An object moves uniformly around a circular path of radius 23.5 cm, making one complete revolution every 1.95 s. (a) What is the translational speed of the object? (b) What is the frequency of motion in hertz? (c) What is the angular speed of the object?

The person must exert a force of 8.32 N to accelerate the lawn mower from rest to 1.3 m/s in 2.5 seconds.

When an object is moving at a constant speed, the net force acting on it is zero.

In this case, the force applied by the person along the handle is balanced by the frictional force opposing the motion.

So, the frictional force [tex](\(F_{\text{friction}}\))[/tex] is equal in magnitude but opposite in direction to the force applied by the person.

Given:

- Force applied by the person  = 87.5 N

- The angle of the handle = 45.0 degrees to the horizontal

To find the horizontal component of the force applied by the person, use trigonometry:

[tex]\[F_{\text{person-horizontal}} = F_{\text{person}} \cdot \cos(\theta)\][/tex]

[tex]\[F_{\text{person-horizontal}} = 87.5 \, \text{N} \cdot \cos(45.0^\circ)\][/tex]

                          [tex]= 87.5 \, \text{N} \cdot 0.707 = 61.29 \, \text{N}[/tex]

So, the person exerts a horizontal force of 61.29 N.

Since the lawn mower moves at a constant speed, the frictional force must be equal in magnitude and opposite in direction to this horizontal force:

[tex]\[F_{\text{friction}} = -61.29 \, \text{N}\][/tex]

To calculate this force [tex](\(F_{\text{acceleration}}\))[/tex], we can use Newton's second law:

[tex]\[F = m \cdot a\][/tex]

Given:

Mass of the lawn mower = 16.0 kg

Acceleration = [tex]\(\frac{\Delta v}{\Delta t}\)[/tex],

Now, calculate [tex]\(\Delta v\):[/tex]

[tex]\[\Delta v = 1.3 \, \text{m/s} - 0 \, \text{m/s} = 1.3 \, \text{m/s}\][/tex]

Now, using Newton's second law:

[tex]\[F_{\text{acceleration}} = m \cdot a = (16.0 \, \text{kg}) \cdot \left(\frac{1.3 \, \text{m/s}}{2.5 \, \text{s}}\right)\][/tex]

[tex]\[F_{\text{acceleration}} = 8.32 \, \text{N}\][/tex]

Thus, the person must exert a force of 8.32 N.

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To accelerate a 16.0 kg lawn mower from rest to 1.3 m/s in 2.5 seconds, a person needs to exert a force of 136 N along the handle, assuming that the same friction force is present as when the mower is pushed at constant speed.

In order to find the force that the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, we can use the second law of motion that states that the net force acting on an object is equal to its mass times its acceleration: F = m*a.

In this case, the acceleration would be the change in speed over time, which is (1.3 m/s - 0 m/s) / 2.5 s = 0.52 m/s². Consequently, F = (16.0 kg) * (0.52 m/s²) = 8.3 N.

It's important to mention this force is in addition to the friction force and the force to move the mower horizontally. Asume that the 87.5 N force previously exerted was just overcoming friction, then the total horizontal force required would be 87.5 N (for friction) + 8.3 N (for acceleration) = 95.8 N. However, since this force makes an angle of 45.0 with the horizontal, the person must exert a larger force along the handle, about 135.5 N. Hence, the correct answer, expressed to three significant figures, is 136 N.

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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:

544.65 ohm and 179.84 ohm

Explanation:

Hello

Let

Resistor 1 (R1) and Resistor 2 (R2)

in series [tex]R_{eq} =R1+R2[/tex]      

in parallel

[tex]\frac{1}{R_{eqp} } =\frac{1}{R1} +\frac{1}{R2}\\\\R1=R_{eqs} -R2\\\\R1 =724.5-R2 (equation 1)\\ R_{eqp}}=\frac{R1*R2}{R1+R2}\\\\\ 135.2 = \frac{R1*R2}{R1+R2}(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data[/tex]

