Two tiny conducting spheres are identical and carry charges of -23.0C and +67.2C. They are separated by a distance of 3.18 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.18 cm. Determine the magnitude of the force that each sphere now experiences.

Answer:

Magnitude of change in momentum = 4.65 kg.m/s

Magnitude of impulse = 4.65 kg.m/s

Magnitude of the average force applied by the bat = 1550 N

Explanation:

Mass of the cricket ball, m = 0.155 kg

Initial velocity of the ball, u = 35.0 m/s

final velocity of the ball after hitting the bat, v = 65.0 m/s

Time of contact, t = 2.00 ms = 2.00 × 10⁻³ s

Now,

Magnitude of change in momentum = Final momentum - Initial momentum

or

Magnitude of change in momentum = ( m × v ) - ( m × u )

or

Magnitude of change in momentum = ( 0.155 × 65 ) - ( 0.155 × 35 )

or

Magnitude of change in momentum = 10.075 - 5.425 = 4.65 kg.m/s

Now, Magnitude of impulse = change in momentum

thus,

Magnitude of impulse = 4.65 kg.m/s

Now,

magnitude of the average force applied by the bat = [tex]\frac{\textup{Impulse}}{\textup{Time}}[/tex]

or

magnitude of the average force applied by the bat = [tex]\frac{\textup{4.65}}{\textup{3}\times\textup{10}^{-3}}[/tex]

or

Magnitude of the average force applied by the bat = 1550 N

Answer:

T = (5 ×3) + (5 × 9.8 ) N    =   64 N

so option D is correct

Explanation:

given data

mass = 5 kg

acceleration = 3 m/s²

to find out

tension

solution

we know acceleration on bucket is upward so

we say

T - mg = ma

here T is tension , m is mass and g is acceleration due to gravity = 9.8 m/s² and a is acceleration so

T = ma + mg

put here value

T = (5 ×3) + (5 × 9.8 ) N = 64 N

so option D is correct

Answer:

velocity of second billiard after collision = 1.4 m/sec  

Explanation:

We have mass of billiard let [tex]m_1=0.5kg[/tex]

Velocity of billiard let [tex]v_1=1.5m/sec[/tex]

Mass of second second billiard let [tex]m_2=0.5kg[/tex]

Velocity of second billiard before collision 0 m/sec  billiard

Velocity of first billiard after collision  0.1 m/sec

Now according conservation of momentum

Momentum before collision and after collision will be same

So momentum before collision = momentum after collision

[tex]0.5\times 1.5+0.5\times 0=0.5\times 0.1+0.5\times velovity\ of\ second\ biliard\ after\ collision[/tex]

So velocity of second billiard after collision = 1.4 m/sec  

Answer:

b)determining the electric field due to each charge and adding them together as vectors.

Explanation:

The electric Field is a vector quantity, in other words it has a magnitude and a direction. On the other hand, the electric field follows the law of superposition. The electric field produced by two elements is equal to the sum of the electric fields produced by each element when the other element is not present. in other words, the total electric field is solved determining the electric field due to each charge and adding them together as vectors.

The electric field due to two-point charges is found by determining the electric field of each charge and adding them together as vectors, considering both their magnitudes and directions.

The correct way to find the electric field due to two-point charges is option (b), determining the electric field due to each charge and adding them together as vectors. The electric field is a vector quantity, meaning it has both magnitude and direction. Thus, when two or more electric fields exist in the same region, you add them as vectors, taking into account both their magnitudes and directions. For example, if you have two positive charges, the fields they generate will be in different directions. Therefore, you would add these fields together as vectors, resulting in the net electric field in that region.

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Answer:

Yes

Explanation:

Yes, the orbital direction of a planet affect its Synodic period as seen from earth. The synodic period of mercury is about 116 days on Earth. Whereas the sidereal period of mercury as seen from Earth is 88 days.  This difference is caused by the simultaneous motion of Earth in its orbit. Hence we can say that the  orbital direction of a planet affect its synodic period as seen from earth.

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, [tex]q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=2.4\ \mu C=2.4\times 10^{-6}\ C[/tex]

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

[tex]U=k\dfrac{q_1q_2}{r}[/tex]

[tex]r=k\dfrac{q_1q_2}{U}[/tex]

[tex]r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}[/tex]

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

The 2.40 μC point charge must be placed approximately 0.34 meters away from the -7.80 μC point charge for their electric potential energy to be -0.500 J.

