A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11-0m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

Answer:

a)[tex]r_1=1.36R[/tex]

b)[tex]r_2=2.083R[/tex]

Explanation:

Given:

a) when the initial velocity of the projectile is 0.520 times the escape velocity from the earth.

Let r be the radial distance from the earth's surface Let M be the mass of the Earth and R be the radius of the Earth

Now using conservation of Energy at earths surface and at distance r we have

[tex]\dfrac{-GMm}{R}+\dfrac{m(0.52V_e)^2}{2}=\dfrac{-GMm}{r_1}\\\dfrac{-GMm}{R}+\dfrac{m\times 0.52^2\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_1}\\r_2=1.36\ R[/tex]

b) when the Initial kinetic Energy of the projectile is 0.52 times the Kinetic Energy required to escape the Earth

Conservation of Energy we have

[tex]\dfrac{-GMm}{R}+0.52\times KE_{escape}=\dfrac{-GMm}{r_2}\\\dfrac{-GMm}{R}+0.52\times\dfrac{m\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_2}\\r_2=2.083\ R[/tex]

Answer:

15.579 kW

Explanation:

current passing through the wire

[tex]I = \dfrac{P}{V} = \dfrac{690 \times 10^3}{12000} = 57.5 A[/tex]

power loss

[tex]P_{loss} = I^2R = (57.5)^2\times 5 = 16531.25 W[/tex]

the current transmission for 50,000 V is

[tex]I' = \dfrac{P}{V'} = \dfrac{690 \times 10^3}{50000} = 13.8 A[/tex]

power loss

[tex]P_{loss} = I^2R = (13.8)^2\times 5 = 952.2 W[/tex]

wasted power

   =  16531.25 W - 952.2 W

   = 15579.05 W = 15.579 kW

Hence, the power wasted is equal to 15.579 kW

Electric power delivers from power station is the rate of energy transfer per unit time. The less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.

The electric power is the amount of electric energy transferred per unit time. It can be given as,

[tex]P=IV[/tex]

Here, (I) is the current and (V) is the electric potential difference.

Given information-

The power delivers by the power station is 690 kW.

The voltage of the power is 12,000 V.

The resistance of the wire is 5.0 Ω.

As the power delivers by the power station is 690 kW and the voltage of the power is 12,000 V. Thus the current flowing through the wire is,

[tex]I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{12000}\\I=57.5 \rm A[/tex]

The power loss due to the resistance of the wire is 5.0 Ω is,

[tex]P_{loss}=I^2R\\P_{loss}=(57.5)^2\times5\\P_{loss}=16531.25\rm W\\[/tex]

Now the electricity is delivered at 50,000 V rather than 12,000 V.

The power delivers by the power station is 690 kW and the The voltage of the power is 50,000 V. Thus the current flowing through the wire is,

[tex]I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{50000}\\I=13.8 \rm A[/tex]

The power loss due to the resistance of the wire is 5.0 Ω is,

[tex]P_{loss}=I^2R\\P_{loss}=(13.8)^2\times5\\P_{loss}=952.9\rm W\\[/tex]

As when the voltage of the power is 12,000 V the power loss is 16531.25 W and when the voltage of the power is 50,000 V the power loss is 952.9 W.

Thus less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is,

[tex]P_{saved}=16531.25-952.2\\P_{saved}=15579\rm W\\P_{saved}=15.579\rm kW[/tex]

Hence, the amount of less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.

Learn more about the electric power here;

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Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

[tex]v=r\omega[/tex]

[tex]v = 4.50\times110\times\dfrac{2\pi}{60}[/tex]

[tex]v=51.83\ m/s[/tex]

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

The average speed of the helicopter blade tip is 51.83 m/s, and its average velocity over one revolution is 0 m/s because it returns to its starting point.

To calculate the average speed of the blade tip, we use the formula for the circumference of a circle (C = 2πr) and the given rotational speed.

