What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant figure.

Answer:

They all hit at the same time

Explanation:

Let the time of flight is T.

The maximum height is H and the horizontal range is R.

The formula for the time of flight is

[tex]T=\frac{2uSin\theta }{g}[/tex] ..... (1)

Te formula for the maximum height is

[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]    .... (2)

From equation (1) and (2), we get

[tex]\frac{T^{2}}{H}=\frac{\frac{4u^{2}Sin^{2}\theta }{g^{2}}}{\frac{u^{2}Sin^{2}\theta }{2g}}[/tex]    

[tex]\frac{T^{2}}{H}=\frac{8}{g}[/tex]

[tex]T=\sqrt{\frac{8H}{g}}[/tex]

here, we observe that the time of flight depends on the maximum height and according to the question, the maximum height for all the three balls is same so the time of flight of all the three balls is also same.

Answer:

Approximately [tex]\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right][/tex].

Explanation:

Consider this [tex]45^{\circ}[/tex] slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin [tex](0, 0)[/tex].

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at [tex]45^{\circ}[/tex] to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as [tex]y = -x[/tex].

Convert the initial speed of this diver to SI units:

[tex]\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}[/tex].

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at [tex]g[/tex] ([tex]g \approx \rm -9.81\; m\cdot s^{-2}[/tex] near the surface of the earth.) At [tex]t[/tex] seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

To eliminate [tex]t[/tex] from this expression, solve the equation between [tex]t[/tex] and [tex]x[/tex] for [tex]t[/tex]. That is: express [tex]t[/tex] as a function of [tex]x[/tex].

[tex]x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}[/tex].

Replace the [tex]t[/tex] in the equation of [tex]y[/tex] with this expression:

[tex]\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}[/tex].

Plot the two functions:

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of [tex]y[/tex] (right-hand side of the equation, the part where [tex]y[/tex] is expressed as a function of [tex]x[/tex].)

[tex]-0.0052535\;x^{2} = -x[/tex],

[tex]\implies x = 190.35[/tex].

The value of [tex]y[/tex] can be found by evaluating either equation at this particular [tex]x[/tex]-value: [tex]x = 190.35[/tex].

[tex]y = -190.35[/tex].

The position vector of a point [tex](x, y)[/tex] on a cartesian plane is [tex]\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right][/tex]. The coordinates of this skier is approximately [tex](190.35, -190.35)[/tex]. The position vector of this skier will be [tex]\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right][/tex]. Keep in mind that both numbers in this vectors are in meters.

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a horizontal ramp. Assuming that the skier is in a free-fall motion, the distance down the slope that he will land is 190.3 m and the total displacement vector from the point of lift-off is 269.15 m

From the given information;

To convert the speed into m/s, we have:

[tex]\mathbf{= 110 \ \times (\dfrac{1000 \ m}{3600 \ s})}[/tex]

= 30.556 m/s

Since the speed race down the steep hill before flying horizontally into the air, then we can assert that the;

Now, for a motion under free-fall, to determine the distance(S) of the slope using the second equation of motion, we have:

The distance along the vertical axis to be:

[tex]\mathbf{S_v= ut + \dfrac{1}{2} gt^2}[/tex]

where the initial velocity = 0 m/s

[tex]\mathbf{S_v= (0)t + \dfrac{1}{2} gt^2}[/tex]

[tex]\mathbf{S_v= \dfrac{1}{2} gt^2}[/tex]

[tex]\mathbf{S_v= \dfrac{1}{2} (9.81)t^2}[/tex]

[tex]\mathbf{S_v=4.905t^2}[/tex]

The horizontal distance [tex]\mathbf{S_x = ut }[/tex]

From the angle at which the ground begins to slope, taking the tangent of the angle with respect to the distance, we have:

[tex]\mathbf{tan \ \theta = \dfrac{opposite}{adjacent}}[/tex]

where;

[tex]\mathbf{tan \ 45^0 = \dfrac{S_v}{S_x}}[/tex]

[tex]\mathbf{tan \ 45^0 = \dfrac{4.905t^2}{30.556t}}[/tex]

1 × 30.556t = 4.905t²

Divide both sides by t, we have:

30.556 = 4.905t

[tex]\mathbf{t = \dfrac{30.556}{4.905}}[/tex]

t = 6.229 sec

However, since the time at which the slope will land is known, we can easily determine the value of the vertical and horizontal distances.

i.e.

The horizontal distance [tex]\mathbf{S_x = 30.556t }[/tex]

The total displacement vector from the point of lift-off is:

=  [tex]\mathbf{190.314 \sqrt{2}\ m}[/tex]  

= 269.15 m

Therefore, we can conclude that the distance down the slope that he will land is 190.3 m and the displacement down the slope that the skier will land is 269.15 m

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Answer:

Explanation:

mass of block, m = 20 kg

let teh angle of inclination of the ramp with the horizontal is θ.

