The vertical velocity of a projectile launch at 35 degree : O continuously decreases O continuously increases O sometimes decreases and sometimes increases O is zero O stays constant.

Answer:

370.6 nm

Explanation:

wavelength in vacuum = 494 nm

refractive index of water with respect to air = 1.333

Let the wavelength of light in water is λ.

The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.

By using the definition of refractive index

[tex]n = \frac{wavelength in air}{wavelength in water}[/tex]

where, n be the refractive index of water with respect to air

By substituting the values, we get

[tex]1.333 = \frac{494}{\lambda }[/tex]

λ = 370.6 nm

Thus, the wavelength of light in water is 370.6 nm.

Answer:

The rock will rise 3.3 m above the top of the window.

Explanation:

The equations used to find the height and velocity of the rock at any given time are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the rock at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

v = velocity of the rock at time t

If we place the frame of reference at the bottom of the window, we can say that at time t = 0.19 s the height of the rock is 1.7 m. That will allow us to find the initial velocity needed to find the time at which the rock is at its maximum height.

y = y0 + v0 · t + 1/2 · g · t²

1.7 m = 0 m + v0 · 0.19 s - 1/2 · 9.8 m/s² · (0.19 s)²

1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²  = v0 · 0.19 s

(1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²) / 0.19 s = v0

v0 = 9.9 m/s

With the initial velocity, we can find at which time the rock reaches its max- height. We know that at maximum height, the velocity of the rock is 0. Then, using the equation of velocity:

v = v0 + g · t

0 = 9.9 m/s - 9.8 m/s² · t

-9.9 m/s / -9.8 m/s² = t

t = 1.0 s

Now calculating the position at time t = 1.0 s, we will find the maximum heigth:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 9.9 m/s · 1.0 s - 1/2 · 9.8 m/s² · (1.0 s)²

y = 5.0 m  (this is the max-height meassured from the bottom of the window)

Then, the rock will rise (5.0 m - 1.7 m) 3.3 m above the top of the window.

Answer:

The distance traveled by Tina before passing David is 900 m

Given:

Initial speed of David, [tex]u_{D} = 30 m/s[/tex]

Acceleration of Tina, [tex]a_{T} = 2.0 m/s^{2}[/tex]

Solution:

Now, as per the question, we use 2nd eqn of motion for the position of David after time t:

[tex]s = u_{D}t + \frac{1}{2}at^{2}[/tex]

where

s = distance covered by David after time 't'

a = acceleration of David = 0

Thus

[tex]s = 30t[/tex]

Now, Tina's position, s' after time 't':

[tex]s' = u_{T}t + \frac{1}{2}a_{T}t^{2}[/tex]

where

[tex]u_{T} = 0[/tex], initially at rest

[tex]s' = 0.t + \frac{1}{2}\times 2t^{2}[/tex]

[tex]s' = t^{2}[/tex]                     (1)

At the instant, when Tina passes David, their distances are same, thus:

s = s'

[tex]30t = t^{2}[/tex]

[tex]t(t - 30) = 0[/tex]

t = 30 s

Now,

The distance covered by Tina before she passes David can be calculated by substituting the value t = 30 s in eqn (1):

[tex]s' = 30^{2}[/tex] = 900 m

The distance covered by Tina before passing David at an acceleration rate of  2 m/s² is 900 meters.

Given to us

Velocity of David, v = 30 m/s

Acceleration of Tina, a = 2 m/s²

Let the time taken by Tina pass David is t.

According to the given information, the distance traveled by Tina will be the same as the distance traveled by David between Tina when she was at rest and when Tina passes her.

Distance traveled by Tina = Distance traveled by David

Distance traveled by David,

[tex]s = v \times t\\\\ = 30 \times t =30t[/tex]

Using the second equation of Motion

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

Substitute,

[tex]30t= (0)t +\dfrac{1}{2}(2)t^2[/tex]

[tex]t = 30\rm\ sec[/tex]

Thus, the time taken by Tina to pass David is 30 seconds.

We have already discussed,

Distance traveled by Tina = Distance traveled by David,

therefore,

[tex]s = v \times t\\\\ = 30 \times 30 =900\rm\ meters[/tex]

Hence, the distance covered by Tina before passing David at an acceleration rate of  2 m/s² is 900 meters.