The radius (r) is 4.50 m, and the helicopter blade spins at 110 revolutions per minute (rpm). First, we need to find the distance one tip of the blade travels in one revolution, which is its circumference:

C = 2πr = 2π(4.50 m) ≈ 28.27 m

Then, to find the distance per minute, we multiply the circumference by the number of revolutions per minute:

Distance per minute = 28.27 m × 110 = 3109.7 m/min

Now, to find the average speed in meters per second, we convert the minute to seconds:

Average speed = 3109.7 m/min × (1 min/60 s) ≈ 51.83 m/s

For part (b), the average velocity over one revolution is 0 m/s since the blade tip returns to its starting point, making the displacement over one revolution zero.

Answer:

The wavelength is [tex]5.591\times10^{-15}\ m[/tex]

Explanation:

Given that,

Speed [tex]v= 7.1\times10^{7}\ m/s[/tex]

Mass of proton [tex]m= 1.67\times10^{-27}\ kg[/tex]

We need to calculate the wavelength

Using formula of de Broglie wavelength

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

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

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

Put the value into the formula

[tex]\lambda=\dfrac{6.63\times10^{-34}}{1.67\times10^{-27}\times7.1\times10^{7}}[/tex]

[tex]\lambda=5.591\times10^{-15}\ m[/tex]

Hence, The wavelength is [tex]5.591\times10^{-15}\ m[/tex]

Final answer:

The de Broglie wavelength of a proton traveling at the same speed as electrons in a TV picture tube can be calculated by using the equation λ = h / p, where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the proton. The momentum of the proton can be calculated by multiplying its mass by its velocity. Once the momentum is found, the de Broglie wavelength can be calculated. In this case, the de Broglie wavelength of the proton would be 5.61 * 10^-15 m.

Explanation:

The de Broglie wavelength (λ) of a particle is given by the equation:

λ = h / p

where h is the Planck's constant

(6.62607015*10^-34 Js) and p is the momentum of the particle. To find the de Broglie wavelength of a proton traveling at the same speed as electrons in a TV picture tube, we need to first calculate the momentum of the proton. The momentum (p) is given by

p = mass * velocity

Plugging in the values, we get:

p = (1.67 * 10^-27 kg) * (7.1 * 10^7 m/s) = 1.18 * 10^-19 kg m/s

Now we can calculate the de Broglie wavelength:

λ = (6.62607015 * 10^-34 Js) / (1.18 * 10^-19 kg m/s) = 5.61 * 10^-15 m

Therefore, the de Broglie wavelength of the proton would be 5.61 * 10^-15 m.

Answer:

a) The student has to run for 8.9 s.

b) The student has to run for 47 m.

c) The bus is traveling at 1.5 m/s when the student reaches the bus.

d) No, she will not catch the bus running at 2.00 m/s.

e) The minimum speed the student must have to catch the bus is 3.7 m/s.

f)  The student has to run for 21.3 s if she runs at 3.7 m/s to catch the bus.

g) She has to run 79 m to catch the bus running at 3.7 m/s

Explanation:

The equations for a straight movement are:

With constant acceleration:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a *t

With constant velocity (a = 0):

x = x0 + v * t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

v = velocity

a = acceleration

t = time

a) When the student catches the bus, the position of the bus and the student are the same:

x student = x bus

The student moves with constant speed while the bus has a constant acceleration. If the origin of the reference system is located where the student starts running, then, x0 student = 0 and x0 bus = 40.7 m. Since the bus starts from rest, v0 = 0.

x student = v * t

x bus =  x0 + v0 * t + 1/2 * a * t² = x0 + 1/2 * a * t²

x student = x bus

v * t = x0 + 1/2 * a * t²

Replacing with data:

5.3 m/s * t = 40.7 m + 1/2 (0.168 m/s²) * t²

0 = 40. 7 m - 5.3 m/s * t + 0.084 m/s²¨* t²

Solving the quadratic equation:

t = 8.9 s and t = 54.1 s

We discard the higher value because if the student catches the bus at 8.9 s, she will not catch it again at 54.1 s.

