A rock is thrown with a velocity v0, at an angle of α0 from the horizontal, from the roof of a building of height h. Ignore air resistance. Calculate the speed of the rock just before it strikes the ground, and show that this speed is independent of α0.

Explanation:

The temperature of the solar nebula was decreasing as it moved away from its center. Therefore, only heavy elements could condense in the inner solar system and terrestrial planets could not form with light elements, such as gases. In the outer solar system, the Jovian planets formed mostly with gases, since temperatures were too low to allow rocky compositions.

Answer:

Explanation:

Given

Power drawn by toaster [tex]P_1=1840\ W[/tex]

Power drawn by Electric fan [tex]P_2=1460\ W[/tex]

Power drawn by lamp [tex]P_3=50\ W[/tex]

Voltage applied [tex]V=120\ V[/tex]

If appliances are applied in parallel then Voltage applied is same

Power is given by [tex]P=\frac{V^2}{R}[/tex]

Resistance of toaster [tex]R_1=\frac{120^2}{1840}=7.82\ \Omega [/tex]

Resistance of electric Fan [tex]R_2=\frac{120^2}{1460}=9.86\ \Omega [/tex]

Resistance of toaster [tex]R_3=\frac{120^2}{50}=288\ \Omega [/tex]

Current is given by

[tex]I=\frac{V}{R}[/tex]

[tex]I_1=\frac{120}{7.82}=15.34\ A[/tex]

[tex]I_2=\frac{120}{9.86}=12.17\ A[/tex]

[tex]I_3=\frac{120}{288}=0.416\ A[/tex]

Answer:

Explanation:

power of toaster, P1 = 1840 W

Power of electric frying pan, P2 = 1460 W

Power of lamp, P3 = 50 W

As they all are connected in parallel, so the voltage is same for all.

Let the current in toaster is i1.

P1 = V x i1

1840 = 120 x i1

i1 = 15.33 A

Let the current in frying pan is i2.

P2 = V x i2

1460 = 120 x i1

i1 = 12.17 A

Let the current in lamp is i3.

P3 = V x i3

50 = 120 x i3

i3 = 0.42 A

Answer: a) 2.5 * 10^14, b) t = 1.2*10^-8 s, c) F = 2.2775 * 10^-15 N

Explanation: Since it starts from rest, initial velocity = 0, final velocity (v) = 3*10^6 m/s, distance covered (s) = 1.80cm = 1.80/100 = 0.018m

Since the force on the electron is constant, it acceleration will be constant too thus making newton's laws of motion valid.

Question a)

To get the acceleration, we use the formulae that

v² = u² + 2as

But u = 0

v² = 2as

(3*10^6)² = 2*a*(0.018)

9* 10^12 = 0.036*a

a = 9 * 10^12 / 0.036

a = 250 * 10^12

a = 2.5 * 10^14 m/s².

Question b)

To get the time, we use

v = u + at

But u = 0

v = at

3*10^6 = 2.5 * 10^14 * t

t = 3*10^6 / 2.5*10^14

t = 1.2*10^-8 s

Question c)

To get the force, we use the formulae below

F = ma

F = 9.11*10^-31 * 2.5 * 10^14

F = 22.775 * 10^-17

F = 2.2775 * 10^-15 N

a) The acceleration of the electron as it travels in a straight line to the grid is 2.5 × 10¹⁴m/s².

b) The time taken for the electron to reach the grid is 1.2 × 10⁻⁸ seconds.

c) The net force that is accelerating the electron is 2.28 × 10⁻¹⁶ Newton.

Given the data in the question

Acceleration; [tex]a = \ ?[/tex]

Time taken; [tex]t = \ ?[/tex]

Net force; [tex]F = \ ?[/tex]

a)

To determine the acceleration of the electron, we use the third equation of motion:

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

Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance traveled.

We substitute our given values into the equation

[tex](3.00 *10^6 m/s)^2 = 0^2 + [ 2 * a * 0.018m ]\\\\9.0 *10^{12} m^2/s^2 = a * 0.036m\\\\a = \frac{9.0 *10^{12} m^2/s^2}{0.036m} \\\\a = 2.5 * 10^{14}m/s^2[/tex]

Therefore, the acceleration of the electron as it travels in a straight line to the grid is 2.5 × 10¹⁴m/s².

b)

To the the time taken to reach the grid, we use the first equation of motion:

[tex]v = u + at[/tex]

Where v is the final velocity, u is initial velocity, a is the acceleration and t is the time taken

We substitute our values into the equation

[tex]3.00 * 10^6 m/s = 0 + [ ( 2.5*10^{14}m/s^2) * t\\\\3.00 * 10^6 m/s = ( 2.5*10^{14}m/s^2) * t\\\\t = \frac{3.00 * 10^6 m/s}{2.5*10^{14}m/s^2} \\\\t = 1.2 * 10^{-8}s[/tex]

Therefore, time taken for the electron to reach the grid is 1.2 × 10⁻⁸ seconds.

c)

To determine the net force of the electron, we use the expression from newton second law of motion:

[tex]F = m * a[/tex]

Where m is mass and a is the acceleration

We substitute our values into the equation

[tex]F = ( 9.11 * 10^{-31} kg ) * ( 2.5 * 10^{14}m/s^2)\\\\F = 2.28 * 10^{-16} kg.m/s^2\\\\F = 2.28 * 10^{-16}N[/tex]

Therefore, the net force that is accelerating the electron is 2.28 × 10⁻¹⁶ Newton.

