A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.6 m/s and at an angle of 40.8° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.73 s, and for the (e) horizontal and (f) vertical components at t = 5.05 s. Assume that the catapult is positioned on a plain horizontal ground.

Answer:

The magnetic field strength is required [tex] 2.84\times10^{-14}\ T[/tex]

Explanation:

Given that,

Speed of proton[tex]v = 3.60\times10^{6}\ m/s[/tex]

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

Charge[tex]q =1.60\times10^{-19}\ C[/tex]

When a proton moves horizontally, at a right angle to a magnetic field .

Then, the gravitational force balances the magnetic field

[tex]mg=Bqv\sin\theta[/tex]

[tex]B = \dfrac{mg}{qv}[/tex]

Here, [tex]\theta = 90^{\circ}[/tex]

Where, B = magnetic field

q = charge

v = speed

Put the value into the formula

[tex]B = \dfrac{1.67\times10^{-27}\times9.8}{1.60\times10^{-19}\times3.60\times10^{6}}[/tex]

[tex]B = 2.84\times10^{-14}\ T[/tex]

Hence, The magnetic field strength is required [tex] 2.84\times10^{-14}\ T[/tex]

Final answer:

To balance the gravitational force on a proton moving at a right angle through a magnetic field, the required field strength is found using the magnetic force formula, resulting in a necessary field strength of 2.86 x 10⁻²³ T.

Explanation:

The student is asking about the magnetic force required to counteract the gravitational force on a proton moving horizontally through a magnetic field. To solve this, we need to use the formula F = qvB, where F is the magnetic force, q is the charge of the proton, v is the velocity of the proton, and B is the magnetic field strength we wish to find.

First, we calculate the weight of the proton using W = mg, where m is the mass of the proton and g is the acceleration due to gravity (9.81 m/s²). This weight is the force we aim to balance with the magnetic force.

Now, let's calculate the weight of the proton: W = (1.67 × 10⁻²⁷ kg) × (9.81 m/s²) = 1.64 × 10⁻²⁶ N.

To keep the proton moving horizontally, the magnetic force needs to equal the proton's weight. So we set F to W and solve the equation for B:

B = W/(qv) = (1.64 × 10⁻²⁶ N) / ((1.60 × 10⁻²⁹ C) × (3.60 × 10⁶ m/s))

B = 2.86 × 10⁻² T

Therefore, a magnetic field strength of 2.86 × 10⁻² Tesla is required to just balance the weight of the proton and keep it moving horizontally.

Answer:

Total number of different selections  = 455

Explanation:

Number of selections from n objects if we select r things [tex]=^nC_r[/tex]            

Here we need to find number off selections of 3 apples from 15 apples.

Number off selections of 3 apples from 15 apples

                    [tex]=^{15}C_3=\frac{15\times 14\times 13}{1\times 2\times 3}=455[/tex]

Total number of different selections  = 455

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, [tex]A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2[/tex]

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, [tex]\rho=11\times 10^{-8}\ \Omega-m[/tex]

We need to find the resistance of the iron rod. It is given by :

[tex]R=\rho\dfrac{L}{A}[/tex]

[tex]R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}[/tex]

[tex]R=0.000077 \Omega[/tex]

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

Answer:

Answer to the question: 91.44m

Explanation:

L=100 yds = 300 ft = 91.44 m

To convert 100 yards to meters for an American football field, you multiply 100 yards by 3 to get 300 feet, then convert feet to meters using the given conversion factor, resulting in approximately 91.4 meters.

To convert the length of an American football field from yards to meters, we first need to understand the conversion factors between these units. We know that 1 yard equals 3 feet, and we are given that 1 meter equals 3.281 feet. Starting with the length of the field in yards, which is 100 yards, we convert yards to feet and then convert feet to meters.

Step-by-Step Conversion

Therefore, the length of an American football field is roughly 91.4 meters, excluding the end zones.

Answer:

The wavelength of x-ray is 0.272 nm.