To find the distance between a -7.80 μC point charge and a 2.40 μC point charge such that their electric potential energy is -0.500 J, we use the formula for electric potential energy:

U = k × (q₁ × q₂) / r

where U is the electric potential energy, k is Coulomb's constant (8.99 × 10⁹ Nm²/C²), q₁ and q₂ are the point charges, and r is the distance between them.

Given:

q₁ = -7.80 μC = -7.80 × 10⁻⁶ C

q₂ = 2.40 μC = 2.40 × 10⁻⁶ C

U = -0.500 J

Rearranging the formula to solve for r, we get:

r = k × (q₁ × q₂) / U

Substituting the given values:

r = (8.99 × 10⁹ Nm²/C²) × (-7.80 × 10⁻⁶ C × 2.40 × 10⁻⁶ C) / -0.500 J

r ≈ 0.34 meters

Therefore, the 2.40 μC point charge must be placed approximately 0.34 meters away from the -7.80 μC point charge for their electric potential energy to be -0.500 J.

Answer:

Time taken, [tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

[tex]T\ cos\theta-mg=0[/tex]

[tex]T=\dfrac{mg}{cos\theta}[/tex]

Sum of forces in x direction,

[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]

[tex]mg\ tan\theta=\dfrac{mv^2}{r}[/tex].............(1)

Also, [tex]r=l\ sin\theta[/tex]

Equation (1) becomes :

[tex]mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}[/tex]

[tex]v=\sqrt{gl\ tan\theta.sin\theta}[/tex]...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

[tex]T=\dfrac{2\pi r}{v}[/tex]

Put the value of T from equation (2) to the above expression:

[tex]T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]

[tex]T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]

On solving above equation :

[tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]

Hence, this is the required solution.

The time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].

As it is given to us that the small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone.

Now since the ball is moving in circular as well as vertical motion, therefore, the sum of vertical forces at any given moment of time can be written as,

[tex]\rm T\ cos \theta = mg\\\\T = \dfrac{mg}{Cos\theta}[/tex]

Also, the sum of the forces in the x-direction,

[tex]\rm T\ sin\theta= \dfrac{mv^2}{r}[/tex]

Substitute the value of T,

[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{r}[/tex]

[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{l\ sin\theta}[/tex]

[tex]v=\sqrt{gl\ tan\theta \cdot sin\theta}[/tex]

We know that the time is taken by the ball to circulate around the axis can be given by the formula,

[tex]T = \dfrac{2\pi r}{v}[/tex]

Substitute the value of v,

[tex]T = \dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta\cdot sin\theta}}[/tex]

[tex]T=2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex]

Hence, the time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].

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Answer:

8.1 m /s

Explanation:

Let the required velocity be v . This is a horizontal velocity so it will cover the horizontal distance of 12.3 m with this velocity without any acceleration .

Time taken t = distance / velocity

t = 12.3 /v

During this period ball also covers vertical distance with initial velocity zero and acceleration of g.

For vertical fall

initial velocity u = 0

Acceleration = g

Time = t

h = ut + 1/2 g t²

11.3 = 0 + .5 x 9.8 x (12.3 / v )²

v² = 65.6

v = 8.1 m /s

Answer:

Ping Pong ball move at velocity 160.4 m/s

Explanation:

given data

mass m1 = 7 kg

velocity v1 = 3 m/s

mas m2 = 2.45 g = 2.45 × [tex]10^{-3}[/tex] kg

same kinetic energy

to find out

How fast Ping Pong ball move (v2)

solution

we know same KE

so

[tex]\frac{1}{2}* m1* v1^{2} = \frac{1}{2}* m2* v2^{2}[/tex]   ...........1

so v2 will be

v2 = [tex]\sqrt{\frac{m1*v1^2}{m2} }[/tex]    .............2

put here value in equation 2 we get v2

v2 = [tex]\sqrt{\frac{7*3^2}{2.45*10^{-3}}}[/tex]

v2 = 160.4

so Ping Pong ball move at velocity 160.4 m/s

Begin with the kinetic energy equation, set the two equal and solve for the unknown velocity of the Ping Pong ball. Substitute the given values into the formula to finalize your answer.

