A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same direction as the car) at a constant 5.65 m/s. The driver applies the brakes (ignore reaction time)of the car but can only accelerate at -2.00 m/s2 because the road is wet. How fast are you moving when you hit the rear of the van?
(A) 16.3 m/s
(B) 22 m/sec
(С) 4 m/sec
(D) 0 m/s

Answer:

vo = 0.175m/s

a = -0.040625 m/s^2

Explanation:

To solve this problem, you will need to use the equations for constant acceleration motion:

[tex]x = \frac{1}{2}at^2 +v_ot+x_o \\v_f^2 - v_o^2 = 2(x-x_o)a[/tex]

In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:

1. [tex]x = \frac{1}{2}at^2 +v_ot+x_o\\0.1m = \frac{1}{2}a(8s)^2 +v_o(8s)+0m \\0.1 = 32a + 8v_o[/tex]

2. [tex]v_f^2 - v_o^2 = 2(x-x_o)a\\(-0.15m/s)^2 - v_o^2 = 2(0.1m-0m)a\\0.0225 - v_o^2 = 0.2a\\a = \frac{0.0225 - v_o^2}{0.2} = 0.1125 - 5v_o^2[/tex]

Replacing in the first equation:

[tex]0.1 = 32(0.1125 - 5v_o^2) + 8v_o\\0.1 = 3.6 - 160v_o^2 + 8v_o\\160v_o^2 - 8v_o - 3.5 = 0[/tex]

[tex]v_0 = \frac{-(-8) +- \sqrt{(-8)^2 - 4(160)(-3.5)}}{2(160)} \\ v_o = 0.175 m/s | -0.125 m/s[/tex]

But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:

[tex]a = 0.1125 - 5v_o^2\\a = -0.040625  m/s^2[/tex]

Answer:

W = 0 :The work done on the wall is zero,because the wall is not moving

Explanation:

Work theory

Work is the product of a force applied to a body and the displacement of the body in the direction of this force.

W= F*d Formula (1)

W: Work (Joules) (J)

F: force applied (N)

d=displacement of the body (m)

The work is positive (W+) if the force goes in the same direction of movement.

The work is negative (W-)if the force goes in the opposite direction to the movement

Data

F= 400-N

d= 0

Problem development

We apply formula (1) to calculate the work done on the wall:

W= 400*0

W=0

Answer:

0.98°

Explanation:

The Sun appears to move across various constellation because of the Earth's revolution around it. This apparent motion is essentially same as the actual motion of the Earth. The orbit of Earth around the Sun is almost circular.

Time taken to complete one revolution = 365 days

Degrees traveled in 365 Days = 360°

The motion per day is

[tex]=\frac{360}{365}[/tex]

= 0.98° per day

Final answer:

To find the total increase in the length of each day over 43 centuries, we use the sum of an arithmetic series: S = n/2(2a1 + (n-1)d). With 4300 days in 43 centuries, an initial increase of 0 ms, and a daily increase of 0.002 ms, the total increase is 18471.9 milliseconds.

Explanation:

The length of the day increases because Earth's rotation is gradually slowing, primarily due to the friction of the tides. To calculate the total increase in the length of each day over 43 centuries, given that the length of a day at the end of a century is 1 ms longer than the day at the start of the century, we use the arithmetic progression formula where the total sum (S) is equal to n/2 times the first term (a1) plus the last term (an). In this case, the difference between each day (common difference, d) is constant at 0.002 ms.

The first term (a1) is 0 ms, as there is no increase on the first day of the first century, and the last term (an) after 43 centuries will be 43 ms, since we're told that each century contributes an additional 1 ms to the length of a day. So, for 43 centuries, we have n=4300 days (number of days in 43 centuries), a1=0 ms, an=43 ms, and d=0.002 ms/day.

Using the formula for the sum of an arithmetic series, S = n/2(2a1 + (n-1)d), we get the following:

S = 4300/2 [2(0) + (4300-1)(0.002 ms)]

S = 2150 [0 + 4299(0.002 ms)]

S = 2150 x 8.598 ms

S = 18471.9 ms

Therefore, over 43 centuries, the total increase in the length of each day adds up to 18471.9 milliseconds.