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Answer:

As potential energy depends on height

P.E=mgh

m and g remains same

just before release it has maximum height,thus option D it has maximum potential energy.

The ball has the most potential energy when it is at the top of the building D. before it is released.

Potential energy is stored energy that an object possesses due to its position, state, or condition. It can be gravitational, elastic, chemical, or electrical. The energy is converted to kinetic energy when the object moves or undergoes a change in its physical or chemical state.

The ball has the most potential energy when it is at the top of the building before it is released. Potential energy depends on an object's height and mass. When the ball is at the top of the building, it has the maximum height and therefore the most potential energy. As the ball falls, its potential energy decreases and is converted into kinetic energy.

Answer:

Time: 0.35 s

Explanation:

The position vector of the particle is

[tex]r=(2.7m/s^2)t^2 i+(5.1 m/s^3)t^3j[/tex]

where the first term is the x-component and the 2nd term is the y-component.

The particle's velocity vector is given by the derivative of the position vector, so:

[tex]v=r'(t)=(2.7\cdot 2)t i + (5.1\cdot 3)t^2 j=5.4t i+15.3t^2j[/tex]

The particle's velocity has an angle with the x-axis of 45 degrees when the x and the y component have same magnitude. Therefore:

[tex]5.4t=15.3t^2\\5.4=15.3t\\t=\frac{5.4}{15.3}=0.35 s[/tex]

The time t at which the angle between the particle's velocity and the x-axis is 45°, given the vectors x = (2.7 m/s^2)t^2 i and y = (5.1 m/s^3)t^3 j, is found to be 0.35 seconds.

To solve this problem, first we need to find the velocity of the particle in both the x and y directions. The velocity in the x-direction, vx, is just the derivative of x with respect to time, dx/dt = (2.7 m/s2)(2t) = 5.4t m/s. The velocity in the y-direction, vy, is the derivative of y with respect to time, dy/dt = (5.1 m/s3)(3t2) = 15.3t2 m/s.

Now, the angle θ between the velocity and the x-axis can be found by taking the tangent inverse of the ratio of vy to vx: θ = tan-1(vy/vx). We want this angle to be 45°, so we set this equation equal to 45° and solve for t. θ = 45° = tan-1((15.3t2 m/s) / (5.4t m/s)) => tan(45) = 15.3t2 / 5.4t => 1 = 15.3t / 5.4 => t = 5.4 / 15.3 s= 0.35s.

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To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

[tex]\lambda = \frac{v}{f}[/tex]

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

[tex]\lambda = \frac{340}{500}[/tex]

\lambda = 0.68m

The distance to move one speaker is half this

[tex]\lambda/2 = 0.34m[/tex]

Therefore the minimum distance will be 0.34m

The minimum distance that one of the speakers should be moved back away from the listener to produce destructive interference at the listener's ear is; Δ = 0.34 m

The formula for wavelength here as it relates to speed and frequency is given as;

λ = v/f

Where;

λ is wavelength

v is speed

f is frequency

We are given;

Frequency; f = 500 Hz

Speed of sound; v = 340 m/s

Thus;

λ = 340/500

λ = 0.68 m

Now, we are told that the line separating them creates a constructive interference at the listeners ear. Thus;

To calculate the minimum distance would one of the speakers be moved back away from the listener to produce destructive interference at the listener's ear we will use the formula;

formula for destructive path length is;

Δ = (m + ½)λ

And m here is zero

Thus;

Δ = ½λ

Δ = 0.68/2

Δ = 0.34 m

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a) Not possible

b) Yes, it's possible (see graph in attachment)

Explanation:

a)

The average velocity of a body is defined as the ratio between the displacement and the time elapsed:

[tex]v=\frac{\Delta x}{\Delta t}[/tex]

where

[tex]\Delta x[/tex] is the displacement

[tex]\Delta t[/tex] is the time elapsed

In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,

[tex]\Delta x = 0[/tex]

And therefore as a consequence,

[tex]v=0[/tex]

which means that the average velocity is zero.

B)

Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking  about average velocity, but we are talking about (instantaneous) velocity.

On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.

