A mass attached to a horizontal spring is stretched by 10 cm and released. It takes 0.2 sec for the mass to reach the equilibrium position. The mass is then stretched to 30 cm and released. How long does it take the mass to reach the equilibrium position?

Answer:

The value of resistance will be 42.08 ohm    

Explanation:

We have given capacitance [tex]C=10\mu F=10\times 10^{-6}F[/tex]

Frequency f = 1 kHz = 1000 Hz

Impedance Z = 45 ohm

Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 1000\times 10^{-5}}=15.92ohm[/tex]

We know that impedance is given by [tex]Z=\sqrt{R^2+X_C^2}[/tex]

[tex]45=\sqrt{R^2+15.92^2}[/tex]

Squaring both side [tex]2025=R^2+253.4464[/tex]

R = 42.08 ohm

Final answer:

To change 350 g of ice at -20 degrees Celsius to water at 20 Celsius, you need to calculate the heat required for two processes: warming the ice and melting the ice. The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, and the specific heat capacity of ice is 2.09 J/g°C. The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.

Explanation:

To calculate the heat necessary to change 350 g of ice at -20 degrees Celsius to water at 20 degrees Celsius, we need to consider the energy required for two processes: warming the ice from -20°C to 0°C and then melting the ice at 0°C to form water at 0°C.

The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of ice is 2.09 J/g°C.

The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.

Let's calculate the heat needed for each process and add them together:

Finally, add both values to find the total heat required.

Answer:

[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]

(negative charge)

Explanation:

Hi!

To solve this problem we use Gauss Law for electric fields, which relates the flux of electric field E through a closed surface S with the charge Q enclosed by that surface:

[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0}[/tex]

[tex]\vec{n} = \text{outwards surface normal}\\\epsilon_0 = 8.85*10^{-12}\frac{C^2}{Nm^2}[/tex]

In this problem S is a cube of side length 2cm. The integral is easy because the electric field is uniform in each face, and normal to the face. The total integral is the sum of the integrals in each of the six faces.

[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0} = 3(-500\frac{N}{C} (2cm)^2) + 3(200\frac{N}{C} (2cm)^2)[/tex]

[tex](-1500*4*10^{-4}\frac{Nm^2}{C} + 600*4*10^{-4}\frac{Nm^2}{C} )=-0.36\frac{Nm^2}{C}[/tex]

[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]

Answer:

displacement is 481.3 m  with 40.88° north east

Explanation:

given data

walk 1 = 300 m north = 300 j

walk 2 = 400 m northwest =400 cos(45) i + sin(45) j

walk 3 = 700 m east southeast  = 700 cos(22.5) i- sin(22.5) j

to find out

displacement and direction

solution

we consider here direction x as east and direction y as north

so

displacement = distance 1 + distance 2 + distance 3

displacement = 300 j + 400 cos(45) i + sin(45) j + 700 cos(22.5) i- sin(22.5) j  

we get here

displacement = 363.9 i + 315 j

so magnitude

displacement = [tex]\sqrt{363.9^{2}+315^{2}  }[/tex]

displacement = 481.3 m

and angle will be = arctan(315/369.9)

angle = 0.713494059 rad

angle is 40.88 degree

so displacement is 481.3 m  with 40.88° north east

Answer:

Magnitude of displacement = 481.24 m

Direction = 40.88 degrees north east.

Explanation:

Let east is along real axis and north is along imaginary axis. So,

First walk = d1 = J300 m

Second walk = d2 = -400Cos(45) + J400Sin(45) = (-282.84 + J282.84) m  

Third walk = d3 = 700Cos(22.5) – J700Sin(22.5) = (646.71 – J267.88) m

Total displacement = d = d1 + d2 + d2 = (363.86 + J314.96)m

Magnitude = √((363.86)^2+(314.96)^2) = 481.24 m

Direction = arctan(314.96/363.86) = 40.88 degrees north east.