Final answer:

The volume of the larger sphere is eight times greater than that of the smaller sphere, and since their masses are proportional to their volumes for objects of uniform density, we can find the radius of the larger sphere to be twice that of the smaller sphere with a radius of 4.10 cm, resulting in a radius of 8.20 cm for the larger sphere.

Explanation:

The question concerns a comparison of the volumes of two spheres made from the same uniform rock. Since the mass of one sphere is eight times greater than the other, and since mass is proportional to volume for objects with uniform density, we can deduce that the volume of the larger sphere is also eight times greater than the smaller sphere. Given that the volume of a sphere (V) is calculated as V = (4/3)πR³, where R is the radius, we understand that because the volumes are proportional to the cube of the radii, we can calculate the radius of the larger sphere if we know the radius of the smaller one. Specifically, if the radius of the smaller sphere is 4.10 cm and its volume is V, then the volume of the larger sphere is 8V, and its radius is 2 times that of the smaller sphere since 2³ equals 8. Therefore, the radius of the larger sphere is 2 × 4.10 cm = 8.20 cm.

Answer:

Option b

Explanation:

An object is said to fall freely when there is no force acting on the object other than the gravitational force. Thus the acceleration of the object is solely due to gravity and no other acceleration acts on the body.

Also the initial velocity of the body in free fall is zero and hence less than the final velocity.

As the body falls down closer to earth, it experiences more gravitational pull and the velocity increases as it falls down and the moment it touches the ground the period of free fall ends at that instant.

Thus the final velocity of an object in free fall is not zero because the final velocity is the velocity before coming in contact with the ground.

Answer:

Frequency is 0.5 Hz and the wave speed is 10 m/s.

Explanation:

As we know that frequency is defined as the how many times the no of cycles repeat in one second so if the wave is vibrating up and down  twice during 1 second then the frequency in 1 second is

[tex]f=\frac{1}{2} hz\\F=0.5hz[/tex]

Therefore frequency is 0.5 Hz.

Now the distance of wawe in each second is,

[tex]d=20m[/tex]

Now the wave velocity is,

[tex]v=fd[/tex]

Here, f is frequency, d is the distance, v is the wave velocity.

Substitute all the variables

[tex]v=0.5\times 20\\v=10m/s[/tex]

Therefore the wave speed is 10 m/s.

In the given question, the wave's frequency is 2 Hertz (Hz), which means it oscillates twice per second. The wave speed, or distance covered by the wave per second, is identified as 20 m/s. These are fundamental concepts in the Physics of wave mechanics.

In this question, we're dealing with the topics of wave frequency and wave speed. The frequency of a wave relates to how many cycles of the wave occur per unit of time - in this case, the wave is oscillating twice per every second, meaning that its frequency is 2 Hertz (Hz).

The wave speed is the speed at which the wave is travelling. Given that the wave travels a distance of 20 m each second, the speed of the wave is 20 m/s (meters per second).

It's important to note that these two properties are interconnected. In general, the speed of a wave (v) is calculated by multiplying its wavelength (λ) by its frequency (f). So, using the relationship v = λf, the properties of a wave can be determined if the other two are known.

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Answer:

v = 75  m/s

Explanation:

given data:

mass of gazelle is 25 kg

length of slope is 3 km

height of slope is 40 m

drag force is 20 N

jump height is 2 m

By work energy theorem we have

[tex]\frac{1}{2} mv^2 = F*L + mg(h_1 +h_2)[/tex]

[tex]\frac{1}{2}*25*v^2 = 20*3000 + 25 *9.8(40 + 2)[/tex]

solving for v

[tex]v = \sqrt{\frac{70290}{12.5}}[/tex]

v = 75  m/s

Answer:

Explained

Explanation:

Equipotential lines cannot cross each other because. The equipotential at a given point in space is has single value of potential throughout. If two equipotential lines intersect with each other, that would mean two values of potential, that would not mean equipotential line. Hence two equipotential lines cannot cross each other.

Equipotential lines are imaginary lines that connect points with the same electric potential. Two equipotential lines cannot cross each other because that would mean two different points on the same line have the same electric potential, which is not possible.

Equipotential lines are imaginary lines that connect points with the same electric potential. Two equipotential lines cannot cross each other because that would mean two different points on the same line have the same electric potential, which is not possible.