Explanation:

Inter planer spacing, d = 0.399 nm = 3.99 × 10⁻¹⁰ m

First order constructive interference occurs when the beam makes an angle of 20° with the planes. We need to find the wavelength of the x-rays. The condition for constructive interference is given by :

[tex]n\lambda=2d\ sin\theta[/tex]

Here, n = 1

[tex]\lambda=2\times 3.99\times 10^{-10} m\ sin(20)[/tex]

[tex]\lambda=2.72\times 10^{-10}\ m[/tex]

[tex]\lambda=0.272\ nm[/tex]

So, the  wavelength of the x-rays is 0.272 nm. Hence, this is the required solution.

Answer:

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

Explanation:

Let unknown velocity be v.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 6i + m x 0i = 6m i

Final momentum = m x (4.8cos 30 i + 4.8sin 30 j) + m x v = 4.16 m i + 2.4 m j + m v

Comparing

4.16 m i + 2.4 m j + m v = 6m i

v = 1.84 i - 2.4 j

Magnitude of velocity

       [tex]v=\sqrt{1.84^2+(-2.4)^2}=3.02m/s[/tex]

Direction,  

        [tex]\theta =tan^{-1}\left ( \frac{-2.4}{1.8}\right )=-53.13^0[/tex]     

Velocity is 3.02 m/s at an angle of 53.13° below X-axis.

To find the final velocity of the first puck, apply conservation of momentum and solve for v'. The final velocity of the first puck is 4.8 m/s.

To find the final velocity of the first puck, we can apply the principle of conservation of linear momentum. The initial momentum of the system is given by m1v1 + m2v2, where m1 is the mass of the first puck, m2 is the mass of the second puck, v1 is the initial velocity of the first puck, and v2 is the initial velocity of the second puck.

Since the collision is elastic, the total momentum before and after the collision is conserved. So, m1v1 + m2v2 = (m1 + m2)v', where v' is the final velocity of both pucks after the collision. We can plug in the given values to find the final velocity of the first puck.

Let's solve for v': m1v1 + m2v2 = (m1 + m2)v' => (m1)(6.0 m/s) + (m2)(0 m/s) = (m1 + m2)(4.8 m/s).

From the given information, we know that the two pucks are identical, so m1 = m2. Substituting this into the equation, we get (m)(6.0 m/s) = 2(m)(4.8 m/s), where m is the mass of each puck. Simplifying this equation, we find the mass cancels out, leaving 6.0 m/s = 2(4.8 m/s). Solving for v', we find v' = 4.8 m/s.

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The amount of work required to pump all of the water over the side of the pool is 471,238.9 Newton Meters or Joules.

The work done to pump water out of a pool involves the concept of physics specifically related to potential energy, gravity, and volume. The work done to move a certain volume of water is given by the formula: Work = Weight x Height.

First, we need to find the volume of the water in the pool. The pool's shape resembles a cylinder, and the volume is given by the formula for a cylinder: Volume = pi × (diameter/2)²× height. Given a diameter of 8 meters and a height of 1.5 meters, the volume to be moved is pi * (8/2)² × 1.5 = 48pi cubic meters.

The weight of this water can be calculated by multiplying its volume by its density. The density of water is 1000 kg/m³. Therefore, the weight of the water is Volume x Density x g (acceleration due to gravity), which is 48pi × 1000 × 9.8 = 471,238.9 kg×m²/s² or Newton Meter (Nm) which is the unit of work.

So, the amount of work required to pump all the water over the side is 471,238.9 Nm or Joules (J).

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The work required to pump all the water over the side of a circular swimming pool of diameter 8 meters and water depth 1.5 meters is calculated using the formula Work = Force x Distance. The Force required is given by the weight of the water, which depends on its volume and density. The result is about 353,429.16 Joules.

To calculate the work required to pump water out of a swimming pool, we can use the formula for work: Work = Force x Distance. The force required is equal to the weight of the water which depends on the volume of the water and its density.