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

[tex]y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2[/tex].

where y(t) represent the height from the ground. For our problem, the initial height will be:

[tex]y_0 \ = \ 1000 m[/tex].

The initial velocity:

[tex]v_0 = - 20 \frac{m}{s}[/tex],

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

[tex]a=-10\frac{m}{s}[/tex]

So, the equation for our problem its:

[tex]y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2[/tex].

Taking t=6 s:

[tex]y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 120 m - 180 m[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 300 m[/tex].

[tex]y(6 \ s) = \ 700 m [/tex].

So this its the height of the ball 6 seconds after being thrown.

Given a ball thrown straight downward from a 1000 m high cliff with an initial speed of 20 m/s and considering an acceleration due to gravity of -10 m/s², it can be calculated using a kinematics equation that the ball will be 760 m above the ground after 6 seconds.

The subject of this question is in the field of Physics, specifically kinematics. The problem is set up to have an object that is thrown downward with an initial speed and is then under the influence of gravity, causing it to accelerate downward. To calculate how far above from the ground the ball will be at a certain time, we use the equation for position in uniformly accelerated motion which is:

y = y₀ + v₀×t + 0.5×a×t²

Here, y₀ represents the initial height, v₀ is the initial speed, a is the acceleration due to gravity, and t is the time. Since the boy is standing at the edge of a 1000 m cliff, y₀ = 1000 m. The ball is thrown downwards with an initial speed of 20 m/s, so v₀=-20 m/s (we take down as the negative direction). On earth, the acceleration due to gravity is approximately -10 m/s² and a = -10 m/s². The time t = 6 seconds is given in the problem.

Substituting these values into our equation gives y = 1000 + (-20×6) + 0.5×(-10) (6²), which simplifies to y = 760 m. Therefore, the ball is 760 meters above the ground 6 seconds after it was thrown.

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Answer:

E the electric field remains unchanged.

Explanation:

potential difference between the plates of a capacitor

V = Q / C  Where Q is charge on the capacitor and C is capacity of capacitor

Here Q is unchanged.

C the capacity has value equal to ε A / d

Here d is distance between plates. when  it is halved, capacitance C becomes double.

V = Q /C , When C becomes double V becomes half

E = V/d , E is electric field between plates having separation of d.

When V becomes half and d also becomes half , there is no change in the value of E.  

Hence E the electric field remains unchanged.

The magnitude of the electric field between two charged parallel plates remains constant when the plate separation is reduced from d to d/2, provided the charge density on the plates stays the same. Thus, the magnitude of the electric field halfway between the plates would still be E.

The student asked what the magnitude of the electric field would be if the separation between two large, flat, parallel, charged plates is reduced from d to d/2, assuming the charge on the plates remains constant.

According to Essential Knowledge 2.C.5, the electric field (E) between two oppositely charged parallel plates with uniformly distributed electric charge is constant in magnitude and direction at points far from the edges of the plates.

The magnitude of the electric field E between parallel plates is given by the equation E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. Since the surface charge density (σ = Q/A, charge per area) does not change when the separation between the plates is altered and the permittivity of free space (ε₀) is a constant, the magnitude of the electric field between the plates remains the same even if the separation is reduced to d/2.

Therefore, the magnitude of the electric field halfway between the plates when the separation is reduced to d/2 would still be E.

Answer:

net electrostatic force = 46.76 N

and direction is vertical downward

Explanation:

given data

charge =  +5μC

corners identical charges = -6 μC

length of side = 0.10 m

to find out

What is the net electrostatic force

solution

here apex is the vertex where the two sides of equal length meet

so upper point A have charge +5μC and lower both point B and C have charge -6 μC

so

force between +5μC and -6 μC is express as

force  = [tex]k \frac{q1q2}{r^2}[/tex]       ..............1

put here electrostatic constant  k = 9 × [tex]10^{9}[/tex] Nm²/C² and q1 q2 is charge given and r is distance 0.10 m

so

force  = [tex]9*10^{9} \frac{5*6*10^(-12)}{0.10^2}[/tex]

force = 27 N

so net force is vector addition of both force

force = [tex]\sqrt{x^{2}+x^{2}+2x^{2}cos60}[/tex]

here x is force 27 N

force = [tex]\sqrt{27^{2}+27^{2}+2(27)^{2}cos60}[/tex]

net electrostatic force = 46.76 N

and direction is vertical downward

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

Answer:

a)X=70.688m

b)v=18.8m/S

Explanation:

the truck has constant speed while the car moves in uniformly accelerated motion

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

\frac{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve .

For the automobile

we use the ecuation number 3

Xo=0

Vo=0

a=2.5

X=0.5(2.5)t^2

X=1.25t^2

for the truck

x=Vt

X=9.4t

the distance is the same we can match the previous equations

1.25t^2=9.4t

t=9.4/1.25=7.52s

using the ecuation for the truck

x=9.4*7.52=70.688m

b) for this point we can use the ecuation number 1 for automobile

Vf=Vo+at

Vf=0+2.5(7.52)

Vf=18.8m/S

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

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

[tex]F=\frac{kQ\cdot Q}{r^2}[/tex]

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

Now the magnitude of charge is 2Q and is at a distance of [tex]\frac{r}{2}[/tex]

[tex]F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}[/tex]

F'=16F

F'=197.504 N

Answer: a) 1.6 * 10 ^-7 N/C (upward) ; b) -2.5*10^-6N (downward)

Explanation: In order to solve this proble we have to use the Coulomb law given by:

F=q*E from this expression we have

E=F/q=4.8*10^-6/30 C= 1.6 * 10 ^-7 N/C

The force on the particle charge by -1.6 X10 C place intead of the initial charge we have

F=q*E= -16 C* 1.6 * 10 ^-7 N/C= -2.5*10^-6N

Answer:

- 273.77 rad/s^2

Explanation:

fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps

f = 0

ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s

ω = 2 π f = 0

θ = 46 revolutions = 46 x 2π radian = 288.88 radian

Let α be the angular acceleration of the centrifuge

Use third equation of motion for rotational motion

[tex]\omega^{2}=\omega _{0}^{2}+2\alpha \theta[/tex]

[tex]0^{2}=397.71^{2}+2 \times \alpha \times 288.88[/tex]

α = - 273.77 rad/s^2

Answer:

a) 10.2 eV

b) 122 nm

Explanation:

a) First we must obtain the energy for each of the states, which is given by the following formula:

[tex]E_{n}=\frac{-13.6 eV}{n^2}[/tex]

So, we have:

[tex]E_{1}=\frac{-13.6 eV}{1^2}=-13.6 eV\\E_{2}=\frac{-13.6 eV}{2^2}=-3.4 eV[/tex]

Now we find the energy that the electron loses when it falls from state 2 to state 1, this is the energy carried away by the emitted photon.

[tex]E_{2}-E_{1}=-3.4eV-(-13.6eV)=10.2eV[/tex]

b) Using the Planck – Einstein relation, we can calculate the wavelength of the photon:

[tex]E=h\nu[/tex]

Where E is the photon energy, h the Planck constant and  [tex]\nu[/tex] the frequency.

Recall that [tex]\nu=\frac{c}{\lambda}[/tex], Rewriting for [tex]\lambda[/tex]:

[tex]E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.13*10^{-15}eV)(3*10^8\frac{m}{s})}{10.2eV}=1.22*10^{-7}m[/tex]

Recall that [tex]1 m=10^9nm[/tex], So:

[tex]1.22*10^{-7}m*\frac{10^9nm}{1m}=122nm[/tex]

Answer:

7.1 m

Explanation:

Given:

Distance traveled by the student in the first attempt = [tex](2.9 \pm 0.1)\ m[/tex]

Distance traveled by the student in the second attempt = [tex](3.9 \pm 0.2)\ m[/tex]

So, the maximum distance that the student could travel in this attempt = [tex](2.9+0.1)\ m = 3.0\ m[/tex]

So, the maximum distance that the student could travel in this attempt = [tex](3.9+0.2)\ m = 4.1\ m[/tex]

Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.