Learn more about the Equation of motion:

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Answer:

25 times greater

Explanation:

Let the mass of the car is m

Initial speed, u = 20 km/h = 5.56 m/s

Final speed, v = 100 km/h = 27.78 m/s

The formula for the kinetic energy is given by

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

So, initial kinetic energy

[tex]K_{i} = \frac{1}{2}m(5.56)^{2}[/tex]

Ki = 15.466 m

final kinetic energy

[tex]K_{f} = \frac{1}{2}m(27.78)^{2}[/tex]

Kf = 385.86 m

Increase in kinetic energy is given by

= [tex]\left ( \frac{K_{f}}{K_{i}} \right )[/tex]

= 385.86 / 15.466 = 25

So, the kinetic energy is 25 times greater.

The kinetic energy of the car increases by a factor of 25.

The increase in kinetic energy of the car can be determined by comparing the initial kinetic energy to the final kinetic energy. Kinetic energy is directly proportional to the square of velocity.

In this case, the velocity is increased from 20 km/hr to 100 km/hr. Let's calculate the ratio of the final kinetic energy to the initial kinetic energy.

The initial kinetic energy is given by 1/2 * (mass of the car) * (initial velocity)^2, and the final kinetic energy is given by 1/2 * (mass of the car) * (final velocity)^2.

Let's substitute the values and calculate the ratio:

Ratio = (1/2 * (mass) * (final velocity)^2) / (1/2 * (mass) * (initial velocity)^2) = (final velocity)^2 / (initial velocity)^2.

Substituting the numbers, Ratio = (100 km/hr)^2 / (20 km/hr)^2 = 10000 / 400 = 25.

Therefore, the factor by which the kinetic energy of the car increases is 25 times greater.

Answer:

Vf = 10.76 m/s

Explanation:

Train kinematics

The train moves with uniformly accelerated movement

[tex]V_f = V_o + a*t[/tex] Formula (1)

Vf: Final speed (m/s)

V₀: Inital speed (m/s)

t: time in seconds (s)

a: acceleration (m/s²)

Movement from t = 0 to t = 5.2s

We replace in formula (1)

4.6 = 0 + a*5.2

a = 4.6/5.2 = 0.88 m/s²

Movement from t = 5.2s to t = 5.2s + 7s = 12.2s

We replace in formula (1)

[tex]V_f = 4.6 + 0.88*7[/tex]

Vf = 10.76 m/s

Answer:

a) [tex]v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s[/tex]

b) [tex]v=+9.97m/s[/tex]

Explanation:

From the exercise we know that

[tex]x_{1} =15m, t_{1}=3s[/tex]

[tex]x_{2} =-3m, t_{1}=1.74s[/tex]

[tex]x_{3} =29m, t_{3}=5.20s[/tex]

From dynamics we know that the formula for average velocity is:

[tex]v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }[/tex]

a) For the three intervals:

[tex]v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s[/tex]

[tex]v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s[/tex]

[tex]v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s[/tex]

b) The average velocity for the entire motion can be calculate by the following formula:

[tex]v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s[/tex]

Answer:

[tex]9.375\times 10^{13}electron[/tex] leave out with a charge of [tex]1.5\times 10^{-5}C[/tex]

Explanation:

We have given total charge [tex]Q=1.5\times 10^{-5}C[/tex]

We know that charge on one electron = [tex]1.6\times 10^{-19}C[/tex]

We have to find the total number of electron in total charge

So [tex]q=ne[/tex], here q is total charge, n is number of electron and e is charge on one electron

So [tex]1.5\times 10^{-5}=n\times 1.6\times 10^{-19}[/tex]

[tex]n=0.9375\times 10^{14}=9.375\times 10^{13}electron[/tex]

So [tex]9.375\times 10^{13}electron[/tex] leave out with a charge of [tex]1.5\times 10^{-5}C[/tex]

Answer:

The distance of c is 56.57

Explanation:

Given that,

Horizontal distance b = 50.071

Angle C = 90.286°

Angle B = 62.253°

We need to calculate the distance of c

Using sine rule

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

Put the value into the formula

[tex]\dfrac{50.071}{\sin62.253 }=\dfrac{c}{\sin90.286}[/tex]

[tex]c= \dfrac{50.071\times\sin90.286}{\sin62.253}[/tex]

[tex]c=56.575[/tex]

Hence, The distance of c is 56.575.