First, let's calculate the volume of the water in the pool. Given that it's a circular pool with a diameter of 8 meters, the radius is 4 meters. The depth of the water is 1.5 meters. So, volume (V) = πr²h = π×(4m)²×(1.5m) = 24π cubic meters.

The density of water is 1000 kg/m³. Therefore, the weight of water = Volume x Density x Gravity = 24π m³ ×1000 kg/m³ × 9.8 m/s² = 235619.44 kg×m/s², or Newtons. This is the force we need to overcome to lift the water.

The distance that we want to lift this mass is the depth of the pool, assuming we are pumping the water over the side of the pool which is 1.5 meters high. So, Work = Force x Distance = 235619.44 N×1.5m = 353429.16 Joules.

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Using the acceleration due to gravity on the planet, we calculate its mass to be 8.97 x 10²⁴ kg. For the mass of the star, using Kepler's third law, we find it to be approximately 1.99 x 10³⁰ kg, assuming the planet's orbit is about 1 AU.

To find the mass of the planet and the star, we can use Newton's form of Kepler's third law and Newton's universal law of gravitation. Let's denote the mass of the star as M and the mass of the planet as m.

Mass of the Planet

The acceleration due to gravity (g) on the surface of the planet is given by:

g = Gm / r²

Where:

G is the gravitational constant (approximately 6.674 x 10⁻¹¹ N m2 kg-²).

m is the mass of the planet.

r is the radius of the planet, which is half of its diameter.

Rearranging the formula to solve for m:

m = g r² / G

Substituting the given values (g = 59.7 m/s² and r = 9 x 106 m), we get:

m = 59.7 x (9 x 106)² / 6.674 x 10⁻¹¹
= 8.97 x 10²⁴ kg

Mass of the Star

Using Newton's version of Kepler's third law, we get:

M = 4π² a3 / G T²

Where: a is the semi-major axis of the planet's orbit, which is the average distance from the planet to the star.

T is the orbital period of the planet around the star.

Assuming that the planet's orbit is approximately 1 Astronomical Unit (AU) since the period is close to 1 Earth year, and converting 402 Earth days to seconds, we find M.

T = 402
days x 24 hours/day x 3600 seconds/hour = 3.47 x 10⁷ seconds

a = 1 AU = 1.496 x 10¹¹ m

Then the mass of the star M is:

M = 4π² (1.496 x 1011)3 / (6.674 x 10⁻¹¹ x (3.47 x 10⁷)²) = 1.99 x 10³⁰ kg

Answer:

focal length of eye lens is 3.28 cm

Explanation:

As we know that the magnification of the microscope is given by the formula

[tex]M = \frac{L}{f_o}\frac{D}{f_e}[/tex]

now we will have

[tex]L = 47.8 cm[/tex]

D = 25 cm

M = 2000

[tex]f_o = 1.82 mm = 0.182 cm[/tex]

now we need to solve above equation to find focal length of eye lens

so here we will have

[tex]f_e = \frac{L}{f_o}\frac{D}{M}[/tex]

[tex]f_e = \frac{47.8}{0.182}\frac{25}[2000}[/tex]

[tex]f_e = 3.28 cm[/tex]

Answer:

Velocity: +ve, Acceleration: -ve

Explanation:

Here I've considered downward direction as positive direction.

Answer:

The answer is -,+ that is minus, plus

Explanation:

In the question, the elevator was described as moving downward, therefore its direction is negative. (-)  

From the question we could tell the elevator is decelerating, so the acceleration vector should be pointing upward, in contrast with the motion of the elevator.(+)

Velocity is a vector quantity that indicates how fast an object is moving and in what direction, it has to do with and object’s displacement, time, and direction. The SI unit of velocity is meter per second (m/s). It should not be confused with speed which is a scalar quantity and measures on distance moved without stating what direction it moves.  

For instance, it would not be enough to say that the car has a velocity of 50 miles/hr. the direction in which the car moves must be included to fully describe the velocity of the car. The correct way would be the car has a velocity of 50 miles/hour East.