So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.

And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.

Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.

Answer:

The wavelength of the particle is [tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]

Explanation:

We know that after accelerating across a potential 'V' the potential energy of the charge is converted into kinetic energy as

[tex]\frac{1}{2}mv^{2}=V\cdot e\\\\\therefore v=\sqrt{\frac{2V\cdot e}{m}}[/tex]

now according to De-Broglie theory the wavelength associated with a particle of mass 'm' moving with a speed of 'v' is given by

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

where

'h' is planck's constant

'm' is the mass of particle

'v' is the velocity of the particle

Applying the values in the above equation we get

[tex]\lambda =\frac{h}{m\cdot \sqrt{\frac{2V\cdot e}{m}}}[/tex]

Thus

[tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]

Answer:

Wavelength of the particle is [tex]\frac{h}{\sqrt{2meV}}[/tex]

Solution:

As per the question:

The particle with mass, m and charge, e accelerates through V (potential difference).

The momentum of the particle, [tex]p_{p}[/tex] if it travels with velocity, [tex]v_{p}[/tex]:

[tex]p_{p} = mv_{p}[/tex]

Now, squaring both sides and dividing by 2.

[tex]\frac{1}{2}p_{p}^{2} = \frac{1}{2}m^{2}v_{p}^{2}[/tex]

[tex]K.E =\frac{1}{2m}p_{p}^{2} = \frac{1}{2}mv_{p}^{2}[/tex]

[tex]K.E = \frac{1}{2}mv_{p}^{2}[/tex]

Also,

[tex]p_{p} = \sqrt{2mK.E}[/tex]          (1)

Now, we know that Kinetic energy of particle accelerated through V:

K.E = eV       (2)

where

e = electronic charge = [tex]1.6\times 10^{- 19} C[/tex]

From eqn (1) and (2):

[tex]p_{p} = \sqrt{2mK.E}[/tex]              (3)

From eqn (2) and (3):

[tex]p_{p} = \sqrt{2meV}[/tex]

From the de-Broglie relation:

[tex]\lambda_{p} = \frac{h}{p_{p}}[/tex]      (4)

where

[tex]\lambda_{p}[/tex] = wavelength of particle

h = Planck's constant

From eqn (3) and (4):

[tex]\lambda_{p} = \frac{h}{\sqrt{2meV}}[/tex]

Final answer:

The final momentum of the car pushed by two men with a net force of 690 N for 6.6 seconds is 4554 N·s. This is calculated using the formula for impulse, which is the product of the net force and the time over which the force is applied.

Explanation:

The question is asking to calculate the final momentum of a car after two men have pushed it with a net force of 690 N for 6.6 seconds. According to Newton's second law of motion, the change in momentum (also known as impulse) is equal to the net force multiplied by the time over which the force is applied. The formula for impulse is:

Impulse (J) = Force (F) × Time (t)

By applying this formula, we can find the final momentum:

Therefore, the impulse is:

J = 690 N × 6.6 s = 4554 N·s

Since the car was initially at rest and assuming there is no external resistance, the final momentum of the car will be equal to the impulse given to it:

Final momentum = 4554 N·s

The final momentum of the car, with an applied force of +690 N for 6.6 seconds, is 4554 Ns.  The initial momentum is zero since the car starts from rest.

To determine the final momentum of the car, we will utilize the relationship between force, time, and momentum. According to the impulse-momentum theorem, the impulse applied to an object is equal to the change in its momentum.

Impulse is given by the formula:

Impulse (J) = Force (F) × Time (t)

Given:

Now, calculate the impulse:

J = 690 N × 6.6 s = 4554 Ns

In this scenario, the car starts from rest, which means its initial momentum (p_initial) is 0. Hence, the final momentum ([tex]p_{final}[/tex]) after 6.6 seconds is equal to the impulse:

[tex]p_{final}[/tex] = 4554 Ns

The final momentum of the car is 4554 Ns.

