An amplitude modulation transmitter radiates 20 KW. If the modulation index is 0.7, find the magnitude of the carrier power?

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

The force on a charged particle in a magnetic field is calculated using the Lorentz force equation. For the given particle with charge -1.24 x 10⁻⁹ C and velocity vector, the force is determined by the cross product of the velocity and the magnetic field, taking the charge into account.

A particle with a charge of -1.24 x 10⁻⁹ C is moving with an instantaneous velocity of (4.19 X 10⁴ m/s)î + (-3.85 X 10⁴ m/s)ᴇ. To find the force exerted on this particle by a magnetic field we use the Lorentz force equation, which is F = q(v x B), where F is the force on the particle, q is the charge, v is the velocity of the particle, and B is the magnetic field.

For part (a) where the magnetic field B is (1.40 T)î, the velocity vector v is perpendicular to B since v has no i-component, thus the force can be found simply by calculating the magnitude as q*v*B since sin(θ) is 1 for θ = 90°. The direction of the force is given by the right hand rule, considering that both v and B are vectors and the charged particle is negatively charged.

For part (b) where the magnetic field B is (1.40 T)ᴅ, the velocity vector v has no k-component, thus v is perpendicular to B and the same principle applies.

The magnitude of the force in both scenarios is computed using the charge, the magnitude of velocity (which is the combination of both the i and j components), and the magnitude of the magnetic field. The direction will differ based on the cross product between v and B.

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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:

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:

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:

216.97 X 10⁻⁵ N

Explanation:

Charge q₁ and q₂ will attract q₃ with force F₁ and F₂ .F₁ and F₂ will be calculated as follows

F₁ = [tex]\frac{9\times10^9\times8\times7.7\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]

F₁ = 82.01 X 10⁻⁵ N

F₂= [tex]\frac{9\times10^9\times8\times15.4\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]

F₂ = 164.02 X 10⁻⁵ N

F₁ and F₂ will act at 60 degree so their resultant will be calculated as follows

R² = (82.01 X 10⁻⁵)² +( 164.02 X 10⁻⁵ )² + 2 X 82.01 X 164.02 X 10⁻¹⁰ Cos 60

R² = 47079.48 X 10⁻¹⁰

R = 216.97 X 10⁻⁵ N

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.

Answer:

A. 30 kg

Explanation:

As we know that,

[tex]1 litre=1000cm^{3}[/tex]

And 1 litre is equivalent to 1kg.

Given that, The volume of the rectangular container is,

[tex]V=20cm\times 30cm\times 50cm\\V=30000cm^{3}[/tex]

And this volume will be equal to [tex]V=30000cm^{3}=30litres[/tex]

And this litres in kg will be equal to,

[tex]V=30litres=30 kg[/tex]

Therefore the mass of this volume of water is 30 kg.

Answer:

[tex]\partial \theta = 0.003[/tex]

Explanation:

we know that

[tex]sin\theta = \frac{3.8}{146.4}[/tex]

[tex]\theta = sin^{-1} \frac{3.8}{146.4}[/tex]

[tex]\theta = 1.484°[/tex]

[tex]\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians[/tex]

as we see that [tex]sin\theta = \theta[/tex]

relative error[tex] \frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}[/tex]

Where X_1 IS HEIGHT OF ROCK

[tex]X_2[/tex] IS THE HEIGHT OF ROAD

[tex]\partial X[/tex] = uncertainity in measuring  distance

[tex]\partial X = 0.05[/tex]

Putting all value to get uncertainity in angle

[tex]\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}[/tex]

solving for [tex]\partial \theta[/tex] we get

[tex]\partial \theta = 0.003[/tex]

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 %.

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:

option C

Explanation:

given,

force act on west  = 20 lb

force act at 45° east of north = 80 lb

magnitude of force = ?

∑ F y  =  80 cos 45⁰

    F y =  56.57 lb

magnitude of forces in x- direction

∑ F x = -20 + 80 sin 45⁰

        = 36.57 lb

net force

F = [tex]\sqrt{F_x^2+F_y^2}[/tex]

F = [tex]\sqrt{56.57^2+36.57^2}[/tex]

F = 67.36 lb≅ 67 lb

hence, the correct answer is option C

Answer:

Number of electrons in the system = 973.

Total mass of the system = [tex]\rm 1.064\times 10^{-24}\ kg.[/tex]

Explanation:

Assumptions:

Given:

Since, the system is comprised of electrons and protons only, therefore,

[tex]\rm N = n_e+n_p\\n_p=N-n_e\ \ \ \ \ \ ................\ (1).[/tex]

The net charge on the system can be written in terms of charges on electrons and protons as

[tex]\rm q=n_eq_e+n_pq_p\ \ \ \ \ ...................\ (2).[/tex]

Putting the value of (2) in (1), we get,

[tex]\rm q=n_eq_e+(N-n_e)q_p\\q=n_eq_e+Nq_p-n_eq_p\\q=n_e(q_e-q_p)+Nq_p\\n_e(q_e-q_p)=q-Nq_p\\n_e=\dfrac{q-Nq_p}{q_e-q_p}\\=\dfrac{-5.3756\times 10^{-17}-1610\times 1.6\times 10^{-19}}{-1.6\times 10^{-19}-1.6\times 10^{-19}}=972.98\\\Rightarrow n_e\approx 973\ electrons.[/tex]

It is the number of electrons in the system.