Weight of the block = mg

There are two components of the weight

The component of weight of block along the plane = mg Sinθ

The component of weight of block normal to the plane = mg Cosθ

Friction force is acting opposite to the motion, i.e., along the plane and upwards, f = μN = μ mg Cosθ

Look at the diagram to find the directions of force.

Answer:

hair grows at rate of 8.166 nm/s

Explanation:

given data:

grow rate=  (1/36)in/day

we know that

1 day = 86400 seconds

therefore we have [tex]= \frac{1}{36}inc / 86400 sec[/tex]

first change equation to [tex]\frac{1}{36}inc / 86400 sec.[/tex]

Then you find x in/s [tex]= \frac{1}{36}inc / 86400 sec[/tex]

[tex]x/s = \frac{1}{36}inc / 86400 sec[/tex]

[tex](86400s)x = (\frac{1}{36}inc) s[/tex]

[tex]x = \frac{\frac{1}{36}\ inc}{86400} = 3.21 x 10^{-7} in/second[/tex]

Now convert in/sec into meters/second.

we know that 1m = 39.37in

[tex]x/1m = \frac{3.21 x10^{-7} in}{39.37\ in}[/tex]

[tex](39.37in)x = (3.21 *10^{-7} in) m[/tex]

[tex]x = \frac{3.21 * 10^{-7}}{39.37}m[/tex]

x = [tex]8.166 * 10^{-9}[/tex] m/second

Now convert [tex]8.166 * 10^{-9}[/tex] m/second into nm/second

we know that [tex]1nm = 10^{-9}m[/tex]

[tex]x/nm = \frac{8.166x 10^{-9} m/second}{10^{-9} m}[/tex]

x =8.166 nm/second

therefore hair grows at rate of 8.166 nm/s

Final answer:

Hair grows at approximately 8.164 nanometers per second or 81.64 atomic layers per second, based on the conversion from 1/36 inches per day.

Explanation:

Converting Hair Growth Rate to Nanometers per Second

To find the rate at which hair grows in nanometers per second, given that hair grows at the rate of 1/36 inches per day, we first convert inches per day to nanometers per day, then to nanometers per second. Since 1 inch equals 25,400,000 nanometers, the hair growth rate per day in nanometers is:

1/36 inch/day × 25,400,000 nm/inch = 705,555.56 nm/day.

Next, we convert this to nanometers per second. There are 86,400 seconds in a day, so:

705,555.56 nm/day ÷ 86,400 seconds/day = 8.164 nm/second.

Considering that the distance between atoms in a molecule is on the order of 0.1 nm, the rate of hair growth can be seen as:

8.164 nm/second ÷ 0.1 nm = 81.64 atomic layers/second.

Therefore, hair grows at approximately 81.64 atomic layers/second.

Answer:

[tex]5.07\cdot 10^{-14} m[/tex]

Explanation:

The velocity of the neutrons is

[tex]v=3.13\cdot 10^7 m/s[/tex]

The mass of a neutron is

[tex]m=1.66\cdot 10^{-27} kg[/tex]

So their momentum is

[tex]p=mv=(1.66\cdot 10^{-27})(3.13\cdot 10^7)=5.20\cdot 10^{-20}kg m/s[/tex]

The relative uncertainty on the velocity is 2 %. Assuming that the mass of the neutron is known with negligible uncertainty, then the relative uncertainty on the momentum of the neutron is equal to the relative uncertainty on the velocity, so 2%. Therefore, the absolute uncertainty on the momentum is

[tex]\sigma_p = 0.02 p =0.02(5.20\cdot 10^{-20})=1.04\cdot 10^{-21} kg m/s[/tex]

Heisenber's uncertainty principle states that

[tex]\sigma_x \sigma_p \geq \frac{h}{4\pi}[/tex]

where

[tex]\sigma_x[/tex] is the uncertainty on the position

h is the Planck constant

Solving for [tex]\sigma_x[/tex], we find the minimum uncertainty on the position:

[tex]\sigma_x \geq \frac{h}{4\pi \sigma_p}=\frac{6.63\cdot 10^{-34}}{4\pi(1.04\cdot 10^{-21})}=5.07\cdot 10^{-14} m[/tex]

The question involves the application of the Heisenberg Uncertainty Principle in Quantum Mechanics to calculate the uncertainty in the position of a particle (in this case, a neutron) given the uncertainty in its velocity.

This is a question regarding the application of the Heisenberg Uncertainty Principle, a fundamental concept in Quantum Mechanics. The uncertainty in position (Δx) can be calculated using this principle which states that the product of the uncertainties in position and momentum of a particle is always greater than or equal to half of Planck's constant.