Answer:

area is  131.9223168 square meters

Explanation:

given data

we have given 1420 square feet

to find out

area in square meters

solution

we know that 1 square feet is equal to  0.09290304 square meter

so for 1420 square feet we will multiply 1420 by  0.09290304 square meter

and we get 1420 square feet will be = 1420 × 0.09290304  square meter

1420 square feet  = 131.9223168 square meter

so area is  131.9223168 square meters

Answer:0.931 s

Explanation:

Given

initial speed=1.1 m/s

height(h)=28 m

after 0.5 sec blue ball is thrown upward

Velocity of blue ball is 24.4 m/s

height with which blue ball is launched is 0.9 m

Total distance between two balls is 28-0.9=27.1 m

Let in t time red ball travels a distance of x m

[tex]x=1.1t+\frac{gt^2}{2}[/tex] --------1

for blue ball

[tex]27.1-x=24.4t-\frac{g(t-0.5)^2}{2}[/tex] -----2

Add 1 & 2

we get

[tex]27.1=24.4t+1.1t+\frac{g(2t-0.5)(0.5)}{2}[/tex]

[tex]27.1=25.5t+g\frac{4t-1}{8}[/tex]

t=0.931 s

after 0.931 sec two ball will be at same height

Answer:

The magnetic field at the center of a circular loop is [tex]3.14\times10^{-5}\ T[/tex].

Explanation:

Given that,

Radius = 4.0 cm

Current = 2.0 A

We need to calculate the magnetic field at the center of a circular loop

Using formula of magnetic field

[tex]B = \dfrac{I\mu_{0}}{2r}[/tex]

Where, I = current

r = radius

Put the value into the formula

[tex]B =\dfrac{2.0\times4\pi\times10^{-7}}{2\times4.0\times10^{-2}}[/tex]

[tex]B =0.00003141\ T[/tex]

[tex]B=3.14\times10^{-5}\ T[/tex]

Hence, The magnetic field at the center of a circular loop is [tex]3.14\times10^{-5}\ T[/tex].

Explanation:

Let i, j and k represents east, north and upward direction respectively.

Velocity due north, [tex]v_a=44j\ mi/hr[/tex]

Velocity of the crosswind, [tex]v_w=44i\ mi/hr[/tex]

Velocity of downdraft, [tex]v_d=-22k\ mi/hr[/tex] (downward direction)

(a) Let v is the position vector that represents the velocity of the plane relative to the ground. It is given by :

[tex]v=44i+44j-22k[/tex]

(b) The speed of the plane relative to the ground can be calculated as :

[tex]v=\sqrt{44^2+44^2+22^2}[/tex]

v = 66 m/s

Hence, this is the required solution.

The speed of the plane relative to the ground is computed as 66 mi/hr by taking the square root of the sum of squares of the components of the velocity vector.

The plane's velocity north is given as 44 mi/hr, eastward crosswind as 44 mi/hr, and downdraft velocity as 22 mi/hr downwards.

We can represent these vectors using a coordinate system where north is the positive y-axis, east is the positive x-axis, and down is the negative z-axis. The position vector V (velocity relative to the ground) can be represented as:

V = vnorthi + veastj + vdownk,

where i, j, and k are the unit vectors in the x, y, and z directions respectively. Substituting the given values, we have:

V = 44j + 44i - 22k

The speed of the plane relative to the ground is the magnitude of this vector, which can be calculated using the Pythagorean theorem:

Speed = √(vnorth^2 + veast^2 + vdown^2),

Substituting the given values results in:

Speed = √(44^2 + 44^2 + (-22)^2)

= √(1936 + 1936 + 484)

= √(4356)

= 66 mi/hr.

Answer:

[tex]6.70\times 10^{-13}\ m[/tex]

Explanation:

Given:

Assumptions:

WE know that the radius of the [tex]n^{th}[/tex] orbit of an atom is given by:

[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\[/tex]

Let us find out the radius of the 1st orbit of the gold atom for which n = 1 and Z = 79.