Answer:

Distance, d = 6.75 meters

Explanation:

Given that,

Time taken by the eyes to shut during a hard sneeze, t = 0.32 s

Speed of the car, v = 76 km/h = 21.11 m/s

We need to find the distance covered by the car during this time. The product of speed and the time taken by the car is called the distance covered by the it. Mathematically,

[tex]d=v\times t[/tex]

[tex]d=21.11\times 0.32[/tex]

d = 6.75 meters

So, the car will cover 6.75 meters during that time. Hence, this is the required solution.

Answer:

3.75s

Explanation:

We can use the equations for constant acceleration motion. Let's call x, the total length of the path, then x/2 will be half of path. After falling from rest and reaching the half of its total path, the velocity of the body will be:

[tex]v_f^2 =v_0^2 + 2a(x/2)[/tex]

vf is the final velocity, v0 is the initial velocity, 0m/s because the body starts from rest. a is the acceleration, gravity = 9.81m/s^2 in this case. Now, clearing vf we get:

[tex]v_f=\sqrt{(0m/s)^2 + 2g(x/2)}\\v_f = \sqrt{g*x}

In the second half:

[tex]x/2 = \frac{1}{2}gtx^{2}  + v_ot[/tex]

[tex]x/2 = \frac{1}{2}g*(t)^2 + \sqrt{g*x}*(t)[/tex]

[tex](\frac{1}{2}\frac{(x-gt^2)}{\sqrt{g}t})^2 = x\\\\\frac{1}{4gt^2}(x^2 - 2xgt^2 + g^2t^4) = x\\\\\frac{1}{4gt^2}x^2 - (\frac{2gt^2}{4gt^2}+1)x + \frac{g^2t^4}{4gt^2} = 0\\ \frac{1}{4gt^2}x^2 - \frac{3}{2}x + \frac{gt^2}{4} = 0\\[/tex]

[tex]0.0211 x^2 - 1.5x + 2.97 = 0\\[/tex]

Solving for x, you get that x is equal to 69.2 m or 2.03m. The total time of the fall would be:

[tex]x = \frac{1}{2}gt^2\\t=\sqrt{(2x/g)}[/tex]

Trying both possible values of x:

[tex]t_1 = 3.75 s\\t_2 = 0.64 s[/tex]

t2  is lower than 1.1s, therefore is not a real solution.

Therefore, the path traveled will be 69.2m and the total time 3.75s

Answer:

[tex]r=6.72\times 10^{-13}\ m[/tex]

Explanation:

Let r is the radius of the n = 1 orbit of the gold. According to Bohr's model, the radius of orbit is given by :

[tex]r=\dfrac{n^2h^2\epsilon_o}{Z\pi me^2}[/tex]

Where

n = number of orbit

h = Planck's constant

Z = atomic number (for gold, Z = 79)

m = mass of electron

e = charge on electron

[tex]r=\dfrac{(6.63\times 10^{-34})^2 \times 8.85\times 10^{-12}}{79\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19})^2}[/tex]

[tex]r=6.72\times 10^{-13}\ m[/tex]

So, the radius of the n = 1 orbit of gold is [tex]6.72\times 10^{-13}\ m[/tex]. Hence, this is the required solution.

Answer:

3.825*10^-6 N

Explanation:

As particle 1 and particle 3 has opposite types of charge, particle 3 will be attracted to particle 1. And as particle 2 and 3 has the same sign, they will repel each other. Due to the position of the particles, both the Force that 1 exerts on 3 and the force that 2 exerts on 3, will have the same direction. Now, you need the magnitude. You can use the following expression:

[tex]F_e = K\frac{q_1*q_3}{r^2}[/tex]

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q2 is the charge of the particles, and r is the distance.