The student has to run for 8.9 s.

b) Using the equation for position of the student:

x = v * t = 5.3 m/s * 8.9 s =47 m

The student has to run for 47 m

c) Using the equation for velocity of the bus:

v = v0 + a * t = 0 m/s + 0.168 m/s² * 8.9 s

v = 1.5 m/s

The bus is traveling at 1.5 m/s when the student reaches the bus

d) The quadratic equation after equalizing the position of the student and the position of the bus would be:

0 = 40. 7 m - 2 m/s * t + 0.084 m/s²¨* t²

If we solve this using the formula to obtain the roots of the parabola we will obtain:

[tex]\frac{-b+\sqrt{b^{2}-4ac} }{2a} = \frac{2+\sqrt{4- 4*40.7*0.084}}{2*0.084}[/tex]

Since the term [tex]\sqrt{4-4*40.7*0.084} = \sqrt{-9,7}[/tex] is not defined in the real numbers, there is no "t" such as the equation of the parabola equals 0. The parabola has no roots. Then, the student will not catch the bus if she runs at 2.00 m/s.

e) The term inside the square root in [tex]\frac{-b+\sqrt{b^{2}-4ac} }{2a}[/tex]

has to be positive or 0, then:

b² - 4* a* c ≥ 0

Notice that "b" is the speed at which the student runs, "a" is 0.084 and "c" is 40. 7 ( see the equation of the parabola obtained in a)). Then:

b² ≥ 4 * a * c

b ≥ √ 4 * a * c  

b ≥ √ 4 * 0.084 * 40.7

b ≥ 3.7 m/s

The minimum speed the student must have to catch the bus is 3.7 m/s.

f) Now we have to solve the quadratic equation obtained in a), but using -3.7 as value of "b". Solving the quadratic equation, we will obtain the values of t = 21.3 s and 22.7 s. Again, we discard the higher value.

The student has to run for 21.3 s if she runs at 3.7 m/s to catch the bus.

g) The distance is given by the equation for the position of the student:

x = v * t = 3.7 m/s * 21.3 s = 79 m.

She has to run 79 m to catch the bus.

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

Answer:

The knights collide 53.0 m from the starting point of sir George.

Explanation:

The equation for the position in a straight accelerated movement is as follows:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

a = acceleration

t = time

The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:

Sir George´s position:

xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²

Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:

xAlfred = 88.0 m + 0 m - 1/2 * 0.200  m/s² * t²

When the knights collide:

xGeorge = x Alfred

1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200  m/s² * t²

0.150 m/s² * t² = 88.0 m - 0.100  m/s² * t²

0.150  m/s² * t² + 0.100  m/s² * t² = 88.0 m

0.250 m/s² * t² = 88.0 m

t² = 88.0 m / 0.250 m/s²

t = 18.8 s

At t = 18.8 s the position of sir George will be

x =  1/2 * 0.300 m/s² * (18.8 s)² = 53.0 m

The question asks for the calculation of the collision point of two knights charging toward each other with different accelerations from a certain distance apart, using principles of kinematics.

The question involves kinematics in one dimension, specifically the calculation of the point of collision of two knights with different accelerations.

Sir George has an acceleration of 0.300 m/s2 and Sir Alfred an acceleration of 0.200 m/s2. They start 88.0 m apart.

To determine the collision point, we set up equations based on the formula for distance covered under constant acceleration from rest, which is s = 0.5 * a * t2, where s is the distance, a is acceleration, and t is time.

Since they begin at the same time and collide at the same time, we have:

For Sir George: sG = 0.5 * 0.300 * t2

For Sir Alfred: sA = 0.5 * 0.200 * t2

As they cover the 88.0 m together, sG + sA = 88.0. Substituting the equations for sG and sA and solving for t can give us the individual distances they covered, and hence, the point of collision relative to Sir George's starting point.

Answer:

D. strengths, weaknesses

Explanation:

Each individual has characteristics that makes them different from one another. Their innate ability or cannot be determined without testing them in various situations.

The various tests which determine a person's aptitude and intelligence. Another aspect to consider is the emotional intelligence of a person i.e., the capability a person has to recognize others' emotions as well as their own emotions.