[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\\Rightarrow r = \dfrac{(1)^2(6.62\times 10^{-34})^2}{4\pi^2\times 9\times 10^9\times 79\times (1.6\times 10^{-19})^2\times 9.1\times 10^{-31}}\\\Rightarrow r =6.70\times 10^{-13}\ m[/tex]

Answer:

speed of the car at the bottom of the driveway is 3.8 m/s

Explanation:

given data

mass = 2.1× 10³ kg

distance = 5 m

angle = 20 degree

average friction force = 4 × 10³ N

to find out

find the speed of the car at the bottom of the driveway

solution

we find acceleration a by  force equation that is

force = mg×sin20 - friction force

ma = mg×sin20 - friction force

put here value

2100a = 2100 ( 9.8)×sin20 - 4000

a = 1.447 m/s²

so from motion of equation

v²-u² = 2as

here u is 0 by initial speed and v is velocity and a is acceleration and s is distance

v²-0 = 2(1.447)(5)

v = 3.8

speed of the car at the bottom of the driveway is 3.8 m/s

Explanation:

Given that,

Initial speed of the missile u = 0

Final speed of the missile, v = 4.5 km/s = 4500 m/s

Time taken by the missile, t = 90 s

Let a is the acceleration of the sports car.  It can be calculated using first equation of motion as :

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

[tex]v=at[/tex]

[tex]a=\dfrac{v}{t}[/tex]

[tex]a=\dfrac{4500\ m/s}{90\ s}[/tex]

[tex]a=50\ m/s^2[/tex]

Value of g, [tex]g=9.8\ m/s^2[/tex]

[tex]a=\dfrac{50}{9.8}\ m/s^2[/tex]

[tex]a=(5.10)\ g\ m/s^2[/tex]

So, the acceleration of the missile is [tex](5.10)\ g\ m/s^2[/tex]. Hence, this is required solution.

Final answer:

The average acceleration of the missile is 50 m/s², and the acceleration in multiples of gravity is approximately 5.10 g.

Explanation:

To calculate the average acceleration of an intercontinental ballistic missile (ICBM) given it goes from rest to a suborbital speed of 4.50 km/s (or 4500 m/s) in 90.0 seconds, you can use the formula for average acceleration, which is the change in velocity (Δ v) divided by the change in time (Δ t). The formula is:

a =  Δv / Δt

Using the given information:
Δ v is (final velocity - initial velocity)
Δ v = 4500 m/s - 0 m/s = 4500 m/s
Δ t = 90.0 s

So the average acceleration, a, is:

a = 4500 m/s / 90.0 s = 50 m/s²

To find the acceleration in multiples of g (9.80 m/s²), divide the average acceleration by the acceleration due to gravity:

Acceleration in multiples of g = a / g

Acceleration in multiples of g = 50 m/s² / 9.80 m/s² ≈ 5.10 g

Therefore, the average acceleration of the missile is 50 m/s² and in multiples of gravity it is approximately 5.10 g.

Final answer:

The acceleration of gravity on the Moon is 1.67 m/s².

Explanation:

The weight of an object is determined by the acceleration due to gravity. On Earth, a 60.0 kg person weighs 588 N (60.0 kg × 9.8 m/s²). However, on the Moon, the person weighs 100.0 N. To find the acceleration of gravity on the Moon, we can rearrange the weight formula:

weight = mass × acceleration due to gravity

Using this formula, we can solve for the acceleration due to gravity on the Moon:

acceleration due to gravity = weight / mass = 100.0 N / 60.0 kg = 1.67 m/s²

The acceleration of gravity on the Moon is approximately 1.625 m/s².

To find the acceleration of gravity on the Moon, we use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a), or F = m * a.

In this case, the force is the weight of the person on the Moon, which is given as 100.0 N, and the mass of the person is 60.0 kg.

We can rearrange the equation to solve for the acceleration due to gravity on the Moon:

[tex]\[ g_{\text{moon}} = \frac{F}{m} \][/tex]

Substituting the given values:

[tex]\[ g_{\text{moon}} = \frac{100.0 \, \text{N}}{60.0 \, \text{kg}} \] \[ g_{\text{moon}} = \frac{100.0}{60.0} \, \text{m/s}^2 \] \[ g_{\text{moon}} = 1.666\ldots \, \text{m/s}^2 \][/tex]

Therefore, the acceleration of gravity on the Moon is approximately 1.625 m/s².