For example, imagine a hill with contour lines representing lines of equal height. If two contour lines were to cross each other, it would mean that two different points on the hill have the same height, which is not possible.

In electricity, the electric potential difference between two points on an equipotential line is zero. If two equipotential lines were to cross, it would create a contradiction, as the potential difference between the two points of intersection would be both zero and non-zero.

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Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let [tex]q_1[/tex] and [tex]q_2[/tex] be the charges such that

[tex]q_1+q_2=4.7[/tex]

and Force between charge particles is given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

[tex]4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}[/tex]

[tex]q_1\cdot q_2=0.522[/tex]

put the value of [tex]q_1[/tex]

[tex]q_2\left ( 4.7-q_2\right )=0.522[/tex]

[tex]q_2^2-4.7q_2+0.522=0[/tex]

[tex]q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}[/tex]

[tex]q_2=0.114 C[/tex]

thus [tex]q_1=4.586 C[/tex]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy [tex]\mathrm{EPE}[/tex].

[tex]\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}[/tex],

where

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

Apply Coulomb's law:

Initial potential energy:

[tex]\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}[/tex].

Final potential energy:

[tex]\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}[/tex].

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

[tex]\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm  -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}[/tex].

Answer:

Answer is c

Explanation:

trust me.

To equalize the charges of two objects, the number of electrons transferred from the first object to the second is calculated by dividing the total charge needed (-4.1 x 10^-6 C) by the charge per electron (-1.602 x 10^-19 C/e).

To balance the charges of the two objects, we must first calculate how many electrons represent the disparity in charge between the two objects. This disparity is of 4.1 µC (-6.1 µC - (-2.0 µC) = -4.1 µC). When we convert this to the fundamental unit of charge (Coulombs), we get -4.1 x 10-6 C.

The charge on an electron is -1.602 x 10-19 C. Therefore, the number of electrons that must be transferred to balance the charges can be found by dividing the total charge needed by the charge per electron. That gives us -4.1 x 10-6 C ÷ -1.602 x 10-19 C/e

The result of this calculation denotes how many electrons must be transferred from the first object to the second to make their charges equal.

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Answer:

t=12600ns

Explanation:

We use the relation between distance and velocity to solve this problem:

d=v*t

d=3.8km=3.8*10^3m

v=3*10^8 m/s^2         light's speed

we solve to find t:

t=d/v=(3.8*10^3)/(3*10^8)=1.26*10^(-5)s=12600*10^(-9)s=12600ns

Answer:

1.27 × 10⁴ ns

Explanation:

Given data

We can find the time (t) that it takes to the light to travel 3.80 km using the following expression.

v = d/t

t = d/v

t = (3.80 × 10³ m)/(3.00 × 10⁸ m/s)

t = 1.27 × 10⁻⁵ s

We know that 1 s = 10⁹ ns. Then,

1.27 × 10⁻⁵ s × (10⁹ ns/1s) = 1.27 × 10⁴ ns

Answer:

Option A decreases with increase in altitude

Explanation:

This can be explained as the value of gravitational acceleration, 'g' is not same everywhere.

It has its maximum value at poles of the Earth and minimum on its equator.

Thus a person will weigh more at poles than equator.

This variation is in accordance to:

[tex]g = \frac{GM_{E}}{radius^{2}}[/tex]

Thus the gravitational acceleration changes as inverse square of the Radius of the Earth.

Thus as we move away from the Earth's center, gravitational acceleration, g decreases.

Answer:

The equation of position and time for a sound wave is [tex]\Delta p=0.270(33.06 x-11342.40 t)[/tex].

Explanation:

Given that,

Wavelength = 0.190 m

Maximum pressure [tex]\Delta P_{max}= 0.270 N/m^2[/tex]

We know that,

The function of position and time for a sound wave,

[tex]\Delta p=\Delta p_{max}(kx-\omega t)[/tex]....(I)

We need to calculate the frequency

Using formula of frequency

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

Put the value into the formula

[tex]f=\dfrac{343}{0.190}[/tex]

[tex]f=1805.2\ Hz[/tex]

We need to calculate the angular frequency

Using formula of angular frequency

[tex]\omega =2\pi f[/tex]

Put the value into the formula

[tex]\omega=2\pi\times1805.2[/tex]

[tex]\omega=11342.40\ rad/s[/tex]

We need to calculate the wave number

Using formula of wave number

[tex]k = \dfrac{2\pi}{\lambda}[/tex]

Put the value into the formula

[tex]k=\dfrac{2\pi}{0.190}[/tex]

[tex]k=33.06[/tex]

Now, put the value of k and ω in the equation (I)

[tex]\Delta p=0.270(33.06 x-11342.40 t)[/tex]

Hence, The equation of position and time for a sound wave is [tex]\Delta p=0.270(33.06 x-11342.40 t)[/tex].