The uncertainty in the momentum (Δp) is given by the product of the uncertainty (0.02) and the average velocity (3.1 × 107 m/s) of the neutron. Thus, Δp = 0.02 × neutron mass (1.675 × 10-27 kg) × average velocity.

Then from the Uncertainty Principle (Δx × Δp ≥ 0.5 × h), we get: Δx ≥ 0.5 × Planck's constant / Δp. By substituting the appropriate values, we can calculate the uncertainity in the position.

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Answer:

4000 J

Explanation:

The energy possessed by virtue of motion of the body is called its kinetic energy.

When a body is in motion , it has kinetic energy.

Mass of the object, m = 80 kg

velocity of the object, v = 10 m/s

The formula for the kinetic energy is given by

[tex]K=\frac{1}{2}mv^{2}[/tex]

K = 0.5 x 80 x 10 x 10

K = 4000 J

Thus, the kinetic energy of the object is 4000 J.

Answer:

f=171.43Hz

Explanation:

Wave frequency is the number of waves that pass a fixed point in a given amount of time.

The frequency formula is: f=v÷λ, where v is the velocity and λ is the wavelength.

Then replacing with the data of the problem,

f=[tex]\frac{120\frac{m}{s} }{0.7m}[/tex]

f=171.43[tex]\frac{1}{s}[/tex]

f=171.43 Hz (because [tex]\frac{1}{s} = Hz[/tex], 1 hertz equals 1 wave passing a fixed point in 1 second).

Answer:

Volume=30mm^3

Explanation:

The equation to find the volume of a cube is to multiply all its sides.

V = a.b.c

where a, b, c correspond to the sides of the cube

a=2mm

b=3mm

c=5mm

V=(2mm)(3mm)(5mm)=30mm^3

the volume of the rectangular cube is 30mm^3

Answer:

d) Sofia.

Explanation:

If they walked down a straight road, their displacement will be 4 miles. However, their position depends on the origin of a coordinate system which could be located anywhere. If the origin of the coordinate system was at the starting point of their walk, then and only then, you can say that their position is also 4 miles.

Answer:

Electric force, [tex]F=1.29\times 10^6\ N[/tex]

Explanation:

Given that,

Charge 1, [tex]q_1=24\ C[/tex]

Charge 2, [tex]q_2=-24\ C[/tex]

Distance between charges, d = 2 km

The electric force is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

k is the electrostatic constant

[tex]F=8.98755\times 10^9\times \dfrac{(24)^2}{(2\times 10^3)^2}[/tex]

F = 1294207.2 N

or

[tex]F=1.29\times 10^6\ N[/tex]

Hence, this is the required solution.

Answer:

Answer:

77.58°

Explanation:

Let the projectile is projected at an angle θ and the velocity of projection is u.

At the maximum height, the projectile has only horizontal component of velocity.

Let the velocity at maximum height is v.

According to the question, the velocity at maximum height is 21.5% of initial velocity.

v = 21.5 % of u

u Cos θ = 21.5 % of u

u Cos θ = 0.215 u

Cos θ = 0.215

θ = 77.58°

Thus, the angle of projection is  77.58°.

Explanation:

Answer:12.206 cm,[tex]\theta =54.99^{\circ}[/tex]

Explanation:

Given

Insect walks 15 cm to the right

so its position vector is[tex]r_1=15i[/tex]

Now it moves 10 cm up so its new position vector

[tex]r_2=15i+10j[/tex]

Now it moves 8 cm left so its final position vector is

[tex]r_3=15\hat{i}+10\hat{j}-8\hat{i}=7\hat{i}+10\hat{j}[/tex]

so its displacement is given by

[tex]|r_3|=\sqrt{7^2+10^2}=\sqrt{149}=12.206 cm[/tex]

For direction, let \theta is the angle made by its position vector with x axis

[tex]tan\theta =\frac{10}{7}=1.428[/tex]

[tex]\theta =54.99^{\circ}[/tex]

Answer:

[tex]t = 3.49 s[/tex]

[tex]a = -5.02 m/s^2[/tex]

Explanation:

As we know that initial speed of the cheetah is given as

[tex]v_i = 17.5 m/s[/tex]

finally it comes to rest so final speed is given as

[tex]v_f = 0[/tex]

now we know that distance covered by cheetah while it stop is given as

[tex]d = 30.5 m[/tex]

now by equation of kinematics we know that

[tex]d = (\frac{v_f + v_i}{2})t[/tex]

here we have

[tex]30.5 m = (\frac{0 + 17.5}{2}) t[/tex]

[tex]30.5 = 8.75 t[/tex]

[tex]t = 3.49 s[/tex]

Now in order to find the acceleration we know that

[tex]v_f - v_i = at[/tex]

[tex]0 - 17.5 = a(3.49)[/tex]

[tex]a = -5.02 m/s^2[/tex]

To determine the time it takes for a cheetah running at 17.5 m/s to stop over a distance of 30.5 m, kinematic equations are used. The acceleration is calculated to be approximately -4.98 m/s², and the time to come to rest is roughly 3.49 seconds.