Testing children in different types of tests which measures their ability in each of them is the only way to know their strength and weaknesses.

Answer:

Explanation:

[tex]x = k\times a^m\times t^n[/tex]

k is constant , it is dimension is zero. Using dimensional unit , we cal write the relation as follows

L = [tex](LT^{-2})^m(T)^n[/tex]

= [tex]L^mT^{-2m+n}[/tex]

Equating power of like items

m=1

-2m+n = 0

n = 2

Answer:

A ) displacement=75.69km

B) Angle[tex]= 28.92^o[/tex]

Explanation:

This is a trigonometric problem:

in order to answer A and B , we first need to know the total displacement on east and north.

the tricky part is when the car goes to the direction northeast, but we know that:

[tex]sin(\alpha )=\frac{opposite}{hypotenuse} \\where:\\opposite=north side\\hypotenuse=distance[/tex]

North'=9.86km

and we also know:

[tex]cos(\alpha )=\frac{adjacent}{hypotenuse} \\where:\\adjacent=east side\\hypotenuse=distance[/tex]

East'=18.54km

So know we have to total displacement

North=28km+North'=37.86km

East=50km+East'=68.54km

To calculate the total displacement, we have to find the hypotenuse, that is:

[tex]Td=\sqrt{North^2+East^2} =75.69km[/tex]

we can find the angle with:

[tex]\alpha = arctg (\frac {North} {East}) \\\\ \alpha = arctg (\frac {37.86} {68.54})= 28.92^o[/tex]

Answer:

Average velocity of the bird is 2.54 m/s.

Explanation:

Given that,

Initial speed of the bird, u = 0

It experiences a displacement of, d = 28 m

Time taken, t = 11 s

We need to find the average velocity of the bird. Let v is the average velocity. Mathematically, it is given by :

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

[tex]v=\dfrac{28\ m}{11\ s}[/tex]

v = 2.54 m/s

So, the average velocity of the bird is 2.54 m/s. Hence, this is the required solution.

Final answer:

The average velocity of the bird is calculated by dividing the total displacement of 28 m by the total time of 11 s, resulting in an average velocity of 2.545 m/s.

Explanation:

To find the average velocity of a bird that accelerates from rest and experiences a displacement of 28 m in 11 s, we use the formula for average velocity, which is total displacement divided by total time. Since the bird starts from rest and moves in one direction, its average velocity will be the same as its average speed.

Average velocity = Total displacement / Total time = 28 m / 11 s = 2.545 m/s.

Therefore, the average velocity of the bird is 2.545 m/s.

Answer: d. 5 m/s^2

Explanation:

Acceleration is the change in velocity in a given time.

a = (30-20)/2 = 5

The ball is travelling in horizontal direction with an acceleration of [tex] \bold{5\ m/ s^{2}} [/tex]

Answer: Option D

Explanation:

The rate at which velocity of an object changes is acceleration i.e., if the velocity of an object is changing, the object is accelerating. The unit of measurement of acceleration is Meter per Seconds Square. Acceleration is the derivative of the changing velocity wrt to time,

[tex]a=\frac{\Delta v}{\Delta t}-----{Eqn} 1[/tex]

Given,  

Initial velocity = 20 m/s [tex](v_{i})[/tex]  

Final velocity = 30 m/s [tex](v_{f})[/tex]

Initial time = 0 s [tex](t_{i})[/tex]

Final time = 2 s [tex](t_{f})[/tex]

From the equation 1,

[tex]\begin{aligned} a &=\frac{\Delta v}{\Delta t}=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} \\ a &=\frac{30 \mathrm{m} / \mathrm{s}-20 \mathrm{m} / \mathrm{s}}{2 \mathrm{s}}=5 \mathrm{m} / \mathrm{s}^{2} \end{aligned}[/tex]

Hence, the acceleration of ball moving horizontally is [tex] \bold{5\ m/s^{2}} [/tex]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

[tex]F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N[/tex]

[tex]T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N[/tex]