This question involves the concept of kinetic energy, that is equal to 1/2 mass x velocity2 for an object in motion. Let's denote the mass of the bowling ball as m1 (7kg), its velocity as v1 (3m/s), the mass of the Ping Pong ball as m2 (0.00245kg, which is 2.45g in kg), and its velocity as v2, which we need to find. If we assume that the two balls have the same kinetic energy, then:

1/2 * m1 * v12 = 1/2 * m2 * v22

By calculating the left-hand side and then rearranging the equation to solve for v2, we will get the velocity of the Ping Pong ball:

v2 = sqrt((m1 * v12) / m2)

Now substitute the given values into the formula to find the velocity of the Ping Pong ball.

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Answer:

Because of that but also because of its magnetic properties.

Explanation:

But its abundance in water and fat, since Magnetic Resonance Imaging uses our own body's magnetic properties to produce the images, and for this is much better to use something that is abundant in our own body and has magnetic properties, that is, and hydrogen nucleus ( single proton), being useful since it behaves like a small magnet.

Answer:

a) 21.8 mts

b) 10.2 seconds

c) 4.28 m/s

Explanation:

Because the blimp is filled with helium, the acceleration won't be the gravity. We have to calculate the new acceleration:

[tex]a=\frac{Vf-Vo}{t}\\\\a=\frac{0-4.28}{`10.2}\\\\a=0.420 m/s^2[/tex]

in order to obtain the height we have to use the formulas of accelerated motion problems:

[tex]X=Vo*t+\frac{1}{2}*a*t^2\\\\X=4.28*(10.2)+\frac{1}{2}*(-0.420)*(10.2)^2\\X=21.8mts[/tex]

we can calculate the time with the same formula:

[tex]-21.8=0*t+\frac{1}{2}*(-0.420)*t^2\\solving\\t=10.2 seconds[/tex]

the velocity at the same height is given by:

[tex]Vf^2=Vo^2+2*a*x\\Vf=\sqrt{2*(-0.420)*(-21.8)} \\Vf=-4.28m/s[/tex]

the speed would be 4.28m/s because is a scalar value.

Answer:

735 J

Explanation:

mass of box, m = 15 kg

Height, h = 5 m

Work is defined as the product of force and the displacement in the  direction of force.

W = F x S

Work is a scalar quantity and its SI unit is Joule.

Here, Force, F = m x g = 15 x 9.8 = 147 N

So, W = 147 x 5 = 735 J

Thus, the work is 735 J.

Answers:

a) [tex]t=0.311 s[/tex]

b) [tex]a=86.847 m/s^{2}[/tex]

c) [tex]y=1.736 m[/tex]

d) [tex]V=17.369 m/s[/tex]

Explanation:

For this situation we will use the following equations:

[tex]y=y_{o}+V_{o}t+\frac{1}{2}at^{2}[/tex] (1)  

[tex]V=V_{o} + at[/tex] (2)  

Where:  

[tex]y[/tex] is the height of the model rocket at a given time

[tex]y_{o}=0[/tex] is the initial height of the model rocket

[tex]V_{o}=0[/tex] is the initial velocity of the model rocket since it started from rest

[tex]V[/tex] is the velocity of the rocket at a given height and time

[tex]t[/tex] is the time it takes to the model rocket to reach a certain height

[tex]a[/tex] is the constant acceleration due gravity and the rocket's thrust

The average velocity of a body moving at a constant acceleration is:

[tex]V=\frac{V_{1}+V_{2}}{2}[/tex] (3)

For this rocket is:

[tex]V=\frac{27 m/s}{2}=13.5 m/s[/tex] (4)

Time is determined by:

[tex]t=\frac{y}{V}[/tex] (5)

[tex]t=\frac{4.2 m}{13.5 m/s}[/tex] (6)

Hence:

[tex]t=0.311 s[/tex] (7)

Using equation (1), with initial height and velocity equal to zero:

[tex]y=\frac{1}{2}at^{2}[/tex] (8)  

We will use [tex]y=4.2 m[/tex] :

[tex]4.2 m=\frac{1}{2}a(0.311)^{2}[/tex] (9)  

Finding [tex]a[/tex]:

[tex]a=86.847 m/s^{2}[/tex] (10)  