Final answer:

The mathematical model for the pressure variation of a sound wave involves sine functions with specific parameters. To find the frequency and speed of the sound wave, formulas relating wavelength, speed, and frequency are utilized.

Explanation:

The mathematical model for the pressure variation of a sound wave is given by the equation: ΔP = ΔPmax * sin((2π / λ) * x - (2πf) * t). In this equation, ΔPmax is the maximum pressure variation, λ is the wavelength, x is the position, t is the time, and f is the frequency.

To find the frequency of the sound wave, you can use the formula: f = v / λ, where v is the speed of sound. Substituting the given values allows you to calculate the frequency.

The speed of the sound wave can be determined using the formula: v = f * λ. By substituting the frequency and wavelength values, you can find the speed of the sound wave.

Answer:

[tex]q=9.83\times 10^{-8}\ C[/tex]

Explanation:

Given that,

Mass of the two spheres, m₁ = m₂ = 1 g = 0.001 kg

Distance between spheres, d = 2.2 cm = 0.022 m

Acceleration of the spheres when they are released, [tex]a=180\ m/s^2[/tex]

Let q is the charge on each spheres. The force due to motion is balanced by the electrostatic force between the spheres as :

[tex]ma=k\dfrac{q^2}{d^2}[/tex]

[tex]q=\sqrt{\dfrac{mad^2}{k}}[/tex]

[tex]q=\sqrt{\dfrac{0.001\times 180\times (0.022)^2}{9\times 10^9}}[/tex]

[tex]q=9.83\times 10^{-8}\ C[/tex]

So, the magnitude of charge on each sphere is [tex]9.83\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

To find the magnitude of the charge on the spheres, we utilize Coulomb's Law and Newton's Second Law, set up an equation, and solve for the charge q. Ensure consistency in the units while solving.

To find the magnitude of the charge on the spheres, we start by using Coulomb's Law, which is expressed as F = k*q1*q2/r^2. Here, F is the force, k is Coulomb's constant (8.99 * 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the separation. Given that the spheres are charged equally (q1 = q2), we can refer to them just as q.

The spheres start accelerating once released, and the only force in operation is the electrostatic force. Thus, according to Newton's second law (F = ma), the force can also be expressed as F = 2*(mass*acceleration) because two spheres are involved.

By equating both expressions for F, we have 2*(mass*acceleration) = k*q^2/r^2. From this equation, we can solve for q = sqrt((2*mass*acceleration*r^2)/k). Substituting given values, we have q = sqrt((2 * 1.0 g * 180 m/s^2 * (2.2 cm)^2)/(8.99 * 10^9 N m^2/C^2)).

Remember to convert grams to kilograms and centimeters to meters to maintain consistency in the units while solving.

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Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average  force exerted by the ball on the glove. It is solved using the conservation of energy as :

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

F = mg

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

[tex]F=\dfrac{mv^2}{2h}[/tex]

[tex]F=\dfrac{0.14\times (32)^2}{2\times 0.25}[/tex]

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

The false statement among the provided ones about conservative forces is 'None of these statements are true'. Conservative forces rely on the starting and ending points, not the path taken, and the work done on any closed path is always zero. These forces also have a close relationship with potential energy.

The statements provided about conservative forces generally hold true however the false statement is 'None of these statements are true', as all other statements presented are indeed accurate. A conservative force is a type of force where the work done is independent of the path. That is, the work it does only depends on the starting and ending points, not the route taken. This implies that during any closed path or loop, the total work done by a conservative force is zero.

Consider a simple example, such as the gravitational force. If you lift an object to a certain height and then back to its original position, the total work done by the gravitational force is zero as the starting and ending points are the same.