To find out how long it takes for a cheetah running at 17.5 m/s to come to rest and what is the cheetah's acceleration while doing so, we can use the following kinematic equation:

[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 covered. Given that the cheetah comes to rest, v = 0 m/s, u = 17.5 m/s, and s = 30.5 m.

We can rearrange the formula to solve for a:

[tex]0 = (17.5 m/s)^2 + 2a(30.5 m)[/tex]

[tex]a = - (17.5 m/s)^2 / (2 * 30.5 m)[/tex]

a = - 4.98 m/s2 (The acceleration is negative because it's a deceleration)

Now, using the formula v = u + at to find the time, t, it takes to come to rest:

[tex]0 = 17.5 m/s - (4.98 m/s^2 * t)\\t = 17.5 m/s / 4.98 m/s^2[/tex]

t = 3.51 s (approximately 3.49 s)

Answer:

[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]

Explanation:

As we know that electric field inside any conducting shell is always zero

so we can say that

[tex]E_{charge} + E_{polarization} = 0[/tex]

here we know that

[tex]E_{charge} = \frac{kq}{r^2} \hat r[/tex]

here we know that

[tex]\hat r = \frac{(0 - -0.6)\hat i + (0.08 - 0)\hat j}{\sqrt{0.6^2 + 0.08^2}}[/tex]

[tex]\hat r = 0.99\hat i + 0.132\hat j[/tex]

now we will have

[tex]E_{polarization} = - E_{charge}[/tex]

[tex]E_{polarization} = - \frac{(9\times 10^9)(9 \times 10^{-8})}{0.6^2 + 0.08^2}(0.99\hat i + 0.132\hat j)[/tex]

[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]

We have that the electric field contributed by the polarization charges on the surface of the metal sphere is

From the question we are told

Generally the equation for the electric field inside any conducting shell    is mathematically given as

E charge+E polarization=0

Therefore

[tex]E charge=\frac{kq}{r^2}r'\\\\E charge=\frac{0--0.6i+0.08-0j}{\sqrt{0.6^2+0.08^2}}\\\\r'=0.99i+0.132j[/tex]

Therefore

E polarization=-E charge

Therefore

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Answer:

(a):  [tex]\rm meter/ second^2.[/tex]

(b):  [tex]\rm meter/ second^3.[/tex]

(c):  [tex]\rm 2ct-3bt^2.[/tex]

(d):  [tex]\rm 2c-6bt.[/tex]

(e):  [tex]\rm t=\dfrac{2c}{3b}.[/tex]

Explanation:

Given, the position of the particle along the x axis is

[tex]\rm x=ct^2-bt^3.[/tex]

The units of terms [tex]\rm ct^2[/tex] and [tex]\rm bt^3[/tex] should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of [tex]\rm ct^2=meter[/tex]

Therefore, unit of [tex]\rm c= meter/ second^2.[/tex]

(b):

Unit of [tex]\rm bt^3=meter[/tex]

Therefore, unit of [tex]\rm b= meter/ second^3.[/tex]

(c):

The velocity v and the position x of a particle are related as

[tex]\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.[/tex]

(d):

The acceleration a and the velocity v of the particle is related as

[tex]\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.[/tex]

(e):

The particle attains maximum x at, let's say, [tex]\rm t_o[/tex], when the following two conditions are fulfilled:

Applying both these conditions,

[tex]\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.[/tex]

For [tex]\rm t_o = 0[/tex],

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c[/tex]

Since, c is a positive constant therefore, for [tex]\rm t_o = 0[/tex],

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0[/tex]

Thus, particle does not reach its maximum value at [tex]\rm t = 0\ s.[/tex]

For [tex]\rm t_o = \dfrac{2c}{3b}[/tex],

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.[/tex]

Here,

[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}<0.[/tex]

Thus, the particle reach its maximum x value at time [tex]\rm t_o = \dfrac{2c}{3b}.[/tex]

The units of constant c are in meters per second squared per second squared (m/s^2). The units of constant b are in meters per second cubed (m/s^3). The velocity can be found by taking the derivative of the position function, and the acceleration can be found by taking the derivative of the velocity function. To find the time when the particle reaches its maximum x value, set the velocity equal to zero and solve for t.

(a) The units of constant c can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t2 will be in meters per second squared. Therefore, the units of constant c will be in meters per second squared per second squared (m/s2).