Using again [tex]y=\frac{1}{2}at^{2}[/tex] but for [tex]t=0.2 s[/tex]:

[tex]y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}[/tex] (11)

[tex]y=1.736 m[/tex] (12)

We will use equation (2) remembering the rocket startted from rest:

[tex]V= at[/tex] (13)  

[tex]V= (86.847 m/s^{2})(0.2 s)[/tex] (14)  

Finally:

[tex]V=17.369 m/s[/tex] (15)  

Answer:

(a) 14.88 ohm

(b) 1.88 A

(c)  -74.4°

Explanation:

Vo = 28 V

f = 12 kHz = 12000 Hz

R = 4 ohm

L = 30 micro henry = 30 x 10^-6 H

C = 800 nF = 800 x 10^-9 F

(a)

The inductive reactance,

XL = 2 π f L = 2 x 3.14 x 12000 x 30 x 10^-6 = 2.26  ohm

The capacitive reactance

[tex]X_{c}=\frac{1}{2\pi fC}=\frac{1}{2 \times 3.14 \times 12000 \times 800 \times 10^{-9}}[/tex]

Xc = 16.59 ohm

Let the impedance is Z.

[tex]Z=\sqrt{4^{2}+\left ( 2.26-16.59 \right )^{2}}[/tex]

Z = 14.88 ohm

(b)

The formula for the amplitude of current

[tex]I_{o}=\frac{V_{o}}{Z}=\frac{28}{14.88}[/tex]

Io = 1.88 A

(c)

Let the phase difference is Ф

[tex]tan\phi =\frac{X_{L}-X_{C}}{R}=\frac{2.26-16.59}{4}=-3.5825[/tex]

Ф = -74.4°

Answer:

10945.9 kg/m^3

Explanation:

mass of cylinder, m = 1 kg

Height of cylinder, h = 55 mm = 0.055 m

Diameter of cylinder = 46 mm

Radius of cylinder, r = 23 mm = 0.023 m

The formula of the volume of the cylinder is given by

[tex]V = \pi r^{2}h[/tex]

V = 3.14 x 0.023 x 0.023 x 0.055

V = 9.136 x 10^-5 m^3

Density is defined as mass per unit volume .

[tex]Density = \frac{1}{9.136 \times 10^{-5}}[/tex]

Density = 10945.9 kg/m^3

Answer:

[tex]N=1.56*10^{20}[/tex] protons

Explanation:

The total charge Q+=+25.0C is the sum of the charge of the N protons, with charge q :

[tex]Q_{+}=N*q_{p}[/tex]

q_{p}=1.6*10^{-19}C    charge of a proton

We solve to find N:

[tex]N=Q_{+}/q_{p}=25/(1.6*10^{-19})=1.56*10^{20}[/tex]

Answer:

(a) d = 15.6 m

(b) v' = - 4.3 m/s

Given:

Initial velocity, v = 9.50 m/s

Acceleration, a = [tex] - 2.30 m/s^{2][/tex] or deceleration = 2.30[tex]m/s^{2}[/tex]

Solution:

(a) For the calculation of the distance covered, we use eqn (2) of motion:

[tex]d = vt + \frac{1}{2}at^{2}[/tex]

where

d = distance covered in time t

t = 6 s (Given)

Now,

[tex]d = 9.50\times 6 - \frac{1}{2}\times 2.30times 6^{2}[/tex]

d = 15.6 m

(b) For the calculation of her final velocity, we use eqn 1 of motion:

v' = v + at

v' = 9.50 + (- 2.30)(6) = - 4.3 m/s

(c) Since, the final velocity after the body slows down comes out to be negative and the distance covered and displacement of the body are different, it does not make sense.

Answer:

a) Vx = -31.95 km/h       b) Vy = -45.07 km/h

c) t = 0.083 h                 d) d = 0.22 km

Explanation:

First we have to express these values as vectors:

ra = (2.5, 3.9) km      rb = (0,0)km

Va = (0, - 21) km/h    Vb = (31.95, 24.07) km/h

Now we can calculate relative velocity:

[tex]V_{A/B} = V_{A} - V_{B} = (0, -21) - (31.95, 24.07) = (-31.95, -45.07) km/h[/tex]

For parts (c) and (d) we need the position of A relative to B and the module of the position will be de distance.