[tex]T_x = T*sin(50) = 0.0234 N[/tex]

The electric force is given by the expression:

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

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

[tex]r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m[/tex]

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

[tex]F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}[/tex]

[tex]q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C[/tex]

O 0.247 μC

Answer:

(a) 20 kA (B) [tex]diameter=5.0035\times 10^{-2}m[/tex] (c) length = 29.94 m

Explanation:

We have given power P = 100 KW

Voltage V = 5 V

Current density [tex]10A/m^2=10\times 10^{6}A/m^2[/tex]

(A) We have to calculate current

We know that power P = voltage × current

So [tex]100\times 1000=5\times i[/tex]

[tex]i=20kA[/tex]

(b) We know that current density is given by

Current density [tex]=\frac{current}{area}[/tex]

So [tex]10\times 10^{6}A/m^2=\frac{20\times 10^3}{area}[/tex]

[tex]area=2\times 10^{-3}m^2[/tex]

We know that [tex]a=\pi r^2[/tex]

[tex]2\times 10^{-3}=3.14\times r^2[/tex]

[tex]r^2=6.34\times 10^{-4}[/tex]

[tex]r=2.517\times 10^{-2}m[/tex]

[tex]d=2 r=2\times 2.517\times 10^{-2}=5.035\times 10^{-2}m[/tex]

(c) We have to find the length of the cable

We know that resistance [tex]R=\frac{V}{I}=\frac{5}{20\times 10^3}=0.25mohm[/tex]

Resistance is given by [tex]R=\rho \frac{l}{A}[/tex]

So [tex]0.25\times 10^{-3}=1.67\times 10^{-8}\frac{l}{2\times 10^{-3}}[/tex]

l = 29.94 m

Answer:

[tex]C=\epsilon_{o}\frac{\pi R^_{2}}{d}[/tex]

This result is a very good approximation, because R>>d

Explanation:

The capacitance for two parallel flat plates, with surface S, space between them d, is:

[tex]C=\epsilon_{o}\frac{S}{d} =\epsilon_{o}\frac{\pi R^_{2}}{d}[/tex]

This calculation is based on the fact that the electric field is constant between the plates. This only happens if the area of the plates is much larger than the distance between them: S>>d, i.e. R>>d. If we assume these factors, the result has a very good approximation.

Answer:

Fx = 22N

Explanation:

There are 2 possible scenarios for this problem:

1.- The 10N force is in the same direction of the acceleration. In this case the other force would be:

[tex]F1 - Fx = m*a[/tex]     where F1 = 10N, m=6kg, a = 2m/s2

[tex]Fx = F1 - m*a = -2N[/tex]    The negative result tells us that this is not possible.

2.- The 10N force is in the opposite direction of the acceleration. In this case the other force would be:

[tex]Fx - F1 = m*a[/tex]     [tex]Fx = m*a + F1 = 22N[/tex]

The greatest magnitude of the other force, by which the object accelerates under the application of two forces is 22 N.

Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

[tex]\sum F=ma[/tex]

Here, (m) is the mass of the body and (a) is the acceleration.

One of the forces is 10.0N. . Let suppose the magnitude of the other force is [tex]F_x[/tex],

The maximum force we get when both the forces acting on the opposite direction. For this system the summation of force will be,

[tex]\sum F=F_x+(-10)\\\sum F=F_x-10[/tex]

The mass of the object is 6 kg. As the body is accelerating in at 2 m/s. Thus, plug in the values in the above formula as,

[tex]F_x-10=2\times6\\F_x=22\rm N[/tex]

Thus the greatest magnitude of the other force, by which the object accelerates under the application of two forces is 22 N.

Learn more about the Newton’s second law of motion here;

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Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5

Explanation:  In order to solve this problem  we have to use the gaussian law in the mentioned regions.

Region 1; 0<r<2

∫E.ds=Qinside the gaussian surface/ε0

inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.

Region 2; 2<r<4;

E.4*π*r^2=8,84/ε0

E=8,84/(4*π*ε0*r^2)

Region 3; 4<r<5

E=0 because is inside the conductor.