Answer:

The two joggers will pass each other after 1 minute and 4 seconds at 3:01:04 PM.

Explanation:

The situation is analogous to two joggers running in opposite direction in a straight line where one jogger starts at the beginning of the line and the other starts at the other end, 300 m ahead.

The equation for the position of the joggers will be:

x = x0 + v · t

Where:

x = position of the jogger at time t

x0 = initial position

v = velocity

t = time

When the joggers pass each other, their position will be the same. Let´s find at which time both joggers pass each other:

x jogger 1 = x jogger 2

0 m + 2.15 m/s · t = 300 m - 2.55 m/s · t

(notice that the velocity of the joggers has to be of opposite sign because they are running in opposite directions).

2.15 m/s · t + 2.55 m/s · t = 300 m

4.70 m/s · t = 300 m

t = 300 m / 4.70 m/s = 63.8 s

The two joggers will pass each other after 1 minute and 4 seconds at 3:01:04 PM.

Answer:

Explanation: This is done using the equation:

[tex]\frac{4}{3} π R^{3}[/tex]

Because the Radius is a know value. We have the following.

[tex]\frac{4}{3} π (10mm)^{3}[/tex]

Which is:

4188.7902 mm

Answer:

0.5 rad / s

Explanation:

Moment of inertia of the disk I₁ = 1/2 MR²

M is mass of the disc and R is radius

Putting the values in the formula

Moment of inertia of the disc  I₁  = 1/2 x 100 x 2 x 2

= 200 kgm²

Moment of inertia of man about the axis of rotation of disc

mass x( distance from axis )²

I₂  = 40 x 1.25²

= 62.5 kgm²

Let ω₁ and ω₂ be the angular speed of disc and man about the axis

ω₂ = tangential speed / radius of circular path

= 2 /1.25 rad / s

= 1.6 rad /s

ω₁ = ?

Applying conservation of angular moment ( no external torque is acting on the disc )

I₁ω₁ = I₂ω₂

200 X ω₁ = 62.5 X 1.6

ω₁ =  0.5 rad / s

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, [tex]q_1=-6.29\times 10^{-6}\ C[/tex]

Second charged particle, [tex]q_2=5.23\times 10^{-6}\ C[/tex]

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

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

[tex]F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}[/tex]

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

Answer:

The angle θ is 6.1° below the horizontal.

Explanation:

Please, see the figure for a description of the situation.

The vector "r" gives the position of the ball and can be expressed as the sum of the vectors rx + ry (see figure).

We know the magnitude of these vectors:

magnitude rx = 11.6 m

magnitude ry = 2.23 m - 0.99 m = 1.24 m

Then:

rx = (11. 6 m, 0)

ry = (0, -1.24 m)

r = (11.6 m + 0 m, 0 m - 1.24 m) = (11.6 m, -1.24 m)

Using trigonometry of right triangles:

magnitude rx = r * cos θ = 11. 6 m

magnitude ry = r * sin θ = 2.23 m - 0.99 = 1.24 m

where r is the magnitude of the vector r

magnitude of vector r:

[tex]r = \sqrt{(11.6m)^{2} + (1.24m)^{2}} = 11.667m[/tex]

Then:

cos θ = 11.6 m / 11.667 m

θ = 6.1°

Using ry, we should obtain the same value of θ:

sin θ = 1.24 m/ 11.667 m

θ = 6.1°

( the exact value is obtained if we do not round the module of r)

After the battery is disconnected, the potential difference between the capacitor plates remains at 9.00V. After a sheet of Teflon is inserted between the plates, the potential difference decreases due to increased capacitance.

Part A: When a parallel-plate capacitor is charged by a battery and then the battery is disconnected, the potential difference between the plates remains the same as it was before disconnecting the battery. In this case, the potential difference would remain 9.00V, as this potential difference is determined by the charge on the capacitor and the capacitance, neither of which changes when the battery is disconnected.