(b) Similar to part (a), the units of constant b can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t3 will be in meters per second cubed. Therefore, the units of constant b will be in meters per second cubed (m/s3).

(c) The velocity v can be found by taking the derivative of the position function x with respect to time t. In this case, v = d/dt(ct2 - bt3) = 2ct - 3bt2.

(d) The acceleration a can be found by taking the derivative of the velocity function v with respect to time t. In this case, a = d/dt(2ct - 3bt2) = 2c - 6bt.

(e) To find the time when the particle reaches its maximum x value, we can set the velocity v equal to zero and solve for t. 2ct - 3bt2 = 0. Solving this equation will give us the time t when the particle reaches its maximum x value.

Answer:

Explanation:

Initial velocity of train u = 3.4 m⁻¹ , Acceleration a = .065 ms⁻² ,

time t = 9.75 x 60 = 585 s.

v = u + at

= 3.4 + .065 x 585

= 41.425 m / s

distance travelled during the acceleration ( s )

s = ut + 1/2 at²

= 3.4 x 585 + .5 x .065 x 585²

= 1989 + 11122.31

= 13111.31 m .

again for slowing process

u = 41.425 m / s

v = u -at

0 = 41.425 - 0.625 t

t = 66.28 s

v² = u² - 2as

0 = 41.425² - 2 x .625 s

s = 1372.82 m

The final velocity of the freight train is approximately 21.2 m/s. The train takes approximately 33.92 seconds to come to a stop. The train travels approximately 15,087 meters during the acceleration and 3,655 meters during the deceleration.

To find the final velocity of the freight train, we can use the equation:

v = u + at

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

Plugging in the given values, we have:

v = 3.4 m/s + (0.065 m/s2)(9.75 min)(60 s/min)

Solving this equation, we get the final velocity of the freight train to be approximately 21.2 m/s.

To find the time it takes for the train to stop, we can use the same equation:

v = u + at

But this time, the initial velocity is the final velocity from part (a), and the acceleration is the deceleration rate of 0.625 m/s2.

Plugging in the given values, we have:

0 = 21.2 m/s + (0.625 m/s2)t

Solving this equation, we find that it takes approximately 33.92 seconds for the train to come to a stop.

To find the distance traveled during the acceleration, we can use the equation:

s = ut + (1/2)at2

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the given values, we have:

s = 3.4 m/s(9.75 min)(60 s/min) + (1/2)(0.065 m/s2)(9.75 min)(60 s/min)2

Solving this equation, we find that the train traveled approximately 15,087 meters during the acceleration.

To find the distance traveled during the deceleration, we can use the same equation:

s = ut + (1/2)at2

But this time, the initial velocity is the final velocity from part (a), and the acceleration is the deceleration rate of 0.625 m/s2.

Plugging in the given values, we have:

s = 21.2 m/s(33.92 s) + (1/2)(-0.625 m/s2)(33.92 s)2

Solving this equation, we find that the train traveled approximately 3,655 meters during the deceleration.

The electric potential at the center (C) of the circle formed by the rod is zero due to the symmetry and cancellations of charges. Calculating the potential at point P requires a detailed approach considering the distance D and the charge distribution.

The question involves calculating the electric potential at two different points with reference to a specially configured plastic rod. For part (a), the total charge along the circumference is given by the sum of Q1 and Q2, where Q2 = -6Q1. However, due to the uniform distribution and the symmetry of the setup, the electric potential at the center of the circle (point C) would effectively be zero because the contributions from the positively and negatively charged segments cancel each other out.

For part (b), calculating the electric potential at point P, which lies on the axis and is a distance D from the center, would require the application of principles of superposition and the integration of electric potential contributions from each infinitesimal segment of the charged parts of the rod. Considering the arrangement and the charges' nature, this part of the solution involves more intricate calculations and requires knowing the specific distribution of the charges along the rod.

Answer:

Maximum speed, v = 36 m/s                                                          

Explanation:

Given that,

The radius of the curved road, r = 120 m

Road is at an angle of 48 degrees. We need to find the maximum speed of stay on the curve in the absence of friction. On a banked curve, the angle at which it is cant is given by :

[tex]tan\theta=\dfrac{v^2}{rg}[/tex]

g is the acceleration due to gravity  

[tex]v=\sqrt{rg\ tan\theta}[/tex]

[tex]v=\sqrt{120\times 9.8\times \ tan(48)}[/tex]

v = 36.13 m/s

or

v = 36 m/s

So, the maximum speed to stay on the curve in the absence of friction is 36 m/s. Hence, this is the required solution.

The proton's acceleration is calculated using the electric field strength and the electron charge. The time to reach 1.1 Mm/s is computed using this acceleration and final speed. The displacement of the proton is derived using the time and acceleration.

This question involves analyzing the motion of a proton in an electric field, using basic principles from physics, specifically the concept of acceleration and kinematics.