An important concept related to the conservative force is potential energy. When work is done against a conservative force, potential energy is accumulated. Conversely, the component of a conservative force, in a particular direction, equals the negative of the derivative of the potential energy for that force, with respect to a displacement in that direction.

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Answer:

[tex]\lambda=8.006\times 10^{-11}\ m[/tex]

Explanation:

Given that,

The speed of an electron, [tex]v=9.1\times 10^6\ m/s[/tex]

We need to find the wavelength of this electron. It can be calculated using De -broglie wavelength concept as :

[tex]\lambda=\dfrac{h}{mv}[/tex]

h is the Planck's constant

[tex]\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 9.1\times 10^6}[/tex]

[tex]\lambda=8.006\times 10^{-11}\ m[/tex]

So, the wavelength of the electron is [tex]8.006\times 10^{-11}\ m[/tex]. Hence, this is the required solution.

Answer:

8.953 s

Explanation:

Here a time of 8953 ms is given.

Some of the prefixes of the SI units are

mili = 10⁻³

micro = 10⁻⁶

nano = 10⁻⁹

kilo = 10³

The number is 8953.0

Here, the only solution where the number of significant figures is mili

1 milisecond = 1000 second

[tex]1\ second=\frac{1}{1000}\ milisecond[/tex]

[tex]\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second[/tex]

So 8953 ms = 8.953 s

Answer:

a) 4.325*10^10 V/m

b) 1.689*10^10 V/m

c) 0

d) 6.384 C/m^3

Explanation:

Hello!

The electric field of a sphere of uniform charge is a piecewise function, let a be the raius of the sphere

For r<a:

  [tex]E = kQ\frac{r}{a^{3}}[/tex]

For r>a:

[tex]E=kQ/r^{2}[/tex]

Since a=0.26m and k= 8.987×10⁹ N·m²/C²

 a)  

[tex]E= 8.987\times10^{9}\frac{0.47C \times0.18m}{(0.26m)^{3}}[/tex]

     E=4.325*10^10 V/m

b)

[tex]E= 8.987\times10^{9}\frac{0.47C}{(0.5m)^{2}}[/tex]

   E=1.689*10^10 V/m

c)

Since r --> ∞         1/r^2 --> 0

 E(∞)=0

d)

The charge density may be obtained dividing the charge by the volume of the sphere:

[tex]\rho = \frac{Q}{V} =\frac{0.47C}{\frac{4}{3} \pi (0.26m)^{3}}=6.384 C/m^{3}[/tex]

Answer:

The rock thrown from the top of the cliff.

Explanation:

This is more of a conceptual question. The rock thrown from the top of the cliff will be accelerated downwards, that means, accelerated towards the bird nest, will the rock thrown from the bottom will be accelerated downwards too, but, in this case, this means that it will be accelerated against the direction of the bird nest.

The rock thrown downwards from the top of the cliff will hit the bird nest located at H/2 first, as it is continuously accelerated by gravity, unlike the upward-thrown rock which initially decelerates.

Which Rock Hits the Nest First?

When considering the rocks thrown by the two children - one upwards from the ground and one downwards from the top of the cliff - the rock that hits the bird nest located at H/2 first would be the one thrown downwards. This is based on the principles of kinematics in physics, which describe the motion of objects without considering the forces that cause the motion. The initial velocities of both rocks are the same in magnitude but opposite in direction; however, gravity only decelerates the upward-thrown rock and accelerates the downward-thrown rock, resulting in the latter reaching the nest quicker.

The rock thrown upwards will slow down as it reaches a height of H/2 and has to combat gravity's pull, which is not the case for the downward-throwing rock. Although both rocks are subjected to the same acceleration due to gravity, the downward-thrown rock will cover the distance to the nest faster due to its uninterrupted acceleration. Ignoring air resistance, we don't have to consider any opposing forces which might otherwise affect the time it takes for each rock to reach the nest.

Explanation:

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Wavelength  is:

Wavelength  = c / Frequency

Given:

Frequency = 5.5 × 10¹⁴ Hz

So,

Wavelength  = 3×10⁸ m/s  / 5.5 × 10¹⁴ Hz = 545 × 10⁻⁹ m = 545 nm  (1 nm = 10⁻⁹ m )

This wavelength corresponds to green color.

