

Given that,

The distance between two successive minima of a transverse wave is wavelength.

Wavelength = 3.10 m

Time = 14.7 s

(I). We need to calculate the frequency

[tex]f=\dfrac{number\ of\ wave\ propagated\ per\ second }{times}[/tex]

[tex]f=\dfrac{5}{14.7}[/tex]

[tex]f=0.34\ Hz[/tex]

(II). We need to calculate the wave speed

Formula of wave speed

[tex]v= \lambda\times f[/tex]

Where,

[tex]\lambda = wavelength[/tex]

[tex]f = frequency[/tex]

Put the value into the formula

[tex]v=3.10\times0.34[/tex]

[tex]v=1.05\ m/s[/tex]

Hence, The frequency is 0.34 Hz and the wave speed is 1.05 m/s.

The frequency of the wave is 0.34 Hz and the wave speed is 1.054 m/s.

The distance between two successive minima in a transverse wave is the same as the wavelength (λ). Given that the question provided this distance as 3.10 m, we can use this as the wavelength. On the other hand, the frequency (f) of the wave is the number of wave cycles that pass a point per unit time. The information from the question gives us five crests (complete cycles) over 14.7 seconds. Thus, we can calculate the frequency by dividing the total cycles by the total time, i.e., f = 5 cycles / 14.7 seconds = 0.34 Hz.

The speed of the wave (v) can be obtained from the equation v = λf. Substituting the values we have, v = (3.10 m) * (0.34 Hz) = 1.054 m/s. Therefore, the frequency of the wave is 0.34 Hz, and the wave speed is 1.054 m/s.

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