Force 1 on 3 is equal to:

[tex]F_e = K\frac{q_1*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.3*10^{-9}C * 7.5*10^{-9}C}{(-0.6m - (-0.3m))^2} = 1.725 * 10^{-6} N[/tex]

Force 2 on 3:

[tex]F_e = K\frac{q_2*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.8*10^{-9}C * 7.5*10^{-9}C}{(0m - (-0.3m))^2} = 2.1 * 10^{-6} N[/tex]

The magnitude of the resultant force is the addition of both forces:

[tex]F_e = 1.725*10^{-6} N + 2.1*10^{-6} N= 3.825 * 10^{-6} N[/tex]

The magnitude of the net force exerted by the two charges on the third charge is 1.38 N.

The magnitude of the net force exerted by q1 and q2 on q3 can be found using Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k(q1 * q2) / r^2

Where F is the force, k is Coulomb's constant (k = 9 x 10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the distance between them.

In this case, q1 = -2.30 nC, q2 = 2.80 nC, q3 = 7.50 nC, y1 = -0.600 m, y3 = -0.300 m, and y2 = 0 m.

Using the formula and substituting the values, we can calculate the force:

F = (9 x 10^9 N * m^2 / C^2) * (q1 * q3) / ((y3 - y1)^2)

F = (9 x 10^9 N * m^2 / C^2) * (-2.30 nC * 7.50 nC) / ((-0.300 m + 0.600 m)^2)

F = -1.38 N

Therefore, the magnitude of the net force exerted by the two charges on the third charge is 1.38 N.

Final answer:

In a tug-of-war where forces are exerted in opposite directions (10 N south and 17 N north), the net force is calculated by subtracting the smaller force from the larger, resulting in a 7 N force towards the north. The correct answer is D.

Explanation:

The net force in a tug of war scenario is the vector sum of all the forces acting on the object. Since force is a vector quantity, it has both magnitude and direction. In this case, the force exerted by one team (17 N north) and the force exerted by the other team (10 N south) are in opposite directions along the same line, so we subtract the smaller force from the larger one to find the net force.

|17 N| - |10 N| = 17 N - 10 N = 7 N north.

Answer:

2.97m

Explanation:

Hi!

To solve this problem we must consider the directions of the electric fields, since both charges are positive each field will point outwards form the sources meaning that the field will point yo the -x direction formpoints along the x-axis to the left of the charges and the field will point to the +x direction for points along the x-axis to the rigth of the sources, having said this, the only place where the fields point to oposite ditections is between them:

In the following diagram X is the 17nC charge and O the 8nC charge, the arrows represent the direction of their respective electric fields and d is an arbitrary distance measured from the origin.

|-------d------|-----5-d------|

X-----------> | <------------O

|---------------5--------------|

At d the electric fields of the charges are:

[tex]E_{X}=k\frac{17nC}{d^{2}} \\E_{O}=-k\frac{8nC}{(5-d)^{2}}[/tex]

And the total field is the sum of both. Since we are looking for a position d at which the total or net electric field is zero we must solve the following equation:

[tex]0=k(\frac{17}{d^{2}} - \frac{8}{(5-d)^2} )[/tex]

Reorganizing terms:

17*(5-d)^2-8*d^2=0

17*25-170d+9d^2=0

Solving for d, we find two solutions

  d1 = 2.9655     d2=15.923

The second solution implies a distance outside the region between the charges for which the equations are no longer valid, since the direction of the 8nC field would be wrong.

So the solution is

 d=2.97 m

Answer:

The value of acceleration equals [tex]0.0556m/s^{2}[/tex]

Explanation:

The  distances covered in each of the three phases are calculated as under

1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)

Using second equation of kinematics

[tex]s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}[/tex]

2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as

[tex]v=u+at\\\\v=0+a\times 300\\\\v=300a[/tex]

Thus distance traveled equals

[tex]s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a[/tex]

3)

The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as

[tex]v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}[/tex]

Now the sum of [tex]s_{1}+s_{2}+s_{3}[/tex] equals 10 kilometers or 10000 meters.