In physics, acceleration is defined as the rate of change of velocity. By altering an object’s speed or direction which changes its velocity hence its acceleration. Just like velocity, acceleration is a vector quantity. The SI unit of acceleration is meter per second squared (m/s^2)

Answer:

The car moves from start to stop 328.88 m in total.

Explanation:

Vo= 0 m/s

V1= 21.1 m/s

V2= 0 m/s

a1= 2,03 m/s²

a2= -1.015 m/s²

Speed Up:

Speed up time:

V1= Vo + a1 * t1

t1= V1/a1

t1= 10.39 sec

total distance of speed up:

d1= Vo * t1 + (a1 * t1²)/2

d1= 109.57m

Slow Down:

V2= V1 - a2 * t2

t2= V1/a2

t2= 20.78 sec

total distance of slow down:

d2= V1 * t2 - (a2 * t2²)/2

d1= 219.31m

Total Distance:

TD= d1+d2= 109.57m + 219.31m

TD= 328.88 m

Answer:

65.75 deg

Explanation:

v = initial speed of launch of projectile

θ = initial angle of launch

H = maximum height of the projectile

maximum height of the projectile is given as

[tex]H=\frac{v^{2}Sin^{2}\theta }{2g}[/tex]         eq-1

D = horizontal range of the projectile

horizontal range of the projectile is given as

[tex]D=\frac{v^{2}Sin{2}\theta }{g}[/tex]                     eq-2

It is also given that

D = 1.8 H

using eq-1 and eq-2

[tex]\frac{v^{2}Sin{2}\theta }{g} = (1.8) \frac{v^{2}Sin^{2}\theta }{2g}[/tex]

[tex]Sin{2}\theta = (1.8) \frac{Sin^{2}\theta }{2}[/tex]

[tex]2 Sin\theta Cos\theta= (0.9) Sin^{2}\theta[/tex]

[tex]2 Cos\theta = (0.9) Sin\theta[/tex]

tanθ = 2.22

θ = 65.75 deg

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let [tex]\rho_{m}[/tex] be the density of metal and [tex]\rho_{w}[/tex] be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

[tex]W =0.10400\ N[/tex]

[tex]mg=0.10400[/tex]

[tex]\rho V g=0.10400[/tex]

[tex]Vg=\dfrac{0.10400}{\rho_{m}}[/tex].....(I)

The weight of metal in water is

Using buoyancy force

[tex]F_{b}=0.10400-0.08400[/tex]

[tex]F_{b}=0.02\ N[/tex]

We know that,

[tex]F_{b}=\rho_{w} V g[/tex]....(I)

Put the value of [tex]F_{b}[/tex] in equation (I)

[tex]\rho_{w} Vg=0.02[/tex]

Put the value of Vg in equation (II)

[tex]\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02[/tex]

[tex]1000\times\dfrac{0.10400}{0.02}=\rho_{m}[/tex]

[tex]\rho_{m}=5200\ kg/m^3[/tex]

Hence, The density of the metal is 5200 kg/m³.

The density of the metal bracelet can be determined through the principles of Archimedes using the difference of its weight in air and water. After calculating the buoyant force, the volume of water displaced was determined, which was also the volume of the bracelet. Dividing its weight in the air by this volume gave us the density, which is approximately 5,434 kg/m³.

In order to find the density of the metal bracelet, we use Archimedes' principle, which states that the buoyant force (force exerted on a submerged object) is equal to the weight of the fluid displaced by the object.

The first step is to calculate the difference in weight in air and water, which gives the buoyant force, i.e., the weight of the water displaced. So, the difference is 0.10400 N - 0.08400 N = 0.02000 N.

Then, we can find the volume of the water displaced by using the formula for weight, w = mg. Here, w is the buoyant force, and g is the acceleration due to gravity (9.8 m/s² on Earth). So, m = w / g = 0.02000 N / 9.8 m/s² = 0.00204 kg. This is the mass of the water displaced, which is also the volume since 1 kg of water is 1 liter (or 1000 cm³).