In all of this, we notice that the total displacement of the object is zero:

[tex]\Delta x = 0[/tex]

However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.

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Zero displacement and nonzero average velocity can occur on an x-t graph, but it is not possible to have zero displacement and nonzero velocity simultaneously.

Zero displacement and nonzero average velocity can occur when an object moves back and forth over a certain distance and returns to its original position, but with a change in time. For example, if you consider a person running in a circular track, after each lap the person returns to the starting point, resulting in zero displacement. However, if the person completes the laps in different time intervals, their average velocity will be nonzero.

On the other hand, it is not possible to have zero displacement and nonzero velocity simultaneously. Velocity is defined as the rate of change of displacement with respect to time. If the displacement is zero, then the velocity must also be zero

In an x-t graph, zero displacement would be represented by the horizontal line at the x-axis, or the line parallel to the x-axis. Nonzero average velocity would be represented by a sloped line that is above or below the x-axis, indicating movement in either the positive or negative direction.

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Answer:

Explanation:

Given

F_net(x)  = (3 x²) N

m v dv / dt = 3 x²

m v dv = 3 x² dx

integrating on both sides and taking limit from x = 2 to 4 m

m v² / 2 - 0 = 3 x³ / 3

mv² / 2 = 4³ - 2³

mv² / 2 = 64 -  8

3 x v² /2 = 56

v = 6.11 m / s

linear momentum

= m v

= 3 x 6.11

= 18.33 kgm/s

Answer:

[tex]p=m.v=37\ kg.m.s^{-1}[/tex]

Explanation:

Given:

WE know from the Newton's second law:

[tex]\frac{d}{dt} (p)=F[/tex]

where:

[tex]p=[/tex] momentum

[tex]F=[/tex] force

[tex]t=[/tex] times

Now put the values

[tex]\frac{d}{dt} (mv)=3\cdot x^2[/tex]

[tex]m.\frac{d}{dt} (v)=4\times x^2[/tex]

Now integrate both sides from final limit to initial:

[tex]m.v=\int\limits^4_2 {3x^2} \, dx[/tex]

[tex]m.v=[\frac{3x^3}{3} ]^4_2[/tex]

[tex]m.v=4^3-2^3[/tex]

[tex]p=m.v=56\ kg.m.s^{-1}[/tex]

Answer:

(a) 69.52 x 10⁻⁹ C

(b) 6.25 N/C

Explanation:

(a) The electric field (E) due to a point charge is directly proportional to the magnitude of the point charge (Q) and inversely proportional to the square of the distance (r) between the point charge and the point where the electric field is. This can be represented as follows;

E = k Q / r²         ---------------------------(i)

Where;

k = constant of proportionality called electric constant = 8.99 x 10⁹Nm²/C²

From the question, the following are given;

E = 10000N/C

r = 0.250m

Substitute these values into equation (i) as follows;

=> 10000 = 8.99 x 10⁹ x Q / 0.250²

=> 10000 = 8.99 x 10⁹ x Q / 0.0625

=> 10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000 / (143.84 x 10⁹)

Q = 69.52 x 10⁻⁹ C

Therefore, the magnitude of the point charge is 69.52 x 10⁻⁹ C.

(b) By the same token, to calculate the magnitude of the electric field at 10.0m, substitute the values of Q = 69.52 x 10⁻⁹ C, k = 8.99 x 10⁹ and r = 10.0m into equation (i) as follows;

=> E = 8.99 x 10⁹ x 69.52 x 10⁻⁹ / 10.0²

=> E = 8.99 x 69.52 / 100.0

=> E = 6.25 N/C

Therefore, the electric field at 10.0m is as large as 6.25 N/C

Final answer:

The minimum separation r_min between two electrons moving towards each other and stopping due to repulsion is found by conserving mechanical energy, transforming the initial kinetic energy into electrostatic potential energy at the point of closest approach.

Explanation:

The minimum separation r_min that two electrons reach when moving directly toward each other and slowing down due to electric repulsion is determined by the conservation of energy. The total mechanical energy (kinetic plus potential) of the system must be conserved. At the maximum approach, both electrons will momentarily be at rest before repelling each other again, so all their kinetic energy is converted into electrical potential energy. The initial kinetic energy of both electrons, which is due to their motion, is transformed into electrostatic potential energy at the point of minimum separation.

The concept involves the calculation of this potential energy and setting it equal to the initial kinetic energy to find r_min. One might use the relationship between kinetic energy (1/2)m(v^2), electrostatic potential energy (k(q^2)/r), and the conservation of energy to solve for the unknown r_min. It's important to note that in such a scenario, relativistic effects are not considered as long as the velocity of the electrons is significantly less than the speed of light (c).

Answer:

c. Friction will be in the opposite direction of the block's motion because molecular interactions between the block and the floor will deform the block in the opposite direction of motion.