Thus we get

[tex]0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}[/tex]

To solve for the acceleration, we can use the constant acceleration equation and find the distance traveled during each phase.

The constant acceleration equation can be used to solve this problem. The total distance traveled by the train is equal to the sum of the distances traveled during each phase.

In the first phase, the train starts at rest and accelerates for 5 minutes. The distance traveled during this phase can be found using the equation s = 0.5at^2, where s is the distance, a is the acceleration, and t is the time. Since the initial velocity is 0, we can simplify the equation to s = 0.5at^2.

In the second phase, the train travels at a constant speed for 5 minutes. The distance traveled during this phase is equal to the product of the speed and the time.

In the third phase, the train decelerates with a negative acceleration. The distance traveled during this phase can be found using the equation s = ut + 0.5at^2, where u is the initial velocity, a is the acceleration, and t is the time. Since the final velocity is 0, we can simplify the equation to s = ut.

The total distance traveled is given as 10 km. By plugging in the values for the distances and using the given times for each phase, we can solve for the acceleration, a, and find that a = 0.014 km/s^2.

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That statement is false.

Angular velocity is the rate at which that angle changes. As long as the reference line doesn't move, it doesn't matter where it is. The rate at which the angle changes is still the same, constant number.

The angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.

When anything is moving, its velocity tells us how quickly that something's location is changing from a certain vantage point and as measured by a particular unit of time.

When a point travels down a path and completes a certain distance in a predetermined amount of time, its average speed over that time is equal to the distance covered divided by the travel time. A train traveling 100 kilometers in two hours, for instance, is doing it at an average speed of 50 km/h.

The pace at which that angle change is known as the angular velocity. It doesn't matter where the reference line is, as long as it stays put. The rate of change of the angle remains constant and the same.

Therefore, the angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.

To know more about velocity:

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Answer:

[tex]y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}[/tex]

Explanation:

Kinematics equation, with gravity as deceleration

[tex]y =y_{o}-v_{o}*sin(\alpha )*t - 1/2*gt^{2}[/tex]

If the base of the building is taken to be the origin of the coordinates, then :

[tex]y_{o} [/tex] is the initial coordinates of the ball

for t=5.00, the ball strikes the ground ⇒ y=0m

then:

[tex]y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}[/tex]

Final answer:

The initial coordinates of the ball below the horizontal are (0, y0), with y0 representing the height from which the ball was tossed and 0 representing the horizontal origin.

Explanation:

The student's question asks for the initial coordinates of a ball tossed from a window with a given initial velocity and angle relative to the horizontal. Since we're dealing with projectile motion, we need to resolve the initial velocity into its horizontal (x-direction) and vertical (y-direction) components.

The initial horizontal velocity (vx0) can be found using trigonometry: vx0 = v0 × cos(θ), where v0 is the initial velocity of 8.40 m/s and θ is the launch angle of 15.0°. However, since the angle is below the horizontal, the vertical component of the initial velocity (vy0) will be negative: vy0 = v0 × sin(θ), pointing downward.

Given the vertical direction is positive upwards, and taking the base of the building to be at the origin (0,0), the initial coordinates of the ball will be (0, y0), where y0 represents the height of the window from which the ball was tossed.

Final answer:

To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4.

Explanation:

To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4, where T is the temperature, I is the intensity of radiation, σ is the Stefan-Boltzmann constant, and A is the surface area.

Plugging in the given values, we have T = (0.055W/m² / (5.670 x 10^-8 W/(m²K^4) * 4π(5.0 x 10^8m)^2))^1/4.

Answer:

The person should use a convex lens of power -0.625 D.

Explanation:

Given that, the far point of the person, with myopic eye, is 160 cm.

It means that the image of an object, which is at infinity, is formed at this point 160 cm distance away from the person's eye.

If the object is infinity, then the object distance, u = -[tex]\infty[/tex].

Image distance, v = -160 cm.

u and v are taken to be negative because the object and the image of the object are on the side of the eye from where the light is coming.