Finally, the density of the metal bracelet can be calculated by dividing its weight in air by the volume of water displaced, and by gravity. That is, density = 0.10400 N / (0.00204 kg x 9.8 m/s²) = 5,434.4 kg/m³. So the density of the metal bracelet is approximately 5,434 kg/m³.

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Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal =  30 degree

tension in the string =  4.5 N

WE KNOW THAT

Work =  force * distance

horizontal force =  [tex]Tcos\theta = 4.5*cos30 = 3.89 N[/tex]

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

Answer:

a) V = 195.70 m/s

b) f=3.02 × 10⁻⁴ Hz

c) T = 3311.25 seconds

Explanation:

Given:

Wavelength, λ = 646 Km = 646000 m

Distance traveled = 3410 Km = 3410000 m

Time = 4.84 h = 4.84 × 3600 s = 17424 seconds

a) The speed (V) of the wave is given as

V = distance / time

V = 3410000 m/ 17424 seconds

or

V = 195.70 m/s

b) The frequency (f) of the wave is given as:

f = V / λ

f= 195.70 / 646000

f=3.02 × 10⁻⁴ Hz

c) The time period (T)  is given as:

T = 1/ f

T = 1/ (3.02 × 10⁻⁴) Hz

T = 3311.25 seconds

Answer:

F = 3482.9 N

Explanation:

Change in velocity of the Parachutist is given as

[tex]v_f = 15 mph = 6.675 m/s[/tex]

[tex]v_i = 120 mph = 53.4 m/s[/tex]

now it is given as

[tex]\Delta v = v_f - v_i [/tex]

[tex]\Delta v = 120 - 15 = 105 mph[/tex]

[tex]\Delta v = 46.7 m/s[/tex]

now the acceleration of the parachutist is given as

[tex]a  = \frac{v_f^2 - v_i^2}{2d}[/tex]

distance moved by the parachutist is given as

[tex]d = 120 ft = 36.576 m[/tex]

now we have

[tex]a = \frac{6.675^2 - 53.4^2}{2(36.576)}[/tex]

[tex]a = - 38.4m/s^2[/tex]

Now the mass of parachutist is given as

[tex]m = 200 lb = 90.7 kg[/tex]

now we have

[tex]F = ma[/tex]

[tex]F = (90.7 kg)(38.4) = 3482.9 N[/tex]

Answer:

The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]

Explanation:

Given that,

Mass of object = 6.22 kg

Distance = 1.02 m

We need to calculate the number of electron

Using formula of electric force

[tex]F_{e}=\dfrac{k(q)^2}{r^2}[/tex]....(I)

We know that,

[tex]q = Ne[/tex]

Put the value of q in equation (I)

[tex]F_{e}=\dfrac{k(Ne)^2}{r^2}[/tex].....(II)

Using gravitational force

[tex]F_{G}=\dfrac{Gm^2}{r^2}[/tex].....(III)

Equating equation (II) and (III)

[tex]F_{e}=F_{G}[/tex]

[tex]\dfrac{k(Ne)^2}{r^2}=\dfrac{Gm^2}{r^2}[/tex]

[tex]N=\sqrt{\dfrac{G}{k}}\times\dfrac{m}{e}[/tex]....(IV)

Put the value in the equation(IV)

[tex]N=\sqrt{\dfrac{6.67\times10^{-11}}{9\times10^{9}}}\times\dfrac{6.22}{1.6\times10^{-19}}[/tex]

[tex]N=3.35\times10^{9}[/tex]

Hence, The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]

Answer:

6.97 m/s, 344 degree

Explanation:

mass of car, m = 900 kg, uc = 15 m/s, vc = 5 m/s, θ = 40 degree

mass of truck, M = 1500 kg, uT = 0, vT = ?, Φ = ?

Here, vT be the velocity of truck after collision and Φ its direction above x axis.