(a): Acceleration of the proton

: The electric force exerted on the proton can be calculated by F=qE where q is the proton's charge (1.6 x 10^-19 C), and E is the supplied electric field (700N/C). Knowing that acceleration a=F/m (F=force, m=mass of proton = 1.67 x 10^-27 kg), we can find the acceleration.

(b): Time taken to reach the speed

: Using the formula v=u+at (u=initial speed=0, a=acceleration from part (a), v=final speed=1.1 x 10^6 m/s) we can solve for t, the time taken.

(c): Displacement of the proton

: Displacement can be calculated by using the formula s= ut + 1/2at²(where u=initial speed, t=time taken from part (b), a=acceleration from part a).

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Answer:

9.9 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s[/tex]

If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.

Hence, the speed of the diver just before striking the water is 9.9 m/s

The speed of the diver just before striking the water is 11.88 m/s.

To calculate the speed of the diver just before striking the water, we can use the concept of conservation of energy. At the top of the dive, the diver has potential energy, which is converted to kinetic energy as the diver falls. When the diver reaches the surface of the water, all of the potential energy has been converted to kinetic energy. We can use the equations for potential and kinetic energy to solve for the speed of the diver.

Given the height of the diving board (5.0 m) and the depth the diver stops (2.2 m below the surface), we can calculate the total distance the diver falls (5.0 m + 2.2 m = 7.2 m). Using the equations for potential energy (PE = mgh) and kinetic energy (KE = 0.5mv^2), we can set the two equal to each other and solve for the speed (v) of the diver:

PE = KE

mgh = 0.5mv^2

mg(7.2 m) = 0.5mv^2

gh = 0.5v^2

v^2 = 2gh

v = sqrt(2gh)

Substituting the values for g (acceleration due to gravity) and h (height of fall), we can calculate the speed of the diver:

v = sqrt(2 * 9.8 m/s^2  * 7.2 m) = sqrt(141.12) m/s = 11.88 m/s

Answer:

19.2*10^6 s

Explanation:

The equation for time dilation is:

[tex]t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

[tex]t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s[/tex]

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.

Answer and Explanation:

The clothing after spinning in the dryer cling together. This is because in the dryer they are rubbed against each other and due to this rubbing, electrons are transferred from one to the other clothes and acquire charge as a result of charging by friction thus producing static electricity.

As the material of the clothes in the dryer is different, clinging will be more.

The sticking of these clothes together is known as Static cling.

In case, the clothing are of same material, the static electricity produced as a result of frictional charging would be less and hence less static cling would occur.

Clothing clings together in the dryer due to static electricity. Clinging would be expected to be more common if the clothing were made of the same material. Static electricity occurs when different materials rub against each other and build up opposite charges.

The clothing tends to cling together after going through the dryer due to static electricity. When fabrics rub against each other in the dryer, they become charged with static electricity. The charged clothes then attract and stick to one another, causing them to cling together.

If all the clothing were made of the same material like cotton, you would expect more clinging compared to drying different kinds of clothing together. This is because different materials have different tendencies to build up static electricity. When different fabrics with different tendencies to build up static electricity are dried together, they can counteract each other's charges, reducing the overall amount of clinging.

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Answer:

Explanation:

Electrons will create negative and protons will create positive charge . 5.29 x 10⁶ electrons will neutralize same no of protons . So

( 7.07 - 5.29 ) x 10⁶ protons will create net positive charge

magnitude of this charge can be calculated as follows

   no of protons x charge on one proton

(7.07 - 5.29)x 10⁶ x 1.6 x 10⁻¹⁹ C

2.848 X 10⁻¹³ C is the net charge on the sphere. Ans

b )

Protons and electrons have equal but opposite charges .

115 protons will neutralize the charge of 115 electrons .

225-115 = 110 electrons will remain un-neutralized .

Charge on these un-neutralized electrons will be calculated as follows .

no of electrons x charge on a single electrons

110 x 1.6 x 10⁻¹⁹ C

176 X 10⁻¹⁹C is the net charge on the sphere.   Ans

The net charge on a sphere is derived from the difference between its protons and electrons, multiplied by the elemental charge. For case (a), the charge is 2.848 x 10⁻¹³ C, and for case (b), it is -1.76 x 10⁻¹⁷ C.

he net charge on a sphere can be calculated by considering the difference between the number of protons and electrons, then multiplying by the fundamental charge of an electron or proton, which is approximately 1.6 x 10⁻¹⁹ Coulombs (C).

Case (a):

Case (b):

Therefore, the net charge on each sphere is 2.848 x 10⁻¹³ C for (a) and -1.76 x 10⁻¹⁷ C for (b).