[tex]r_{A/B} = (2.5, 3.9) + (-31.95, -45.07) * t[/tex]

[tex]d = |r_{A/B}| = \sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}[/tex]

In order to find out the minimum distance we have to derive and find t where it equals zero:

[tex]d' = \frac{-2*(2.5-31.95*t)*(-31.95)-2*(3.9-45.07*t)*(-45.07)}{2*\sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}} =0[/tex]

Solving for t we find:

t = 0.083 h

Replacing this value into equation for d:

d = 0.22 km

Answer:

Frequency, [tex]\nu=1.96\times 10^{20}\ Hz[/tex]

Explanation:

Given that,

Energy of a gamma rays, [tex]E=0.815\ MeV=0.815\times 10^6\ eV[/tex]

Since, [tex]1\ eV=1.6\times 10^{-19}\ J[/tex]

[tex]0.815\times 10^6\ eV=0.815\times 10^6\times 1.6\times 10^{-19}\ J[/tex]

[tex]E=1.304\times 10^{-13}\ J[/tex]

The energy of a wave is given by :

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

[tex]\nu[/tex] is the frequency of such a photon

[tex]\nu=\dfrac{E}{h}[/tex]

[tex]\nu=\dfrac{1.304\times 10^{-13}\ J}{6.63\times 10^{-34}\ Js}[/tex]

[tex]\nu=1.96\times 10^{20}\ Hz[/tex]

So, the frequency of such a photon is [tex]1.96\times 10^{20}\ Hz[/tex]. Hence, this is the required solution.

Answer:

C

Explanation:

the correct answer is c) Pre Schematic.

It is in the Pre-schematic stage of development that the children learn to draw human figures by sketching a circle with dangly lines for legs. It remains till the age of six. After pre schematic stage comes the schematic stage, where in children can draw complex structures.  

Children drawing human figures with a circle and lines for legs are in the Pre-schematic stage of art development, indicating an early attempt at representational drawing before achieving greater accuracy and detail in later stages.

When children draw human figures by sketching a circle with dangly lines for legs, they are demonstrating the Pre-schematic developmental stage in art. This stage is distinct from the earlier scribbling phase, signaling that children have begun to recognize and attempt to represent the human form in a more organized albeit simplified manner. However, this representation lacks the complexity and accuracy found in later stages, such as the Schematic stage, where more detailed and proportionate figures are drawn.

The Pre-schematic stage typically occurs in children aged between 4 and 7 years old. During this time, their cognitive and motor skills are developing rapidly, but they are still learning how to translate what they see and imagine onto paper with greater fidelity. Thus, the circle and lines approach is a developmentally appropriate method for them to begin exploring human figures in their art.

Answer

Work done will be [tex]14.7\times 10^{-6}J[/tex] and it will be positive

Explanation:

We have given charge [tex]2.1\times 10^{-6}C[/tex]

We have to find work done in moving the charge from 15 volt to 8 volt

Let [tex]V_1=15V\ and\ V_2=8volt[/tex]

So potential difference [tex]V=V_1-V_2=15-8=7volt[/tex]

We know that work done [tex]W=QV[/tex], here Q is charge and V is potential difference

So work done [tex]W=QV=2.1\times 10^{-6}\times 7=14.7\times 10^{-6}J[/tex]

It will be positive work done because work is done in moving charge from higher potential to lower potential  

The magnitude of the minimum deceleration needed to avoid the accident is –0.55 m/s²

To solve the question given above, we'll begin by calculating the distance travelled during the reaction time. This can be obtained as follow:

Speed = 18 m/s

Time = 0.45 s

Speed = distance / time

18 = distance / 0.45

Cross multiply

Distance = 18 × 0.45

Thus, the engineer travelled a distance of 8.1 m during the reaction time.

Next, we shall the distance between the engineer and the car. This can be obtained as follow:

Total distance = 300 m

Distance during the reaction time = 8.1 m

= 300 – 8.1

Finally, we shall determine the magnitude of the deceleration needed to avoid the accident. This can be obtained as follow:

Initial velocity (u) = 18 m/s

Final velocity (v) = 0 m/s

Distance (s) = 291.9 m

0² = 18² + (2 × a × 291.9)

0 = 324 + 583.8a

Collect like terms

0 – 324 = 583.8a

–324 = 583.8a

Divide both side by 583.8

a = –324 / 583.8

Therefore, the magnitude of the deceleration needed to avoid the accident is –0.55 m/s²

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To avoid an accident, the engineer must decelerate the train at a minimum rate of approximately -0.55 m/s^2, taking into account the reaction time which contributed an additional 8.1 m to the stopping distance.