Explanation:

When a block is put on the floor , there is interaction at molecular level between the molecules of block and floor . This mutual interaction

( attraction ) pulls them together. When the body is pushed forward, the distance between molecules is increased . Due to mutual attraction , block is deformed at the interface in the opposite direction of motion . This strain causes restoring stress which is also called friction . This elastic  stress creates force on the block in backward direction and force on the floor in forward direction .

Final answer:

The correct statement is C. Friction acts in the opposite direction of the block's motion due to molecular interactions between the block and the floor, and it always opposes the motion to slow down or stop the block.

Explanation:

The correct answer to the question, "Which statement best explains the direction of the force that friction applies on the moving block is that friction will be in the opposite direction of the block's motion. This occurs because friction always opposes the direction of motion to prevent or slow down the motion of the block. The encounter between the block and the floor's surface creates molecular interactions that resist the block's movement, and thus, the force of friction acts in the opposite direction to the motion of the block.

Moreover, it's essential to clarify that while friction does generate thermal energy due to the interactions at the interface between the block and the floor, this thermal energy does not add kinetic energy to the block or convert kinetic energy into potential energy in the context of motion. Instead, it's a byproduct of the frictional force's work against the block's motion.

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

[tex]\omega = 30 rad/s[/tex]

The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

[tex]\theta = (\frac{\omega_0 + \omega}{2})t[/tex]

where:

[tex]\theta[/tex] is the angular displacement of the cd during the time interval t

[tex]\omega_0[/tex] is the initial angular velocity of the cd

[tex]\omega[/tex] is the final angular velocity

Here we have:

t = 1.36 s

[tex]\omega_0 = 0[/tex] (assuming the cd starts from rest)

Therefore, the angular displacement of the cd during this time is:

[tex]\theta=(\frac{0+30}{2})(1.36)=20.4 rad[/tex]

And since [tex]1 rev = 2 \pi rad[/tex], we can convert into number of revolutions completed:

[tex]\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev[/tex]

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Answer:

 ΔP = 97.93 Pa ,     T₂-T₁ = 71.5° C

Explanation:

For this problem let's use Bernoulli's relationship, as point 1 we will take the plane and as point 2 the air

            P₁ + ½ ρ g v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

As the whole process occurs at the same height y₁ = y₂ = 11 km. We will consider that the air goes in the opposite direction to the plane

            P₂ –P₁ = ½ ρ (v₁² - v₂²)

Let's reduce the magnitudes to the SI system

           v₁ = 935 km / h (1000 m / 1 km) (1h / 3600s) = 259.72 m / s

           v₂ = 280 m / s

The density of air at 11000 m is

           Rho = 0.3629 kg / m

           P₂-P₁ = ½ 0.3629 (259.72 + 280)

           ΔP = 97.93 Pa

The variation of the temperature with the altitude is 0.65 per 100 m

            T₂ –T₁ = (0.65 / 100) 1000

             T₂-T₁ = 71.5° C

The temperature has decreased this value

Explanation:

For a pipe with one end open ,we have the formula for fundamental frequency as

[tex]f_1= \frac{v}{4L}[/tex]

v= velocity of sound in air =340 m/s

L= length of pipe

hence, [tex]L= \frac{v}{4f_1}[/tex]

given f_1 = 440 Hz

Substituting , L = 340/(4×440) = 0.193 m

a)  Let the bottle be filled to height , h.

Given that height of bottle ,H = 25 cm = 0.25 m

Also we found out that length of pipe L = 0.193 m

So, we have   h = H - L

= 0.25 - 0.193 = 0.057 m = 5.7 cm.

Answer:

(a). The bottle filled at the height is 5.5 cm

(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.

Explanation:

Given that,

Frequency = 440.0 Hz

Height of bottle = 25.0 cm

Suppose,

Let the bottle be filled to height h.

For a pipe with one end open,

We need to calculate the length of the pipe

Using formula of fundamental frequency

[tex]F=\dfrac{v}{4L}[/tex]

[tex]L=\dfrac{v}{4f}[/tex]

Where, L = length

v = speed of sound

Put the value into the formula

[tex]L=\dfrac{343}{4\times440}[/tex]

[tex]L=0.195\ m[/tex]

(a). We need to calculate the height

Using formula of height

[tex]h=H-l[/tex]

Where, H = height of bottle

l = length of pipe

Put the value into the formula

[tex]h=25.0\times10^{-2}-0.195[/tex]

[tex]h=0.055\ m[/tex]

[tex]h=5.5\ cm[/tex]

(b). We need to calculate the frequency of the nest higher harmonic of this bottle

Using formula of frequency

[tex]f_{next}=nf_{1}[/tex]

Where, [tex]f_{1}[/tex]=fundamental frequency

Put the value into the formula

[tex]f_{2}=2\times440[/tex]

[tex]f_{2}=880\ Hz[/tex]

Hence, (a). The bottle filled at the height is 5.5 cm

(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.