Frequency = 7 × 10¹⁴ Hz

So,

Wavelength  = 3×10⁸ m/s  / 7 × 10¹⁴ Hz = 428 × 10⁻⁹ m = 428 nm  (1 nm = 10⁻⁹ m )

This wavelength corresponds to  Violet color.

Frequency = 8 × 10¹⁴ Hz

So,

Wavelength  = 3×10⁸ m/s  / 8 × 10¹⁴ Hz = 375 × 10⁻⁹ m = 375 nm  (1 nm = 10⁻⁹ m )

It does not fall in visible region. So no color. It is in UV region.

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

[tex]Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}[/tex]

[tex]Vout = 5 \frac{2000}{5000}[/[tex]

[tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V[/tex]

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200[/tex]

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

[tex]Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V[/tex]

[tex]Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V[/tex]

[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140[/tex]

[tex]R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260[/tex]

so.

[tex]V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}[/tex]

Answer:

The maximum height that the rocket reaches is 645.5 m.

Explanation:

Given that,

Mass = 10000 kg

Acceleration = 2.25 m/s²

Distance = 525 m

We need to calculate the velocity

Using equation of motion

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

Put the value in the equation

[tex]v^2=0+2\time2.25\times525[/tex]

[tex]v=\sqrt{2\times2.25\times525}[/tex]

[tex]v=48.60\ m/s[/tex]

We need to calculate the maximum height with initial velocity

Using equation of motion

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

[tex]h=\dfrac{v^2-u^2}{-2g}[/tex]

Put the value in the equation

[tex]h=\dfrac{0-(48.60)^2}{-2\times9.8}[/tex]

[tex]h=120.50\ m[/tex]

The total height  reached by the rocket is

[tex]h'=s+h[/tex]

[tex]h'=525+120.50[/tex]

[tex]h'=645.5\ m[/tex]

Hence, The maximum height that the rocket reaches is 645.5 m.

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, [tex]u_{x} = 10 m/s[/tex]

The distance between the buildings, [tex]d_{x} = 2.0 m[/tex]

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

[tex]h = ut + \frac{1}{2}gt^{2}[/tex]

[tex]h = \frac{1}{2}gt^{2}[/tex]

[tex]2.5 = \frac{1}{2}\times 10\times t^{2}[/tex]

t = 0.707 s

Now,

When the policeman was chasing across:

[tex]d_{x} = u_{x}t + \frac{1}{2}gt^{2}[/tex]

[tex]d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m[/tex]

The distance they will meet at:

9.57 - 2.0 = 7.57 m

The policeman will fall a distance of 0.2 meters while jumping to the next building, calculated using the kinematic equation for vertical displacement under gravity.

The question relates to the topic of physics, specifically to the calculation of the distance fallen by an object under the influence of gravity, which is a fundamental kinematic concept. We want to find out how far the policeman will fall when jumping across to another building. We know that he's attempting to jump across a gap that is 2 meters wide, and the other building is 2.5 meters lower. Given that the acceleration due to gravity is approximated as 10 m/s² and assuming no air resistance, we can calculate the time it will take for the policeman to travel horizontally the 2 meters distance. Since the policeman's horizontal velocity is 10 m/s, the time to cross 2 meters is 2 meters / 10 m/s = 0.2 seconds.

Next, we need to calculate the vertical displacement during this time. Using the formula for vertical displacement s under constant acceleration a due to gravity, with initial vertical velocity u = 0 (since the policeman is only moving horizontally initially), s = u⋅t + 0.5⋅a⋅t². Substituting the values, we get s = 0⋅(0.2 s) + 0.5⋅(10 m/s²)⋅(0.2 s)². Simplifying this, we find s = 0.5⋅(10)⋅(0.04) = 0.2 meters.

Therefore, the policeman will fall a distance of 0.2 meters during the time it takes to jump across to the other rooftop.