Use conservation of momentum in X axis

900 x 15 + 1500 x 0 = 900 x 5 Cos 40 + 1500 x vT Cos Φ

13500 - 3447.2 = 1500 vT CosΦ

vT CosΦ = 6.7 .....(1)

Use conservation of momentum in y axis

0 + 0 = 900 x 5 Sin 40 + 1500 vT SinΦ

vT SinΦ = - 1.928 .....(2)

Squarring both the equations and then add

vT^2 = 6.7^2 + (-1.928)^2

vT = 6.97 m/s

Dividing equation 2 by 1

tan Φ = - 1.928 / 6.7

Φ = - 16 degree

Angle from + X axis = 360 - 16 = 344 degree

Answer:

Explanation:

We cannot walk on a surface which has no friction.

To move across the surface, take a stone and throw it in the opposite direction of motion so that you get a reaction in the direction of motion and then you move across the surface.

Answer:The sled slides 16.875m before rest.

Explanation:

[tex]a=\frac{F}{m} =\frac{12N}{20kg}[/tex]

a=0.6 m/s²

[tex]Vf=0=Vi-a.t[/tex]

[tex]t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg[/tex]

[tex]d= Vi.t - \frac{a.t^{2}}{2}[/tex]

[tex]d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m[/tex]

Explanation:

Electric field intensity E is defined as the electric (Coulomb) force on a unit test (1 C) charge. Mathematically, it is given by :

[tex]E=\dfrac{F}{q}[/tex]

The electric force is given by :

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

Where

Q and q are electric charges

d is the distance between charges

The electric field intensity at a distance d from the center is given by :

[tex]E=\dfrac{k\dfrac{qQ}{d^2}}{q}[/tex]

So,

[tex]E=\dfrac{kQ}{d^2}[/tex]

Hence, this is the required solution.

Answer:

The distance between the screen and slit is 1.08 m

Explanation:

Given that,

Wavelength = 681 nm

Width a= 0.109 mm

Number of fringe n = 14

Distance from the center of the central maximum d= 9.51 cm

We need to calculate the distance between the screen and slit

Using formula of distance

[tex]D=\dfrac{d\times a}{n\lambda}[/tex]

Where, a = width

d = distance from the center of the central maximum

[tex]\lambda[/tex] = wavelength

Put the value into the formula

[tex]D=\dfrac{9.51\times10^{-2}\times0.109\times10^{-3}}{14\times681\times10^{-9}}[/tex]

[tex]D = 1.08\ m[/tex]

Hence, The distance between the screen and slit is 1.08 m

Answer:

The answers are: a)Fcp=0,23N b)As Fcp=0,93N, it increases 4 times when the speedis doubled

Explanation:

Let´s explain what´s the centripetal force about: It´s the force applied over an object moving on a curvilinear path. This force is directed to the rotation center.

This definition is described this way:

[tex]Fcp*r=m*V^2[/tex] where:

Fcp is the Centripetal Force

r is the horizontal circle radius

m is the ball mass

V is the tangencial speed, same as the rotation speed

w is the angular speed

Here we need to note that the information we have talks about 1cycle/0,639s (One cycle per 0,639s). We need to express this in terms of radians/seconds. To do it we define that 1cycle is equal to 2pi, so we can find the angular speed this way:

[tex]w=(1cycle/0,639s)*(2pi/1cycle)[/tex]

So the angular speed is [tex]w=9,83rad/s[/tex]

Now that we have this information, we can find the tangencial speed, which will be the relation between the angular speed and the circle radius, this way:

[tex]V=w*r[/tex] so the tangencial speed is:

[tex]V=(9,83rad/s)*(0,149m)[/tex]

[tex]V=1,5m/s[/tex]

Now we have all the information to find the Centripetal Force:

[tex]Fcp=(m*V^2)/(r)[/tex]

[tex]Fcp=((0,0154kg)*(1,5m/s)^2)/(0,149m)[/tex]

a) So the Centripetal Force is: [tex]Fcp=0,23N[/tex]  

b) If the tangencial speed is doubled, its new value will be 3m/s. replacing this information we will get the new Centripetal Force is:

[tex]Fcp=((0,0154kg)*(3m/s)^2)/(0,149m)[/tex]

The Centripetal Force is: [tex]Fcp=0,93N[/tex]

Here we can see that if the speed is doubled, the Centripetal Force will increase four times.