Answer:

The speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s

Explanation:

qV = 0.5Mv²

where;

V is the potential difference, measured in Volts

q is the charge of the electron in Coulomb's = 1.6 × 10⁻¹⁹ C

Mass is the mass of the electron in kg =  9 × 10⁻³¹ kg

v is the velocity of the electron in m/s

Applying coulomb's law, we determine the Potential difference V

V = kq/r

V = (8.99X10⁹ * 1.6 × 10⁻¹⁹)/(1X10⁻⁹)

V = 14.384 X 10¹⁹ V

The speed of the electron can be determined as follows;

v² = (2qV)/M

v = √(2qV)/M)

v = √(2*1.6 × 10⁻¹⁹* 14.384 X 10¹⁹)/(9 × 10⁻³¹)

v = √(5.1143 X 10³¹) = 7 X 10¹⁵m/s

Therefore, the speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s

Answer:

The time taken for the monkey to reach the coconut, t is = H/v₀

Explanation:

Let the coordinate of the boy's hand be y = 0. The height of the tree above the boy's hand be H.

Coordinate of where the monkey meets coconut = y

Using the equations of motion,

For the monkey, initial velocity = 0m/s, time to reach coconut = t secs and the height at which coconut is reached = H-y

For the coconut, g = -10 m/s², initial velocity = v₀, time to reach monkey = t secs and height at which coconut meets monkey = y

For monkey, H - y = ut + 0.5gt², but u = 0,

H - y = 0.5gt²..... eqn 1

For coconut, y = v₀t - 0.5gt² ....... eqn 2

Substituting for y in eqn 1

H - y = H - (v₀t - 0.5gt²) = 0.5gt²

At the point where the monkey meets coconut, t=t

H - v₀t + 0.5gt² = 0.5gt²

v₀t = H

t = H/v₀

Solved!

Answer: The question is incomplete as some other details are missing. Here is the complete question ; A helium atom falls in a vacuum . The mass of a helium atom is 6.64 x 10-27 kg . (a) From what height must it fall so that its translational kinetic energy at the bottom equals the average translational kinetic energy of a helium molecule at 300 K? (b) At what temperature would the average speed of helium atoms equal the escape speed from the Earth, 1.12 x 104 m/s.

a) Height = 95.336Km

b) Temperature = 20118.87K

Explanation:

The detailed steps and calculations is shown in the attached file.

The helium atom must fall from 95.336 km to achieve average translational kinetic energy at 300 K.

It is the required energy to accelerate the particle or object from the rest to achieve the given velocity.

The height can be calculated by the formula,

[tex]h = \dfrac 23 \times \dfrac {kt}{mg}[/tex]

Where

h - height =?

k - constant =[tex]1.38\times 10^{-23}[/tex] J/k

t - temperature = 300 k

m - mass

g - gravitational acceleration = 9.8 m/s^2

Put the values in the formula,

[tex]h = \dfrac 23 \times \dfrac {1.38\times 10^{-23}\times 300 }{6.64 \times 10^{-27}\times 9.8}\\\\h = 95335.4 {\rm\ m \ \ or}\\\\h = 95.336\rm \ km[/tex]

Therefore, the helium atom must fall from 95.336 km to achieve average translational kinetic energy at 300 K.

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Answer:

Explanation:

Given

Spring constant of spring A is [tex]k_A=100\ N/m[/tex]

Spring constant of spring B is [tex]k_B=200\ N/m[/tex]

If displacement in both the springs is [tex]x=0.2\ m[/tex]

Potential Energy stored in the spring is given by

[tex]U=\frac{1}{2}kx^2[/tex]

where k=spring constant

x=compression or extension

[tex]U_A=\frac{1}{2}\times 100\times (0.2)^2----1[/tex]

[tex]U_B=\frac{1}{2}\times 200\times (0.2)^2----2[/tex]

Divide 1 and 2

[tex]\frac{U_B}{U_A}=\frac{200}{100}=2[/tex]

[tex]U_A=\frac{U_B}{2}[/tex]

So Potential Energy Stored in Spring A is half of Spring B

The spring potential energy of spring A is half that of spring B, if the spring constant of spring A is 100 N/m and that of B is 200 N/m.

Spring constant of spring A, k₁ = 100 N/m

Spring constant of spring B, k₂= 200 N/m

Displacement of both the spring, x = 0.20 m

The potential energy stored in a spring is given by the equation U = 0.5kx²

For spring A, the potential energy is U₁ = 0.5 × 100×(0.20)² = 2 J.

For spring B, the potential energy is U₂ = 0.5 × 200×(0.20)² = 4 J

Therefore, the potential energy stored in spring A is half that of spring B.