Therefore, the number of protons is given by

[tex]\rm n_p = N-n_e=1610-973=637.[/tex]

The total mass of the system is given by

[tex]\rm M=n_em_e+n_pm_p\\=(973\times 9.11\times 10^{-31})+(637\times 1.67\times 10^{-27})\\=1.064\times 10^{-24}\ kg.[/tex]

Final answer:

The system consists of 336 electrons and 1274 protons, given the net charge. To calculate the total mass, multiply the number of each particle by its respective mass and then sum the results., which gives 5192.362×10−31 kg as answer.

Explanation:

To find the number of electrons in a system with a net charge, we can use the formula:

Given that the net charge of the system is -5.376×10−17 C and the charge of an electron is approximately -1.6022×10−19 C, we can calculate the total number of electrons using the formula:

Number of electrons = (-5.376×10−17 C) / (-1.6022×10−19 C/electron)

Number of electrons = 3.3555×102

Since we can’t have a fraction of an electron, we round to the nearest whole number, which is 336 electrons.

To determine the mass of the system, first, we need to find the number of protons, which would be 1610 - 336 electrons = 1274 protons. Now we multiply the number of protons by the mass of a proton and the number of electrons by the mass of an electron to get the total mass:

Mass of electrons = 336 × 9.11×10−31 kg

Mass of protons = 1274 × 1.673×10−27 kg

Adding these two gives the total mass of the system, which is 5192.362×10−31 kg.

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:

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:

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:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

[tex]R =u\cos\theta t [/tex]

Put the value into the formula

[tex]\dfrac{230}{6} = u\cos\theta[/tex]

[tex]u\cos\theta=38.33[/tex].....(I)

We need to calculate the height

Using vertical component

[tex]H=u\sin\theta t-\dfrac{1}{2}gt^2[/tex]

Put the value in the equation

[tex]16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2[/tex]

[tex]u\sin\theta=\dfrac{16+9.8\times18}{6}[/tex]

[tex]u\sin\theta=32.06[/tex].....(II)

Dividing equation (II) and (I)

[tex]\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}[/tex]

[tex]\tan\theta=0.8364[/tex]

[tex]\theta=\tan^{-1}0.8364[/tex]

[tex]\theta=39.90^{\circ}[/tex]

(a). We need to calculate the initial speed

Using equation (I)

[tex]u\cos\theta\times t=38.33[/tex]

Put the value into the formula

[tex]u =\dfrac{230}{6\times\cos39.90}[/tex]

[tex]u=49.96\ m/s[/tex]

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Answer:

19.2 m/s

Explanation:

We set a frame of reference with the origin at the cliff top and the positive X axis pointing down.

Then the initial position is:

X0 = 0

The initial speed is:

V0 = -4 m/s

It is negative because it is speed upwards and the frame of reference is positive downwards.

Since the diver is in free fall, he is affected only by the acceleration of gravity, we can consider him moving under constant acceleration:

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

He will hit the water at X = 18 m, so:

18 = 0 - 4 * t + 1/2 * 9.81 * t^2

4.9 * t^2 - 4 * t - 18 = 0

Solving this equation electronically:

t = 2.37 s

The diver will hit the water 2.37 s after jumping.

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

V(2.37) = -4 + 9.81 * 2.37 = 19.2 m/s

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)[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:

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:

[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:

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:

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]

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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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) 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:

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:

(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.

When a charged rod is brought near a neutral metal sphere, the opposite charges are attracted towards the rod, causing an initial attraction. However, when the sphere touches the rod, the charges redistribute, leading to like charges repelling each other with a stronger force than the attraction between opposite charges, resulting in the sphere being repelled.

When a charged rod is brought near a neutral substance, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction.

Thus, when a positively charged glass rod is brought close to the neutral metal sphere, the opposite charges in the metal sphere are attracted towards the rod. However, when the sphere touches the rod, it suddenly flies away from the rod. This is because the charges in the metal sphere redistribute, and now the like charges repel each other with a stronger force compared to the attraction between the opposite charges.

This repulsion causes the metal sphere to be repelled away from the rod, creating a net repulsive force.

The metal sphere is first attracted to the positively charged glass rod due to induction, and then repelled after touching the rod due to the repulsion between like charges.

When the positively charged glass rod is brought close to the uncharged metal sphere, the sphere experiences electrostatic induction. The electrons in the metal sphere are repelled by the positive charges on the rod and move to the far side of the sphere, leaving the side closest to the rod with a net positive charge. This positive side of the sphere is attracted to the negative charges in the rod (or the electrons in the rod are attracted to the positive side of the sphere), causing the sphere to be drawn toward the rod.

As the sphere gets closer to the rod, the attraction increases until they make contact. When the sphere touches the rod, electrons from the rod flow onto the sphere, as the rod has a surplus of electrons due to its negative induction on the side opposite the sphere. This transfer of electrons results in the sphere acquiring a net negative charge, as it gains more electrons.

Once the sphere has the same negative charge as the side of the rod it touched, the electrostatic force of repulsion between the like charges takes over. The negatively charged sphere is now repelled by the negatively charged side of the glass rod. This repulsion is strong enough to overcome the gravitational force and the nylon thread's tension, causing the sphere to suddenly fly away from the rod.

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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