Finally

Region 4; r>5

E.4*π*r^2=(8,84-2.02)/ε0

Answer: Option (D) is correct.

Explanation:

From the given options, we can state that mass is not a vector quantity. Mass is considered as a scalar quantity. In other words, mass can be referred to as from how much matter a subject is made of. Mass has magnitude but it does not tend to give any indications about the direction of the subject in frame.

The work done on a particle by the force in the +x-direction from x = x0 to infinity can be calculated using the formula Work = ∫(b/x^n) dx. The limits of integration for the integral are from x0 to infinity.

To calculate the work done by the force in the +x-direction, we can use the formula:

Work = ∫F dx = ∫(b/x^n) dx

Since the force is in the +x-direction and the particle moves along the x-axis from x = x0 to infinity, the limits of integration for the integral are from x0 to infinity.

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Using the formula for distance covered in free fall, the coin was determined to have been dropped from a height of 19.6 meters.

To determine the height from which a coin was dropped if it reaches the ground in 2 seconds, we can use the formula for the distance covered in free fall (neglecting air resistance), which is given by:

Distance (d) = ½ * g * t2

where g is the acceleration due to gravity (9.8 m/s2 on Earth), and t is the time in seconds. Plugging in the values, we get:

Distance (d) = ½ * 9.8 * (22) = ½ * 9.8 * 4 = 19.6 meters.

Thus, the coin was dropped from a height of 19.6 meters.

The coin was dropped from a height of 19.62 meters, using the equation [tex]\( s = \frac{1}{2}gt^2 \)[/tex].

To solve this problem, we can use the equation of motion for an object in free fall under gravity:

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

Where:

- [tex]\( s \)[/tex] is the displacement (height) of the object

- [tex]\( u \)[/tex] is the initial velocity (which is 0 because the object is dropped)

- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex] on Earth)

- [tex]\( t \)[/tex] is the time taken

Given that the initial velocity [tex]\( u = 0 \)[/tex], we can simplify the equation to:

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

Substituting the known values:

[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (2 \, \text{s})^2 \][/tex]

[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times 4 \, \text{s}^2 \][/tex]

[tex]\[ s = \frac{1}{2} \times 39.24 \, \text{m} \][/tex]

[tex]\[ s = 19.62 \, \text{m} \][/tex]

So, the height from which the coin was dropped is [tex]\( 19.62 \, \text{m} \)[/tex].

Answer:

The width of the central bright fringe is 7.24 mm.

Explanation:

Given that,

Wavelength = 632.8 nm

Width d= 0.350 mm

Distance between screen and slit D= 2.00 m

We need to calculate the distance

Using formula of distance

[tex]y_{m}=\dfrac{\lambda D}{d}[/tex]

Put the value into the formula

[tex]y_{m}=\dfrac{632.8\times10^{-9}\times2.00}{0.350\times10^{-3}}[/tex]

[tex]y_{m}=3.62\ mm[/tex]

We need to calculate the width of the central bright fringe

Using formula of width

[tex]width = 2\times|y_{m}|[/tex]

Put the value into the formula

[tex]width=2\times3.62[/tex]

[tex]width = 7.24\ mm[/tex]

Hence, The width of the central bright fringe is 7.24 mm.

The width of the central maximum on the screen, achieved by sending a Helium-neon laser light through a 0.350-mm-wide single slit and projected on a screen 2.00 m from the slit, would be approximately 3.61 mm.

To respond to this question, we need to apply the formula for a single-slit diffraction pattern, which is y = Lλ /w, where y represents the distance from the central maximum to the first minimum (i.e., the width of the central maximum), L is the distance from the slit to the screen, λ  is the wavelength, and w is the width of the slit.