Part B: When a dielectric (in this case Teflon) is inserted between the plates of a charged capacitor without a connected battery, the potential difference between the plates decreases. This is because the dielectric increases the capacitance of the capacitor, causing the potential difference to decrease for a fixed charge. The exact amount of the decrease would depend on the dielectric constant of Teflon.

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The potential difference between the capacitor plates remains 9.00 V after disconnecting the battery. After inserting Teflon, the potential difference decreases to around 4.29 V.

A) The potential difference between the plates will remain at 9.00 V after the battery is disconnected as no charge can leave the plates.

B) Introducing a dielectric such as Teflon with a dielectric constant K will reduce the potential difference.

The formula to calculate the new potential difference (V') is given by [tex]V' = V/K[/tex], where

Thus, the new potential difference will be [tex]V' = 9.00\left V / 2.1 \approx 4.29\left V[/tex]. This occurs because Teflon inserts a dielectric constant that reduces the voltage.

Answer:

The reulting electric field at x = 4.0 and y = 0.0 from the dipole is 0.5612 N/C

Solution:

As per the question:

Charges of the dipole, q = [tex]\pm 2\mu C[/tex]

Separation distance between the charges, d = 0.0010 m

Separation distance between the center and the charge, d' = [tex]\frac{d}{2} = 5\times 10^{- 4} m[/tex]

x = 4.0 m

y = 0.0 m

Now,

The electric field due to the positive charge on the right of the origin:

E = [tex]k\frac{q}{(d' + x)^{2}}[/tex]

where

k = Coulomb's constant = [tex]9\times 10^{9} Nm^{2}C^{- 2}[/tex]

Now,

E = [tex](9\times 10^{9})\frac{2\times 10^{- 6}}{(5\times 10^{- 4} + 4)^{2}} = 1124.72\ N/C[/tex]

Similarly, electric field due to the negative charge:

E' = [tex]k\frac{q}{(x - d')^{2}}[/tex]

E' = [tex](9\times 10^{9})\frac{2\times 10^{- 6}}{(4 - 5\times 10^{4})^{2}} = - 1125.28\ N/C[/tex]

Thus

[tex]E_{total} = E' - E = 0.5612 N/C[/tex]

The electric field due to the dipole at the point x = 4.0 m, y = 0.0 m is -0.56 N/C. Option b

In this problem, we will use the formula for the electric field due to a dipole, E = k * 2p / r^3. Here, k is the Coulomb constant, p is the dipole moment, and r is the distance from the center of the dipole to the point where we want to find the electric field.

First, we need to find the dipole moment, p. The dipole moment is the product of the charge and the separation between the charges, so p = q * d = 2 * 10^-6 C * 0.0010 m = 2 * 10^-9 C.m.

The distance to the point where we want to find the electric field is 4.0 m, so r = 4.0 m. Plugging these values into the formula for the electric field gives us E = (9 * 10^9 N.m^2/C^2) * 2 * 2 * 10^-9 C.m / (4.0 m)^3 = 0.56 N/C.

Because the charges are aligned along the x axis with the positive charge to the right of the origin and we are considering a point to the right of both charges, the electric field will point in the negative x direction. Therefore, the correct answer is (B) -.56 i N/C.

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The question deals with the Doppler effect which occurs with the relative motion between the source of the wave and the observer. The observed change in frequency, as the police car with a siren sounding at 500 Hz passes the truck moving in the opposite direction, is calculated to be approximately 67.5 Hz.

The question you're asking involves the concept of the Doppler effect, which is observed when the frequency of a wave changes because of relative movement between the source of the wave and the observer.

Here, I will explain how to use the formula for the Doppler effect when the source is moving towards the observer:

f' = f0 * (v + v0) / v

And here is the formula when the source is moving away from the observer:

f' = f0 * v / (v + vs)

In these formulae, f' is the observed frequency, f0 is the source frequency (500 Hz), v is the speed of sound (343 m/s), v0 is the observer's speed towards the source (truck's speed = 36 m/s), and vs is the source's speed away from the observer (police car's speed = 45 m/s).

Firstly, as the police car approaches the stationary observer (which is the truck), the formula becomes :

f' = 500 * (343 + 36) / 343

Calculating this gives us an observed frequency of approximately 530.5 Hz.