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

[tex]\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}[/tex]

Put the value into the formula

[tex]525=\sqrt{\dfrac{c+v}{c-v}}\times950[/tex]

[tex]\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2[/tex]

[tex]\dfrac{c+v}{c-v}=0.305[/tex]

[tex]c+v=0.305\times(c-v)[/tex]

[tex]v(1+0.305)=c(0.305-1)[/tex]

[tex]v=\dfrac{0.305-1}{1+0.305}c[/tex]

[tex]v=−0.532c[/tex]

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle [tex]\angle \mathrm{B\hat{C}D}[/tex] which corresponds to this minor arc.

This angle comes can be split into two parts:

[tex]\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}[/tex].

Also,

[tex]\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}[/tex].

The radius of this circle is:

[tex]\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}[/tex].

The lengths of segment DC, AC, BC can all be found:

In the two right triangles [tex]\triangle\mathrm{DAC}[/tex] and [tex]\triangle \rm BAC[/tex], the value of [tex]\angle \mathrm{B\hat{C}A}[/tex] and [tex]\angle \mathrm{A\hat{C}D}[/tex] can be found using the inverse cosine function:

[tex]\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}} [/tex]

[tex]\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}} [/tex]

[tex]\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}[/tex].

The length of the minor arc will be:

[tex]\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km[/tex].

Answer:

(A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

Explanation:

Given that,

Distance in x axis = 5.00 cm

Electric field = 10.0 N/C

Distance in x axis = 10.0 cm

Electric field = 17.0 N/C

Since, q is the same charge, two formulas can be set equal using the two different electric fields.

(a). We need to calculate the location of the charge

Using formula of force

[tex]F = qE[/tex]....(I)

Using formula of electric force

[tex]F =\dfrac{kq^2}{d^2}[/tex]....(II)

From equation (I) and (II)

[tex]qE=\dfrac{kq^2}{d^2}[/tex]

[tex]E=\dfrac{kq}{d^2}[/tex]

[tex]q=\dfrac{E(x-r)^2}{k}[/tex]...(III)

For both points,

[tex]\dfrac{E(x-r)^2}{k}=\dfrac{E(x-r)^2}{k}[/tex]

Put the value into the formula

[tex]\dfrac{10.0\times(x-5.00)^2}{k}=\dfrac{17.0\times(x-10.0)^2}{k}[/tex]

[tex]10.0\times(x-5.0)^2=17.0\times(x-10.0)^2[/tex]

Take the square root of both sides

[tex]3.162(x-5.0)=4.123(x-10.0)[/tex]

[tex]3.162x-3.162\times5.0=4.123x-4.123\times10.0[/tex]

[tex]3.162x-4.123x=-4.123\times10.0+3.162\times5.0[/tex]

[tex]0.961x=25.42[/tex]

[tex]x=\dfrac{25.42}{0.961}[/tex]

[tex]x=26.45\ cm[/tex]

(B). We need to calculate the charge

Using equation (III)

[tex]q=\dfrac{E(x-r)^2}{k}[/tex]

Put the value into the formula

[tex]q=\dfrac{10.0(26.45\times10^{-2}-5.00\times10^{-2})^2}{9\times10^{9}}[/tex]

[tex]q=5.11\times10^{-11}\ C[/tex]

[tex]q=51.1\ pC[/tex]

Hence, (A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

The charge is located at x = 2.25 cm along the x-axis and has a magnitude of approximately 8.41 x 10⁻¹² C. These were determined using the relationships of electric field magnitude and distance from the point charge. The analysis involved solving the electric field equations at different points based on Coulomb's law.

The electric field due to a point charge follows the equation:

E = k * |q| / r²

where E is the electric field, k is Coulomb's constant (8.99 x 10⁹ Nm²/C²), q is the charge, and r is the distance from the charge.

Part A: Finding the Charge's Location

At x = 5.00 cm, E1 = 10.0 N/C and at x = 10.0 cm, E2 = 17.0 N/C.

Assume the charge q is located at x = x0.

For x = 5.00 cm, the distance r1 is |x0 - 5.00 cm|.

For x = 10.0 cm, the distance r2 is |x0 - 10.0 cm|.

Using the electric field equations:
10.0 = k * |q| / (|x0 - 5.00|²)
17.0 = k * |q| / (|x0 - 10.0|²)

Divide the second equation by the first to eliminate q:

(17.0 / 10.0) = (|x0 - 5.00|² / |x0 - 10.0|²)

Solving this, we get |x0 - 5.00|² = 1.7 * |x0 - 10.0|².

Let u = x0 - 5.00 and v = x0 - 10.0, hence v = u - 5.00.

Substituting and solving gives the location x0 = 2.25 cm.

Part B: Finding the Magnitude of the Charge

Use the equation E = k * |q| / r² with one of the electric field values from Part A.