The values provided in the problem are L = 2.00 m, λ = 632.8 nm = 632.8 x 10-9 m, and w = 0.350 mm = 0.350 x 10-3 m. Substituting these into our formula gives us:

y = (2.00 m)(632.8 x 10-9 m) / (0.350 x 10-3 m) ≈ 0.00361m or 3.61mm

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Answer:

a)[tex]E=2.88*10^{-13}N/C[/tex]

b)[tex]E=1.44*10^{-13}N/C[/tex]

c)[tex]F=4.61*10^{-32}N[/tex]

Explanation:

The definition of a electric field produced by a point charge is:

[tex]E=k*q/r^2[/tex]

a)Electric Field due to the alpha particle:

[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(100)^2=2.88*10^{-13}N/C[/tex]

b)Electric Field  due to the electron:

[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(100})^2=1.44*10^{-13}N/C[/tex]

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force magnitude but with opposite direction:

[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(100)^2=4.61*10^{-32}N[/tex]

Answer:

Metals have more mass per unit of volume than wood.

Explanation:

Density is defined as the amount of matter contained in a unit of volume. A material that is denser than another will therefore have more mass per unit of volume.

The density of a body can affect buoyancy, bodies with low enough densities can float in fluids. For example wood can float in water while metals can't.

Answer:

Explanation:

in the electric field electron will face a force which will create an acceleration

( here - ve ) as follows

Force on electron

= charge on electron x electric field  = Q X E

acceleration=  Force / mass = Q E / m

mass of electron = 9.1 x 10⁻³¹

acceleration a = [tex]\frac{1.6\times10^{-19}\times 9900}{1.67\times10^{-27}}[/tex]

= 17.4 x 10¹⁴ ms⁻² .

Now initial velocity  u = 5 x 10⁶ m/s

Final velocity v = 0

acceleration a = 17.4 x 10¹⁴ ms⁻²

distance of travel = s

v² = u² - 2as

0 = (5 x 10⁶)² - 2 x 17.4 x 10¹⁴ s

s = 7.18 mm

2 ) v = u - at

0 = 5 x 10⁶ - 17.4 x 10¹⁴ t

t = .287 x 10⁻⁸ s

Total time elapsed  = 2 x .287 x 10⁻⁸

= .57 x 10⁻⁸ s .

Answer:

option C

Explanation:

the correct answer is option C

Skater is standing on the frictionless surface when her friend through Frisbee at her he starts moving in the backward direction as there is no friction acting to stop him.  

but when again the skater through Frisbee back to his friend there is backward force which will increase the velocity of the skater.

so, momentum is directly proportional to velocity so, velocity increases momentum also increases.

Final answer:

The largest momentum transfer to the skater happens when she catches and then throws the Frisbee back to her friend, utilizing the conservation of momentum.

Explanation:

The largest momentum transferred to the skater occurs in the scenario where the skater catches the Frisbee, holds it momentarily, and then throws it back to her friend. In this case, not only is the momentum of the incoming Frisbee transferred to the skater when caught, but additional momentum is transferred when she throws it back due to the conservation of momentum.

Answer:

(a). The average speed of the car is 12.75 m/s

(b). The distance is 43.35 m.

Explanation:

Given that,

Initial speed = -25.5 m/s

Final speed = 0

Time = 3.4 s

(a). We need to calculate the average speed

Using formula of average speed

[tex]v_{avg}=\dfrac{v_{f}-v_{i}}{2}[/tex]

Put the value into the formula

[tex]v_{avg}=\dfrac{0-(-25.5)}{2}[/tex]

[tex]v_{avg}=12.75\ m/s[/tex]

(b). We need to calculate the acceleration

Using equation of motion

[tex]v_{f}=v_{i}+at[/tex]

[tex]a =\dfrac{v_{f}-v_{i}}{t}[/tex]

[tex]a=\dfrac{-25.5-0}{3.4}[/tex]

[tex]a=-7.5\ m/s^2[/tex]

We need to calculate the distance

Using equation of motion

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

[tex]s=25.5\times3.4+\dfrac{1}{2}\times(-7.5)\times(3.4)^2[/tex]

[tex]s=43.35\ m[/tex]

Hence, (a). The average speed of the car is 12.75 m/s

(b). The distance is 43.35 m.