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Answer:

v= 14.22 m/s

Explanation:

Given that

u = 12.1 m/s

a=0.35   m/s².

t= 6.07 s

We know v = u +at

v=final velocity

u=initial velocity

t=time

a=acceleration

Now by putting the values in the above equation

v= 12.1 + 0.35 x 6.07 m/s

v= 14.22 m/s

Therefore the final velocity will be 14.22 m/s.

The final velocity of a bicycle racer with an initial velocity of 12.1 m/s and an acceleration of 0.350 m/s² over 6.07 s is approximately 14.22 m/s.

To compute the final velocity of a bicycle racer sprinting towards the end of a race, we can employ the kinematic equation for velocity under constant acceleration:

v = u + at

Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given that the racer has an initial velocity u of 12.1 m/s, accelerates at a rate a of 0.350 m/s² for a duration t of 6.07 s, we substitute these values into the equation:

v = 12.1 m/s + (0.350 m/s² × 6.07 s)

Calculation:

v = 12.1 m/s + (0.350 m/s² × 6.07 s)
v = 12.1 m/s + 2.1245 m/s
v = 14.2245 m/s

The final velocity of the racer is approximately 14.22 m/s.

Answer:

The new force is 10 times the force F

Explanation:

Electric force between charged particles q1 and q2 at distance d is:

[tex]F=k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex] (1)

A new force between two different particles at a different distance is:

[tex]F'=k\frac{\mid q_{1}'q_{2}'\mid}{d'^{2}}=k\frac{\mid 5q_{1}\frac{q_{2}}{2}\mid}{(\frac{d}{2})^{2}}=\frac{5}{\frac{2}{4}}k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex]

[tex]F'=10k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex]

Note that on the right side of the equation the term [tex]k\frac{\mid q_{1}q_{2}\mid}{d^{2}}=F [/tex] on (1), so:

[tex]F'=10F [/tex]

The acceleration of the car is 1.67 m/s². It takes approximately 11.23 seconds for the car to travel the length of the ramp. The traffic travels approximately 224.6 meters while the car is moving the length of the ramp.

(a) To find the acceleration of the car, we can use the equation:

v² = u² + 2as

Where v is the final velocity (20 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the displacement (120 m).

Plugging in the values, we have:

20² = 0² + 2a(120)

Simplifying the equation, we get:

a = (20²)/(2 * 120)

Therefore, the acceleration of the car is approximately 1.67 m/s².

(b) Time can be calculated using the equation:

s = ut + 0.5at²

Where u is the initial velocity (0 m/s), a is the acceleration (1.67 m/s²), and s is the displacement (120 m).

Plugging in the values, we have:

120 = 0 + 0.5(1.67)t²

Simplifying the equation, we get:

t² = (120)/(0.5 * 1.67)

t = √(120/0.835)

t ≈ 11.23 s

Therefore, it takes approximately 11.23 seconds for the car to travel the length of the ramp.

(c) The distance the traffic travels while the car is moving the length of the ramp can be calculated using the equation:

d = v * t

Where v is the velocity of the traffic (20 m/s) and t is the time taken (11.23 s).

Plugging in the values, we have:

d = 20 * 11.23

Therefore, the traffic travels approximately 224.6 meters while the car is moving the length of the ramp.

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The final answers are [tex](a)~{a ~is~ 1.67 \, \text{m/s}^2} \], ~(b)~{t ~is~ 12s} ~and ~~(c)~d ~is~ 240 \, m[/tex].

To solve the problem, we will use the kinematic equations for uniformly accelerated motion. The car starts from rest, which means its initial velocity (u) is 0 m/s, and it reaches a final velocity (v) of 20 m/s over a distance (s) of 120 m.

a) To find the acceleration (a) of the car, we can use the third kinematic equation that relates initial velocity, final velocity, acceleration, and distance:

[tex]\[ v^2 = u^2 + 2as \]\\ Plugging in the known values, we get: \[ (20 \, \text{m/s})^2 = (0 \, \text{m/s})^2 + 2a(120 \, \text{m}) \] \[ 400 \, \text{m}^2/\text{s}^2 = 240a \, \text{m} \] \[ a = \frac{400 \, \text{m}^2/\text{s}^2}{240 \, \text{m}} \] \[ a = \frac{100}{60} \, \text{m/s}^2 \] \[ a = \frac{5}{3} \, \text{m/s}^2 \] \[ a \approx 1.67 \, \text{m/s}^2 \][/tex]

b) To find the time (t) it takes for the car to travel the length of the ramp, we can use the first kinematic equation:

[tex]\[ v = u + at \]\\ Since the initial velocity (u) is 0 m/s, the equation simplifies to: \[ 20 \, \text{m/s} = 0 + \left(\frac{5}{3} \, \text{m/s}^2\right)t \] \[ t = \frac{20 \, \text{m/s}}{\frac{5}{3} \, \text{m/s}^2} \] \[ t = \frac{20}{1} \cdot \frac{3}{5} \, \text{s} \] \[ t = 12 \, \text{s} \][/tex]

c) While the car is accelerating along the ramp, the traffic on the freeway is moving at a constant speed of 20 m/s. The distance (d) the traffic travels in the time (t) the car takes to travel the ramp is given by:

[tex]\[ d = vt \] \[ d = (20 \, \text{m/s})(12 \, \text{s}) \] \[ d = 240 \, \text{m} \][/tex]

So, the traffic on the freeway travels 240 meters while the car is moving the length of the ramp.

The final answers are:

[tex]\[ \boxed{a \approx 1.67 \, \text{m/s}^2} \] \[ \boxed{t = 12 \, \text{s}} \] \[ \boxed{d = 240 \, \text{m}} \][/tex]

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

[tex]dtan13^0 = 0.231d[/tex]

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

[tex]dtan4^0 = 0.07d [/tex]

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

To find the distance to the 200-foot tall monument from a specified point, we use tangent trigonometric functions with the given angles of elevation and depression. The total horizontal distance is calculated by separately finding the distances to the top and the bottom of the monument and combining them.

The question involves solving a problem using trigonometry to find the distance to a monument given the angles of elevation and depression from a certain point. Since the height of the monument is given as 200 feet, and we have the angles of elevation (13°) to the top of the monument and the angle of depression (4°) to the bottom, we can use trigonometric functions to calculate the distances to the top and the bottom of the monument separately and then sum them to find the total distance from the observer to the monument.

To find the horizontal distances (D) from the observer to the top and bottom of the monument, we can use the tangent function (tan) from trigonometry. The tangent of the angle of elevation (13°) is equal to the height of the monument divided by the distance to the top (Dtop), and the tangent of the angle of depression (4°) is equal to the height from the window to the base of the monument divided by the distance to the bottom (Dbottom).

Assuming the window is at the same height as the base of the monument (which is not given explicitly, but implied by the angle of depression), we get two equations:

We can then solve for Dtop and Dbottom using these equations and add the two distances to find the total horizontal distance to the monument.

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Answer:

Explanation:

Maximum vertical distance or height = h = 35.4 m

let's consider the initial speed at the top is zero.

As the roller coaster is coming from top to bottom there is the conversion of gravitational potential energy into kinetic energy. So we will consider the law of conservation of energy.

As in this case,

Loss in potential energy = Gain in Kinetic energy

mgh = 1/2mv²

mass will cancel out will mass.

gh = 1/2 v²

v = √2gh

v = √2×9.8×35.4

v =√693.84

v = 26.34 m/s

The rollar coaster will have the maximum speed of 26.34 m/s when it reaches the bottom if we ignore the frictional forces.

Answer:

26.34 m/s.

Explanation:

Given:

h = 5.4 m

g = 9.81 m/s^2

Change in Potential energy = change in Kinetic energy

mgh = 1/2mv²

gh = 1/2 v²

v = √2gh

= √2×9.8×35.4

=√693.84

v = 26.34 m/s.

Answer:

V = 31.62 m V

Explanation:

Given,

Voltage across thick membrane = ?

Thickness of the membrane, d = 5.5 n m

Electric field = 5.75 M V/m

we know

[tex]E = \dfrac{V}{d}[/tex]

   V = E d

   V = 5.75 x 10⁶ x 5.5 x 10⁻⁹

   V = 31.62 x 10⁻³ V

   V = 31.62 m V

Hence, The Voltage across the membrane is equal to 31.62 m V.

The voltage across the membrane is dependent on the electric field and thickness of the membrane. The voltage across the membrane is 0.0316 V.

The voltage is defined as the difference in electric potential between two points.

Given that the electric field E is 5.75 mv/m and the thickness of the membrane d is 5.5 nm.

The voltage across the membrane is calculated as given below.

[tex]E = \dfrac {V}{d}[/tex]

[tex]V = Ed[/tex]

[tex]V = 5.75 \times 10^6 \times 5.5 \times 10^{-9}[/tex]

[tex]V = 0.0316 \;\rm V[/tex]

Hence we can conclude that the voltage across the membrane is 0.0316 V.

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Answer:

a) ΔV = 0.1875 V , b) v = 9.73 m / s

Explanation:

For this exercise we can use Necton's second law, as the speed is constant the forces on the driver are equal

               [tex]F_{E} - F_{B}[/tex] = 0

               [tex]F_{E}[/tex] = F_{B}

               q E = q v B

               E = v B

The electrical force induced in the conductor is

                ΔV = E l

                ΔV = v B l

Let's calculate

               ΔV = 2.5 0.75 0.10

               ΔV = 0.1875 V

b) If ΔV = 0.73 what speed should it have

             v = DV / B l

             v = 0.73 / 0.75 0.1

             v = 9.73 m / s

Explanation:

When the skydiver accelerates in the downward direction then tends to gain speed with each second. More is the resistance in air more will be the increase accompanied by the skydiver.