Let the focal length of the lens which is to be used for the correction of this myopic eye be f, then using lens equation,

[tex]\rm \dfrac 1f = \dfrac 1v-\dfrac 1u=\dfrac 1{-160 }-\dfrac1{-\infty}=\dfrac 1{-160}-0=\dfrac 1{-160}.\\\Rightarrow f=-160\ cm.[/tex]

The negative focal length indicates that the lens should be convex.

The power of the lens is given by

[tex]\rm P=\dfrac{1}{f}=\dfrac{1}{-160\ cm }=\dfrac{1}{-1.6\ m}=-0.625\ D.[/tex]

So, the person should use a convex lens of power -0.625 D.

Answer:

- 1.428 D

Explanation:

A myopic person is able to see the nearby objects clearly but cannot see the far off objects very clearly. It can be cured by using concave lens of suitable power.

far point = 70 cm

v = - 70 cm (position of image from the lens)

u = ∞ (position of object from lens)

Let f be the focal length of the lens

Use lens equation

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{1}{f}=\frac{1}{-70}-\frac{1}{∞}[/tex]

f = - 70 cm

[tex]P=\frac{1}{f}[/tex]

Where, P is the power of lens

P = - 1.428 Dioptre

Thus, the power of lens used is - 1.428 D.

Answer:

(a) 62.57 L

(b) 801.94 kPa

Explanation:

Given:

[tex]n[/tex] = number of moles of gas = 10.3 mol

[tex]P_1[/tex] = initial pressure of the gas = [tex]519\ kPa = 5.19\times 10^5\ Pa[/tex]

[tex]T_1[/tex] = initial temperature of the gas = [tex]361.38^\circ C = (361.38+273)\K = 598.38\ K[/tex]

[tex]T_2[/tex] = final temperature of the gas = [tex]651.6^\circ C = (651.6+273)\K = 924.6\ K[/tex]

[tex]V[/tex] = volume of the tank

R = universal gas constant = [tex]8.314 J/mol K[/tex]

Part (a):

Using Ideal gas equation, we have

[tex]PV=nRT\\\Rightarrow V = \dfrac{nRT}{P}\\\Rightarrow V = \dfrac{10.3\times 8.314\times 598.38}{5.19\times 10^{5}}\\\Rightarrow V = 6.2567\times10^{-2}\ m^3\\\Rightarrow V = 62.567\ L[/tex]

Hence, the volume of the container is 62.567 L.

Part (b):

As the volume of the container remains constant.

Again using ideal gas equation,

[tex]PV=nRT\\\because V,\ n, R\ are\ constant\\\therefore \dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\Rightarrow P_2 = \dfrac{T_2}{T_1}P_1\\\Rightarrow P_2 = \dfrac{924.6}{598.38}\times 519\\\Rightarrow P_2 = 801.94\ kPa[/tex]

Hence, the final pressure of the gas is now 801.94 kPa.

Answer:

75 m /s

Explanation:

Acceleration of ball a = change in velocity / time

=( 90 - 60)/ 3 = 10 m s⁻²

Initial velocity u = 60 m/s

a = 10 m s⁻²

time t = 3

distance traveled s = ut + 1/2 at²

= 60 x 3 + .5 x 10 x 3 x 3

= 225 m

Average velocity = total distance traveled /time

= 225 / 3 = 75 m/s

Answer:

θ = 3.38 degrees

r = 0.48mi

Explanation:

a) Start by converting 1.7 mi into feet.

1mile = 5280 feet,

so 1.7 mi = 8976 ft

[tex]y=rsin\theta,[/tex]

so [tex]\theta = sin^{1] {530}{8976}[/tex]

and θ = 3.38 degrees

b )Now to gain another 150 ft of elevation, again use the

equation y=rsin(θ)

150 = rsin(3.38)

r = 2544.187 ft.

Convert r = 2544.187 ft into miles by dividing by 5280,

and r = 0.48mi