Given:

initial angular speed, [tex]\omega _{i}[/tex] = 21.5 rad/s

final angular speed, [tex]\omega _{f}[/tex] = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

[tex]\alpha = \frac{\omega_{f} - \omega _{i}}{t}[/tex]

Now, putting the given values in the above formula:

[tex]\alpha = \frac{28.0 - 21.5}{3.50}[/tex]

[tex]\alpha = 1.86 m/s^{2}[/tex]

Therefore, angular acceleration is:

[tex]\alpha = 1.86 m/s^{2}[/tex]

Answer:

[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]

Explanation:

Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.

By using the definition of refractive index

refractive index of the medium is given by

n = speed of light in vacuum / speed of light in medium

n = c / v  ..... (1)

Use Snell's law

n = Sin i / Sin r

Where, r be the angle of refraction

From equation (1)

c / v = Sin i / Sin r

Sin r = v Sin i / c

[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]

Answer:

a) Tmax=827.12N

b) Tmin = 626.22N

c) 776.895 N

Explanation:

Given:

Mass of wrecking ball M1=63.9 Kg

Mass of the chain M2=20.5 Kg

acceleration due to gravity, g=9.8m/s²

Now,

(a)The Maximum Tension generated in the chain,

    Tmax=(M1+M2)×g)

    Tmax=(M1+M2)×(9.8 m/s²)

    Tmax=(63.9+ 20.5)×(9.8 m/s²)  

    Tmax=827.12N

(b) The Minimum Tension Tmin will be due to the weight of the wrecking ball only

Mathematically,

Tmin=weight of the wrecking ball

Tmin = 63.9kg×9.8m/s²

Tmin = 626.22N

(c)Now. the tension at 3/4 from the bottom of the chain

In this part we have to use only 75% of the chain i.e the weight acting below the point of consideration

thus, the tension will be produced by the weight of the 3/4 part of the chain and the wrecking ball

Therefore, the weight of the 3/4 part of the chain = [tex]\frac{3}{4}\times 20.5\times 9.8 N[/tex]

= 150.675 N

Hence, the tension at a point 3/4 of the way up from the bottom of the chain will be  = 150.675 N + (63.9×9.8) N = 776.895 N

Answer:

The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].

Explanation:

Given that,

Mass [tex]M=6.65\times10^{-27}\ kg[/tex]

Temperature = 20.0°C

We need to calculate the root-mean square speed

Using formula of root mean square speed

[tex]v_{rms}=\sqrt{\dfrac{3kTN_{A}}{M}}[/tex]

Where, N = Avogadro number

M = Molar mass

T = Temperature

k = Boltzmann constant

Put the value into the formula

[tex]v_{rms}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times293\times6.022\times10^{23}}{4\times10^{-3}}}[/tex]

[tex]v_{rms}=1351.37\ m/s[/tex]

We need to calculate the de Broglie wavelength

Using formula of de Broglie wavelength

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

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

[tex]\lambda=\dfrac{6.626\times10^{-34}}{6.65\times10^{-27}\times1351.37}[/tex]

[tex]\lambda=7.373\times10^{-11}\ m[/tex]

Hence, The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].

Final answer:

The de Broglie wavelength of helium atoms moving at the root-mean-square speed is approximately 4.779 × 10^-10 meters.

Explanation:

To find the de Broglie wavelength of helium atoms moving at the root-mean-square speed, we can use the equation:

λ = h / (m * v)

where λ is the de Broglie wavelength, h is Planck's constant (6.626 × 10-34 J · s), m is the mass of the helium atom (6.65 × 10-27 kg), and v is the root-mean-square speed.

The root-mean-square speed of helium atoms at room temperature can be found using the equation:

v = √(3 * k * T / m)

where k is the Boltzmann constant (1.38 × 10-23 J/K) and T is the temperature in Kelvin (20.0 + 273 = 293 K).