As a result, a point will come where air resistance force is balanced by gravitational force. Hence, the skydiver will attain terminal velocity.

So, air resistance for 1 [tex]m/s^{2}[/tex] = 0.45 [tex]m/s^{2}[/tex]

Air resistance for v [tex]m/s^{2}[/tex] = 0.45 v[tex]m/s^{2}[/tex]

As acceleration = change in velocity w.r.t time

                 a = [tex]\frac{dv}{dt}[/tex] = 0.45

            [tex]\frac{dv}{V}[/tex] = 0.45t

Now, we will integrate both the sides as follows.

            ln V = 0.45t

                V = [tex]e^{0.45t}[/tex]

Since,         [tex]F_{a} = F_{g}[/tex]   (in the given case)

so,                    ma = mg

On cancelling the common terms the equation will be as follows.

                     0.45 v = 9.8 [tex]m/s^{2}[/tex]

                         v = 21.77 [tex]m/s^{2}[/tex]

Thus, we can conclude that the skydivers terminal speed is 21.77 [tex]m/s^{2}[/tex].

The force between the two charges, according to Coulomb's law, is approximately 1.92 Newtons. Note that the force is attractive since one charge is positive and the other is negative.

The question refers to Coulomb's law, which is fundamental in Physics. The law describes the force between two electric charges. We apply the formula |F|=K|QQ′|/d^2 where the charges are Q1=-13.5nC (converted to C by multiplying by 10^-9) and Q2=35.5nC, the distance d=1.735mm (converted to meters by multiplying by 10^-3), and K=1/4πϵ0≈8.99x10^9 N.m²/C². Inserting these values into the formula we get |F| ≈ 1.92 N. The magnitude of the force is therefore approximately 1.92 Newtons. It's important to note that since one charge is positive and the other negative, the force will be attractive, not repulsive.

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Answer:

E= 50.1*10³ N/C

Explanation:

Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:

[tex]vf^{2} -vo^{2} = 2*a*x[/tex]

We know that v₀ = 0 (it starts from rest),  that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:

[tex]a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2[/tex]

According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.

Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:

F = q*E (1)

In this case, q=e= 1.6*10⁻19 C

But this force, can be expressed in this way, according Newton's 2nd Law:

F = m*a (2) ,

where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.

From (1) and (2), we can solve for E, as  follows:

[tex]E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C[/tex]

⇒ E = 50.1*10³ N/C

Answer:

T

Explanation:

I learned this in 5th grade

It is true that an electron moves to a higher energy level in an atom after absorbing a photon of a specific energy.

The transfer of a bound electron to a more energetic, but still bound, state is known as electron excitation.

This can be accomplished through photoexcitation, in which the electron absorbs a photon and absorbs all of its energy, or through electrical excitation, in which the electron receives energy from another, more energetic electron.

An electron can become excited if it receives additional energy, such as when it absorbs a photon, or light packet, or collides with a nearby atom or particle.

Electrons can absorb energy (discretely) to move from lower to higher energy states.

When they reach higher energy states, they become unstable and emit photons to return to their ground state (of specific energy).

Thus, the given statement is true.

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Answer:

  q= 20 W                

Explanation:

Given that

Initial temperature ,T₁ = 2 ºC

Heat gains ,Q = 18 kJ

The final temperature ,T₂ = 20 ºC

time ,t= 15 min

We know that

1 min = 60 s

t= 15 x 60 = 900 s

The average rate of heat transfer is give as

[tex]q=\dfrac{Q}{t}[/tex]

[tex]q=\dfrac{18}{900}\ kW[/tex]

q=0.02 kW

q= 20 W

Therefore the rate of heat transfer will be 20 W.

A soda is initially at 2 °C and 15 minutes later, after gaining 18 kJ, its temperature is 20 °C. The average rate of heat transfer is 20 W.

Heat transfer describes the flow of heat (thermal energy) due to temperature differences and the subsequent temperature distribution and changes.

A soda is initially at 2 °C and 15 minutes later, after gaining 18 kJ, its temperature is 20 °C.

First, we will convert 15 min to seconds using the conversion factor 1 min = 60 s.

15 min × 60 s/1 min  = 900 s

Then, we can calculate the average rate of heat transfer using the following expression.

q = Q/t = 18 × 10³ J / 900 s = 20 W

where,

A soda is initially at 2 °C and 15 minutes later, after gaining 18 kJ, its temperature is 20 °C. The average rate of heat transfer is 20 W.

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