Let's use E1 = 10.0 N/C and r1 = |2.25 cm - 5.00 cm| = 2.75 cm = 0.0275 m:

10.0 = (8.99 x 10⁹ N*m²/C²) * |q| / (0.0275 m)².

Solving for q, we get q ≈ 8.41 x 10⁻¹² C.

Thus, the charge is located at 2.25 cm along the x-axis with a magnitude of 8.41 x 10⁻¹² C.

Answer:

1109m

Explanation:

The distance h₁ traveled by the rocket during acceleration:

[tex]h_1=\frac{1}{2}at^2[/tex]

The velocity v₁ at this height h₁:

[tex]v_1=at[/tex]

When the rocket runs out of fuel, energy is conserved:

[tex]E=\frac{1}{2}mv_1^2+mgh_1=mgh_2[/tex]

Solving for h₂:

[tex]h_2=\frac{v_1^2}{2g}+h_1=\frac{(at)^2}{2g}+\frac{1}{2}at^2[/tex]

Answer:

A.) t₁= 4.37 s

B.) d₁= 19.09 m

C.) [tex]v_{f1} = 8,74 \frac{m}{s}[/tex]

Explanation:

Bicyclist kinematics :

Bicyclist moves with uniformly accelerated movement:

d₁ = v₀₁*t₁ + (1/2)a₁*t₁²  equation (1)

d₁ : distance traveled by the cyclist

v₀₁: initial speed (m/s)

a₁: acceleration  (m/s²)

t₁:  time it takes the cyclist to catch his friend (s)

Friend  kinematics:

Friend moves with constant speed:

d ₂= v₂*t₂ Equation (2)

d₂ : distance traveled by the friend (m)

v₂: friend speed (m/s)

t₂ : time that has elapsed since the cyclist meets the friend until the cyclist catches him.

Data

v₀₁=0

a₁ = 2.0 m/s²

v₂ = 3.0 m/s

Problem development

A.) When the bicyclist catches his friend, d₁ = d₂=d  and t₂=t₁+2

in the equation (1) :

d = 0 + (1/2)2*t₁²  = t₁²

d = t₁²  equation (3)

in the equation (2) :

d = 3*(t₁+2) ₂ = 3*t₁+6

d =  3t₁+6 equation (4)

Equation (3) = Equation (4)

                t₁²  =  3t₁+6  

t₁² - 3t₁ - 6  = 0 ,we solve the quadratic equation

t₁= 4.37 s : time it takes the cyclist to catch his friend

B.) d₁ : distance traveled by the cyclist

In the Equation (2) d₁= (1/2)2* 4.37² = 19.09 m

C.)  speed of the cyclist when he catches his friend

[tex]v_{f1} = v_{o1} + a*t[/tex]

[tex]v_{f1} =0 + 2* 4.37 [/tex]

[tex]v_{f1} = 8.74 \frac{m}{s}[/tex]

Answer:

A) The time it takes for the T-shirt to reach the fan is 2.3582s

B) The height the fan is from the ground is 10.2438m

Explanation:

First, we need to find the velocity vectors for each axe:

[tex]v_{x}=30*cos(32\textdegree)=25.44\frac{m}{s}\\ v_{x}=30*sin(32\textdegree)=15.90\frac{m}{s}[/tex]

Next, using the movement equation for the x axis, we get the time:

[tex]D=v_{x}*t\\t=\frac{D}{v_{x}}=2.3585s[/tex]

Finally, using the movement equation for the y axis, we find the height (assuming the gravity as 9,8[tex]\frac{m}{s^{2}}[/tex])

[tex]h=v_{y}*t-\frac{g*t^{2}}{2}=10.2438\frac{m}{s^{2}}[/tex]

Answer:

The magnitude of the resultant vector is 13.656 units.

Explanation:

The vector A can be represented vectorially as

 [tex]\overrightarrow{r}_{a}=6.00\widehat{i}-4.40\widehat{j}[/tex]

Similarly vector B can be represented vectorially as

[tex]\overrightarrow{r}_{b}=3.30\widehat{i}-5.60\widehat{j}[/tex]

Thus upon adding the 2 vectors we get

[tex]\overrightarrow{r}_{a}+\overrightarrow{r}_{b}=6.00\widehat{i}-4.40\widehat{j}+3.30\widehat{i}-5.60\widehat{j}\\\\=(6.00+3.30)\widehat{i}-(4.40+5.60)\widehat{j}\\\\=9.30\widehat{i}-10.00\widehat{j}[/tex]

Now the magnitude of the vector is given by:

|r|=[tex]\sqrt{x^{2}+y^{2}}\\\\|r|=\sqrt{9.30^{2}+(-10)^{2}}\\\\\therefore |r|=13.65units[/tex]