Answer:

a) The acceleration is 2.14 m/s^{2}

b) The distance traveled by the car is 65.61 m

c) The average velocity is 18.75 m/s

Explanation:

Using the equations that describe an uniformly accelerated motion:

a) [tex]a=\frac{v_f - v_o}{t} =\frac{22.5m/s - 15.0 m/s}{3.50s}[/tex]

b) [tex]d= d_0 + v_0 t + \frac{1}{2} a t^{2} = 0 +15.0 x 3.5 + \frac{2.14x3.50^{2} }{2} = 65.61 m[/tex]

c) [tex]v_m =\frac{d}{t}=\frac{65.61}{3.5}  =18.75 m/s[/tex]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

[tex]x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration[/tex]

So now we have an equation and unkown value.

for the thrown rock

[tex]\frac{1}{2}(9.8)*t^2+29*t-300=0[/tex]

for the dropped rock

[tex]\frac{1}{2}(9.8)*t^2+0*t-300=0[/tex]

solving both equation with the quadratic formula:

[tex]\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}[/tex]

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

The thrown rock hits the ground approximately 1.89 seconds earlier than the dropped rock when released simultaneously from a height of 300 meters.

To determine how much earlier the thrown rock strikes the ground compared to the dropped rock, we need to use the kinematic equation Projectile motion that relates displacement (Δx), initial velocity (vi), acceleration (a), and time (t): Δx = vi * t + 0.5 * a * t₂.

Since we're dealing with gravity, acceleration due to gravity (a) is 9.8 m/s². For the rock that is dropped, the initial velocity (vi) is 0 m/s, while the rock thrown downward has an initial velocity of 29 m/s.

First, we'll find the time it takes for the dropped rock to hit the ground:

Next, we use the same equation to find the time for the thrown rock:

This forms a quadratic equation in the form of at₂ + bt + c = 0. To solve for t, we use the quadratic formula. We only need the positive root since time cannot be negative:

Finally, the difference in time between when the two rocks hit the ground is:

Therefore, the thrown rock hits the ground approximately 1.89 seconds earlier than the dropped rock.

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Answer:

Q must be placed at 0.53 L

Explanation:

Given  data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at  distance x cm from q1

The force by charge q_1 due to q is

[tex]F1 = \frac{k q q_1}{x^2}[/tex]

[tex]F1 = \frac{k q ( 4.0 μC )}{ x^2}[/tex]                  ----1

The force by charge q_2 due to q is

[tex]F2 =  \frac{k q q_2}{(L-x)^2}[/tex]

[tex]F2 = \frac{kq (3.0 μC)}{(L-x)^2}[/tex]                   --2

we know that net electric force is equal to zero

F_1 = F_2

[tex]\frac{k q ( 4.0 μC )}{x^2}   =\frac{k q ( 3.0 μC )}{(l-x)^2}[/tex]

[tex]\frac{4}{3}*(L-x)^2 = x^2[/tex]

[tex]x = \sqrt{\frac{4}{3}*(L - x)[/tex]

[tex]L-x = \frac{x}{1.15}[/tex]

[tex]L = x + \frac{x}{1.15} = 1.86 x[/tex]

x = 0.53 L

Q must be placed at 0.53 L

The third charge should be placed either between the two charges or on the extended line of the two charges to make the electric force on that charge zero. The exact position depends on the charges involved and can be calculated using the principle of superposition and Coulomb's Law.

The problem deals with the principle of superposition in electrostatics and the force on a charge due to other charges nearby. The force on any charge due to a number of other charges is simply the vector sum of the forces due to individual charges. Starting from this principle, we can try to figure out where the third charge must be placed so that the net force on it is zero.

The two possible positions along the line of the charges are on either side of the two given charges, let's call them 4.0 µC (charge_1) and 3.0 µC (charge_2). These positions can be calculated using the formula of force between two point charges (Coulomb's Law): F = k(q1 x q2)/r² where F is force, k is Coulomb's constant, q1 and q2 are charges and r is distance between charges.

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Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

Explanation:

For 2 quantities A and B represented as

[tex]A\pm \Delta A[/tex] and [tex]B\pm \Delta B[/tex]

The sum is represented as

[tex]Sum=(A+B)\pm (\Delta A+\Delta B)[/tex]

For the the values given to us the sum is calculated as

[tex]Sum=(2.9+3.9)\pm (0.1+0.2)[/tex]

[tex]Sum=6.8\pm 0.3[/tex]

Now the since the uncertainity inthe sum is [tex]\pm 0.3[/tex]

The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

Thus closest distance equals [tex]6.8-0.3=6.5[/tex]meters