An engineer in a locomotive sees a car stuck on the track and needs to calculate the necessary deceleration to avoid an accident. The engineer has a reaction time of 0.45 seconds, and the initial speed of the train is 18 m/s. The total stopping distance must include the distance traveled during the reaction time plus the actual braking distance.

The distance covered during the engineer's reaction time is found using the formula d = vt, where d is distance, v is velocity, and t is time. We get d = 18 m/s * 0.45 s = 8.1 m.

So, the stopping distance minus the reaction distance is 300 m - 8.1 m = 291.9 m. To find the deceleration, we can use the formula v^2 = u^2 + 2as, where v is the final velocity (which is 0), u is the initial velocity, a is the acceleration (deceleration in this case), and s is the distance. Solving for a, we get a = -u^2 / (2s). Substituting the values, a = -(18 m/s)^2 / (2 * 291.9 m), resulting in a deceleration of approximately -0.55 m/s^2.

Answer:

length = 2L, mass = M/2, and maximum angular displacement = 1 degree

Explanation:

We consider only small amplitude oscillations (like in this case), so that the angle θ is always small enough. Under these conditions recall that the equation of motion of the pendulum is:

[tex]\ddot{\theta}=\frac{g}{l}\theta[/tex]

And its solution is:

[tex]\theta=Asin(\omega t + \phi)[/tex]

Where [tex]\omega=\sqrt\frac{g}{l}[/tex] are the angular frequency of the oscillations, from which we determine their period:

[tex]T=\frac{2\pi}{\omega}\\T=2\pi\sqrt\frac{l}{g}[/tex]

Therefore the period of a pendulum will only depend on its length, not on its mass or angle, for angles small enough. So, the answer is the one with the greater length.

The period of a pendulum is only determined by its length and the acceleration due to gravity, and is independent of other factors such as mass and maximum displacement.

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. Therefore, none of the combinations of variables given will result in a greater period for the pendulum. The period is only determined by the length and the value of acceleration due to gravity.

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Answer:

so speed = 165 m/s

Explanation:

given data

speed = (25.0 + A + B) urks/dort

A = 39

B = 18

1 urk = 58 m

to find out

Convert speed to meters per second

solution

we know speed =  (25.0 + A + B)

put A and B

speed =  (25.0 + 39+ 18) = 82 urk/dort

we know that hour is divided into 125 time units name dorts

so we can say

1 hour = 125 × dorts

and we know 1 hour = 3600 seconds

so

3600 = 125 × dorts

dorts = 28.8 seconds

and we have given

1 urk = 58 m

so 82 urk = 82 × 58 = 4756 m

so from speed

speed = 82 urk/dort

speed =  [tex]82 * \frac{58}{28.8}[/tex] m/s = 165.139

so speed = 165 m/s

Final answer:

The speed (82.0 urks/dort) converts to 165 meters per second when using the given conversion factors, with values A=39 and B=18 included.

Explanation:

To convert a speed from urks/dort to meters per second, we first need to express the speed in urks/dort in terms of meters and seconds, using the given conversions: 1 urk = 58.0 meters and 1 hour = 125 dorts. Given the values A=39 and B=18, the speed in urks/dort is 25.0 + 39 + 18 = 82.0 urks/dort.

Firstly, we convert urks to meters:

(82.0 urks/dort) × (58.0 meters/urk) = 4756 meters/dort

Now, we can convert dorts to seconds. Since 1 hour is divided into 125 dorts and there are 3600 seconds in an hour, there are 3600 seconds / 125 dorts = 28.8 seconds/dort.

Speed in meters per second (m/s) = 4756 meters/dort × (1 dort/28.8 seconds)

Speed in m/s = 165 m/s (rounded to three significant figures)

Therefore, the speed in meters per second is 165 m/s, rounded to three significant figures.

Answer: If a force on an object is aimed in the direction of the object’s velocity, the force does positive work (b).