Then, as the police car moves away from the truck, we use the second formula:

f' = 500 * 343 / (343 + 45)

This gives us an observed frequency of about 463 Hz.

Therefore, the total change in frequency, as heard by the observer in the truck, is approximately 530.5 Hz - 463 Hz, which gives us a change in frequency of approximately 67.5 Hz.

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Final answer:

Using the Doppler effect equation for sound with both the observer and the source moving towards each other, the observed frequency is calculated as 636 Hz. The change in frequency heard by the observer in the truck is 136 Hz.

Explanation:

The scenario described involves the application of the Doppler effect, which is an increase or decrease in the frequency of sound, light, or other waves as the source and observer move toward or away from each other. To solve this problem, we will use the Doppler equation for sound when source and observer are moving in opposite directions towards each other:

f' = f((v + vo) / (v - vs))

where:
f' is the observed frequency,
f is the emitted frequency (500 Hz in this case),
v is the speed of sound in air (343 m/s),
vo is the observer's velocity towards the source (36 m/s, as the truck moves in the opposite direction to the police car),
vs is the source's velocity towards the observer (45 m/s).

Plugging in the values:

f' = 500 Hz ((343 m/s + 36 m/s) / (343 m/s - 45 m/s))
  = 500 Hz ((379 m/s) / (298 m/s))
  = 500 Hz * 1.272
  = 636 Hz

The observed frequency is 636 Hz, so the change in frequency is the observed frequency minus the emitted frequency:

Change in frequency = 636 Hz - 500 Hz = 136 Hz.

Answer:

If the potential is zero , the electric field could be different to zero

Explanation:

The relation between the electric field and the potential is:

=−∇

∇: gradient operator

If the electric potential, , is zero at one point but changes in the neighbourhood of this point, then the Electric field, , at that point is different from zero.

Answer: A little more that 5 Kg for a healthy person

Explanation: First, we know the following:

The regular adult has from 9 to 12 pints of blood. This is around 5 liters for a healthy male adult.

The human body is composed mostly on water, around 80%.

Blood is mostly composed on plasma, which makes blood thicker than water.

Knowing that, almost all the body is compose of water, it is safe to think that blood density should be near to that of water but higher.

The density on water is a know value. Which makes the following true:

1 Liter of Water weights 1 Kg

It could be said then, that the total mass of blood for a healthy person should be a little more that 5 kgs.

Final answer:

Vesna Vulovic's fall of 6 miles and 551 yards converts to approximately 10,159.8324 meters, combining both conversions of miles and yards to meters.

Explanation:

The question asks for the conversion of the distance Vesna Vulovic survived falling without a parachute from miles and yards into meters. To convert 6 miles and 551 yards to meters, we first note that 1 mile equals 1,609.34 meters, and 1 yard equals 0.9144 meters. Therefore, 6 miles convert to 9,656.04 meters (6 x 1,609.34) and 551 yards convert to 503.7924 meters (551 x 0.9144). Adding these two distances together yields a total fall of 10,159.8324 meters.

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

Answer: in air 33.33 ps and in the glass 50 ps: so the difference 16.67 ps

Explanation: In order to calculate the time for a pulse travellin in air  and in a glass we have to consider the expresion of the speed given by:

v= d/t  v the speed in a medium is given by c/n where c and n are the speed of light and refractive index respectively.

so the time is:

t=d/v=d*n/c

in air

t=0.01 m*1/3*10^8 m/s= 33.33 ps

while for the glass

t=0.01 m*1.5* 3* 10^8 m/s= 50 ps

Finally the difference is (50-33.33)ps = 16.67 ps

Answer:

[tex]E_{k}=1589.5ftlb[/tex]  

Explanation:

[tex]E_{k}=E_{movement}+E_{rotational}\\[/tex]    

[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}[/tex]     (1)

For this wheel:

[tex]w=\frac{v}{r}[/tex]

[tex]I=mr^{2}[/tex]:    inertia of a ring

We replace (2) and (3) in (1):

[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}(mr^{2})(\frac{v}{r})^{2}=mv^{2}=5.5*17^{2}=1589.5ftlb[/tex]