Plugging the values into the equations and solving for λ:

λ = (6.626 × 10-34 J · s) / (6.65 × 10-27 kg * √(3 * 1.38 × 10-23 J/K * 293 K / 6.65 × 10-27 kg))

λ = 4.779 × 10-10 m

Therefore, the de Broglie wavelength of the helium atoms moving at the root-mean-square speed is approximately 4.779 × 10-10 meters.

Answer:

The mass of neon is 13.534 g.

Explanation:

Given that,

Volume = 3.8 L

Pressure = 6.8 atm

Temperature = 470 K

Atomic mass of neon =20.2 g/mol

Gas constant R = 8.314 = 0.082057 L atm/mol K

We need to calculate the mass of neon

Using equation of gas

[tex]PV=nRT[/tex]

[tex]6.8\times3.8=n\times0.082057\times470[/tex]

[tex]n=\dfrac{6.8\times3.8}{0.082057\times470}[/tex]

[tex]n= 0.670[/tex]

We know that,

[tex]n = \dfrac{m}{molar\ mass}[/tex]

[tex]m = n\times molar\ mass[/tex]

[tex]m=0.670\times20.2[/tex]

[tex]m=13.534\ g[/tex]

Hence, The mass of neon is 13.534 g.

Answer:

Part a)

t = 1.05 s

Part b)

[tex]\theta = 8.78 rad[/tex]

Explanation:

Initial angular speed is given as

[tex]\omega_i = 40 rev/min = 0.66 rev/s[/tex]

[tex]\omega_i = 2\pi (0.66) = 4.19 rad/s[/tex]

angular acceleration is given as

[tex]\alpha = 8 rad/s^2[/tex]

now we have

part a)

final angular speed = 120 rev/min

[tex]\omega_f = 2\pi(\frac{120}{60} rev/s)[/tex]

[tex]\omega_f = 12.57 rad/s[/tex]

now by kinematics we have

[tex]\omega_f = \omega_i + \alpha t[/tex]

[tex]12.57 = 4.19 + 8 t[/tex]

[tex] t = 1.05 s[/tex]

Part b)

Angle turned by the blades is given by

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]

[tex]\theta = 4.19(1.05) + \frac{1}{2}(8)(1.05)^2[/tex]

[tex]\theta = 8.78 rad[/tex]

Answer:

The magnetic field strength, B = 110 T

Explanation:

It is given that,

Cyclotron frequency, [tex]\nu=3\times 10^{12}\ Hz[/tex]

We need to find the magnetic field strength. The formula for cyclotron frequency is given by :

[tex]\nu=\dfrac{qB}{2\pi m}[/tex]

B is the magnetic field strength

q and m are the charge and mass of electron.

[tex]B=\dfrac{2\pi m\nu}{q}[/tex]

[tex]B=\dfrac{2\pi\times 9.1\times 10^{-31}\ kg\times 3\times 10^{12}\ Hz}{1.6\times 10^{-19}\ C}[/tex]          

B = 107.20 T

or

B = 110 T (approx)

So, the magnetic field strength of the electron is 110 T. Hence, this is the required solution.

Using the charge, mass, and cyclotron frequency of the electron, the magnetic field strength is calculated to be approximately 0.084 T.

The frequency of the electron in the magnetic field, called the cyclotron frequency, can be used to find the strength of the magnetic field. The equation for cyclotron frequency is ƒ = qB/2πm, wherein ƒ is the frequency, q is the charge, B is the magnetic field, and m is the mass of the electron. In this case, to solve for B (magnetic field), we can rearrange the equation to B = 2πmƒ/q.

Substituting the given values, ƒ = 3.0 x 10^12 Hz, q = 16 x 10^-19 C, and m = 9.1 x 10^-31 kg, into the formula, and performing the proper calculation, the magnetic field strength is found to be approximately 0.084 T, which corresponds to the option (D).

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