Answer:

Explanation:

A body has zero velocity and non- zero acceleration. It is possible as at the time when a body thrown upwards is at the top of the height. In that case velocity is zero but acceleration is equal to g.

A body travels with northward velocity but southward acceleration . it is also possible as in case when a body is going with northward velocity  but when break is applied acceleration becomes  southward.

A body travels with northward velocity and northward  acceleration . it is also possible as in case when a body is going with northward velocity  and  when accelerator  is applied . Acceleration becomes northward.

A body travels with constant velocity and a constant non zero acceleration . It is not possible . When there is acceleration , there must be a a change in velocity either in terms of magnitude or direction or both.

A body travels with a constant acceleration and a time varying velocity. It is possible. Time varying velocity represents acceleration.

In physics, a constant non-zero acceleration would necessitate a change in velocity over time. Therefore, it's impossible for a body to have a constant velocity and a constant non-zero acceleration at the same time.

The situation that is NOT possible is a body traveling with a constant velocity and a constant non-zero acceleration.

Acceleration is the rate at which an object's velocity changes. If the velocity of the body remains constant, it means there is no change in its speed or direction. Therefore, it is not possible for the body to have a non-zero acceleration while traveling with a constant velocity.

On the other hand, scenarios with zero velocity and non-zero acceleration or a body traveling with a northward velocity and a northward acceleration are possible.

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Answer:

(a) angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side

b] largest order = 4

Explanation:

(a) for diffraction maxima,

[tex]sin \theta =m\times \lambda/d[/tex]

Here, m is the order, [tex]\lambda[/tex] is the wavelength, [tex]\theta[/tex]  is the angle at which maxima occur, d is inter planar spacing.

And we know that lines per mm (N) is related with d as,

[tex]N=\frac{1}{d}[/tex]

Given that the wavelength is,

[tex]\lambda=600.0 nm=600\times 10^{-9}m[/tex]

And [tex]N=\frac{400 lines}{mm} \\N=\frac{400 lines}{10^{-3}m }[/tex]

Now,

[tex]sin \theta =m\times \lambda\times N[/tex]

Therefore,

[tex]sin \theta= m\times600\times 10^{-9} \times 400\times 10^{3}\\sin \theta=0.24m[/tex]

Here, m can be 1,2,3,4 as sin theta has to be less than 1.

[tex]\theta = arcsin 0.24 , arcsin 0.48 , arcsin 0.72 , arcsin 0.96[/tex]

Therefore, angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side

b] largest order = 4

Final answer:

The angles at which the maxima are observed can be determined using dsinθ = mλ. For a grating with 400 lines per mm and light of wavelength 600.0 nm, the largest order that can be seen occurs when the angle of the next order is equal to or greater than the angle of the first order maximum.

Explanation:

The angles at which the maxima are observed can be determined using the formula:

dsinθ = mλ

Where d is the grating spacing, θ is the angle, m is the order of the maximum, and λ is the wavelength of the light.

For a grating with 400 lines per mm and light of wavelength 600.0 nm:

d = 1/400 mm = 0.0025 mm = 0.0025 cm

θ = sin-1((mλ)/d)

To determine the largest order that can be seen, we need to find the angle at which the next order would overlap with the first order. This would be when the angle of the next order is equal to or greater than the angle of the first order maximum.

Answer:

8.75 m/s^2

Explanation:

initial speed, u = 300 m/s

final speed, v = 400 m/s

distance, h = 4 km = 4000 m

Let a be the acceleration of the plane.

Use third equation of motion

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

[tex]400^{2}=300^{2}- 2 \times a \times 4000[/tex]

a = 8.75 m/s^2

Thus, the acceleration of the plane is 8.75 m/s^2.

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume  x density = 1.35 x 10^8 x 10^-3 x 1000

                                                               = 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = [tex]P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W[/tex]

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

[tex]\eta =\frac{Power output}{Power input}[/tex]

[tex]\eta =\frac{716}{83475}=0.858[/tex]

Thus, the efficiency is 85.8 %.