Explanation:

Hi, the answer is oprion b. positive work.

If a force on an object is aimed in the direction of the object’s velocity, the force does positive work.

An object's kinetic energy will only change if the force acting on the object changes the object's speed. This will only happen if there is a component of the force in the direction that the object moves.

So, A force will do work only if the force has a component in the direction that the object moves.

The answer to the question is 'b. positive work.' If the force on an object is in the direction of the object’s velocity, positive work is done as it adds energy to the system.

If a force on an object is aimed in the direction of the object’s velocity, the force does positive work. Work is defined as the product of the force times the displacement times the cosine of the angle between them.

Since the angle between the force and the displacement is zero when the force is in the direction of velocity (cos 0 = 1), the work done is positive. The work done by a force in the direction of an object's motion adds energy to the system. 

An example of this would be pushing a lawn mower forward, where the force applied and the direction of the mower's motion are the same.

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Answer:

The ball hit the pyramid 18.52m down the pyramid face.

Explanation:

As you can see in the image below, the will hit the pyramid at at point where the distance travelled vertically divided by the distance travelled horizontally is equal to tan(50), since it's at this moment where the path of the ball will coincide with the walls of the pyramid.

The horizontal vellocity of the ball will remain constant at a value of 7m/s along the whole journey. This is because there is no horizontal acceleration that can affect the horizontal velocity. On the contrary, the vertical velocity will start at 0m/s and will increase because of gravity.

The distance travelled horizontally will be:

[tex]x = v_x*t = 7t[/tex]

The distance travelled vertically will be:

[tex]y = \frac{1}{2}gt^2+v_o_yt+y_o | v_o_y = 0m/s, y_o=0m\\ y = \frac{1}{2}gt^2[/tex]

So, then:

[tex]\frac{y}{x} = tan(50)\\\frac{\frac{1}{2}gt^2}{v_xt} =tan(50)\\\frac{1}{2}gt = v_xtan(50)\\t= \frac{2v_xtan(50)}{g} = \frac{2*7m/s*tan(50)}{9.81m/s^2}=1.7s[/tex]

At time = 1.7s:

[tex]x = v_x*t = 7t = 7m/s*1.7s = 11.9m[/tex]

[tex]y=\frac{1}{2}gt^2 =\frac{1}{2}*9.81m/s^2*(1.7s)^2=14.19m [/tex]

Using Pythagorean theorem, we can find the distance:

[tex]d = \sqrt{x^2+y^2}=\sqrt{(11.9m)^2+(14.19m)^2} = 18.52m[/tex]

Final answer:

The magnitude of the electric field at the center of a square with four point charges (either all positive or three positive and one negative) each of magnitude 4.50 nC and sides of length 2.80 cm, is zero. This outcome is due to symmetry and the principle of superposition.

Explanation:

To find the magnitude of the electric field at the center of a square due to point charges located at its corners, we use the principle of superposition. This entails calculating the electric field contribution from each charge individually and then vectorially adding these contributions together.

For our case, where all charges have a magnitude of 4.50 nC and the square has sides of length 2.80 cm, the geometry of the problem simplifies the calculations significantly.

When all four charges are positive, they all contribute to the electric field in a symmetrically outward manner relative to the center. Thus, their contributions in terms of magnitude cancel out, making the net electric field at the center zero.

In the scenario where three charges are positive and one is negative, we approach the problem similarly. However, the negative charge introduces an incongruence in the symmetry. Its electric field contribution will 'attract' rather than 'repel' like the other three positive charges.

Despite this, due to the square's symmetry and equal magnitude of the charges, the net electric field at the center still cancels out, resulting in zero magnitude.

This conclusion is based on symmetry and the principle that equal point charges at equal distances in a square configuration result in a net zero electric field at the square's center, irrespective of the sign when the magnitudes are equal.

Answer:

g = 0.98 m

Explanation:

given data:

upper portion weight 438 N and its center of gravity is 1.17 m

upper legs weight 144 N and its center of gravity is 0.835 m

lower leg weight 88 N and its center of gravity is 0.270 m

position of center of gravity of whole body can be determine by using following relation

[tex]g = \frac{\sum ({m_i g_i)}}{\sum m_i}[/tex]

where [tex]m_i[/tex] is mass of respective part and [tex]g_i[/tex] is center of gravity

[tex]g = \frac{\sum{ 438\times 1.17 + 144\times 0.835 + 88\times 0.270}}{438+144+88}[/tex]

g = 0.98 m

Answer:

(a) 6.56 km

(b) [tex]59.68^\circ[/tex] north of west

Explanation:

Given:

[tex]\vec{d}_1= 3.35\ km\ east = 3.35\ km\ \hat{i}\\\vec{d}_2= 9.31\ km\ north = 9.31\ km\ \hat{j}\\\vec{d}_3= 6.66\ km\ west = -6.61\ km\ \hat{i}\\\vec{d}_4= 3.65\ km\ south = -3.65\ km\ \hat{j}\\[/tex]

Let the resultant displacement vector be [tex]\vec{D}[/tex].

As the resultant is the vector sum of all the vectors.

[tex]\therefore \vec{D}=\vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} =(3.35\ km\ \hat{i})+(9.31\ km\ \hat{j})+(-6.61\ km\ \hat{i})+(-3.65\ km\ \hat{j})\\\Rightarrow \vec{D} =-3.31\ km\ \hat{i}+5.66\ km\ \hat{j}\\\textrm{Magnitude of the resultant displacement vector} = \sqrt{(-3.31)^2+(5.66)^2}\ km= 6.56\ km\\[/tex]

[tex]\textrm{Angle with the positive west} = \theta = \tan^{-1}(\dfrac{5.66}{3.31})= 59.68^\circ[/tex]

The magnitude of the resultant vector sum of all the displacements is 6.55 km, and its direction is 59.65° relative to due west.

To find the magnitude of the resultant vector (D = A + B + C + D), we first need to add the displacement vectors head-to-tail. Since each vector points in a cardinal direction, we can add their magnitudes algebraically after designating the direction for each: east and north are positive, while west and south are negative. The resultant in the east-west direction is 3.35 km (east) - 6.66 km (west) = -3.31 km (west). The resultant in the north-south direction is 9.31 km (north) - 3.65 km (south) = 5.66 km (north).

Now we use the Pythagorean theorem to find the magnitude of the resultant:

R = √((-3.31 km)² + (5.66 km)²) = √(10.96 km² + 32.05 km²) = √43.01 km² = 6.55 km

The direction is determined by the tangent of the angle relative to the west. Let φ be the angle, then

tan(φ) = opposite / adjacent = 5.66 km / 3.31 km

φ = arctan(5.66 / 3.31) = 59.65° (north of west)

Therefore, the resultant vector is 6.55 km in magnitude, at an angle of 59.65° relative to due west.

Answer:

24 cm

Explanation:

Given:

Mass of proton = 1.67 × 10⁻²⁷ Kg

kinetic energy = 2.5 × 10⁻¹³ J

magnetic field in the cyclotron, B = 0.75 T

Now,

Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]  = 2.5 × 10⁻¹³ J

where, v is the velocity of the electron

or

[tex]\frac{1}{2}\times1.67\times10^{-27}\times v^2[/tex]  = 2.5 × 10⁻¹³ J

or

v² = 2.99 × 10¹⁴

or

v = 1.73 × 10⁷ m/s

also,

centripetal force = magnetic force

or

[tex]\frac{mv^2}{r}[/tex]  = qvB

q is the charge of the electron

r is the radius of the dipole magnets

on substituting the respective values, we get

[tex]\frac{1.67\times10^{-27}\times1.73\times10^7}{r}[/tex]  = 1.6 × 10⁻¹⁹ × 0.75

or

r = 0.2408 m ≈ 24 cm

Hence, the correct answer is 24 cm

The radius of the dipole magnets in the cyclotron is mathematically given as

r=24 cm

Question Parameter(s):

A cyclotron is designed to accelerate protons (mass 1.67 x 10^-27 kg)

Up to kinetic energy of 2.5 x 10^-13 J.

Generally, the equation for the Kinetic energy is mathematically given as

K.E=0.5mv^2  

Therefore

0.5*1.67*10^{-27}*v^2  = 2.5 × 10^{-13} J

v = 1.73 × 10⁷ m/s

In conclusion

mv^2/r  = qvB

(1.67*10^{-27}*1.73*10^7)/r  = 1.6 × 10^{-19}* 0.75

r=24 cm

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