[tex]135.2=\frac{(724.5-R2)(R2)}{724.5-R2+R2} }\\135.2=\frac{(-R_{2} ^{2}+724.5R2) }{724.5}\\ 135.2*724.5=-R_{2} ^{2}+724.5R_{2}\\\\R_{2} ^{2} -724.5R_{2}  +97952.4 =0\\R_{2} =\frac{724.5 \±\sqrt{(-724.5^{2})-4(1)(97952.4) } }{2(1)}\\R_{2} =\frac{724.5\±\sqrt{(133092.25) } }{2(1)}\\\\R_{2} =\frac{724.5+364.81810}{2} \\R_{2} =544.65 ohm\\\\R_{2} =\frac{724.5-364.81810}{2} =179.84 ohm[/tex]\\\\

let R2=544.65 and replace in equation 1

R1=724.5-544.65

R1=179.85

so, the resistors are 544.65 ohm and 179.84 ohm

Have a  great day

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:

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 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 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:

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:

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:

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)

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:

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 electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

[tex]E=\dfrac{kq}{r^2}[/tex]

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

Answer:

Work one by the force, W = 1378.87 lb-ft

Explanation:

It is given that,

Force acting on the sled, F = 20 lb

Angle between the force and the horizontal, θ = 40°

Distance moved, d = 90 ft

We need to find the work done by the force. We know that the work done can be calculated as :

[tex]W=Fd\ cos\theta[/tex]

[tex]W=20\ lb\times 90\ ft\ cos(40)[/tex]

W = 1378.87 lb-ft

So, the work done by the force is 1378.87 lb-ft. Hence, this is the required solution.

Answer:

[tex] \frac{F}{64}[/tex]

Explanation:

m = product of masses of two objects

r = distance between the two objects = 100 m

F = initial force between the two object

r' = new distance between the two objects = 800 m

F' = new force between the two objects = ?

k = constant of proportionality

initial force between the two object is given as

[tex]F=\frac{km}{r^{2}}[/tex]                                     eq-1

new force between the two objects is given as

[tex]F'=\frac{km}{r'^{2}}[/tex]                                  eq-2

Dividing eq-2 by eq-1

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

Inserting the values

[tex]\frac{F'}{F}=\frac{100^{2}}{800^{2}}[/tex]

[tex]F' = \frac{F}{64}[/tex]

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 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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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:

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:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

The distance the ball will strike the opposite wall is 32.79 m.

The time of motion of the ball from the given height is calculated as follows;

h = vsinθ(t) + ¹/₂gt²

50 = 20 x sin(40)t + 0.5(9.8)t²

50 = 12.86t + 4.9t²

4.9t² + 12.86t - 50 = 0

solve the quadratic equation using formula method,

t = 2.14 s

The horizontal distance of the ball from the initial position is calculated as follows;

X = vcosθ(t)

X = 20 x cos(40) x 2.14

X = 32.79 m

Thus, the distance the ball will strike the opposite wall is 32.79 m.

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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:

6.4 J

Explanation:

For cart A

mA = 0.4 kg, uA = 6 m/s, vA = - 2 m/s

For cart B

mB = 0.8 kg, uB = 0, vB = ?

use the law of conservation of momentum

momentum of system before coliision = momentum of system after collision

mA uA + mB uB = mA vA + mB vB

0.4 x 6 + 0 = 0.4 x (- 2) + 0.8 x vB

vB = 4 m/s

Kinetic energy of Cart B after collision = 1/2 mB vB^2

                                                                = 1/2 x 0.8 x 4 x 4 = 6.4 J

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 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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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.

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]

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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Answer:

0.2 m

Explanation:

Diameter = 0.3 m

radius, r = 0.15 m

Length, H = 0.4 m

density of wood, d = 0.5 g/cm^3 = 500 kg/m^3

density of water, d = 1000 kg/m^3

Let h be the depth of cylinder immersed in water.

By the principle of floatation.

Buoyant force = Weight of cylinder

Volume immeresed x density of water x g = Volume of cylinder x density of wood x g

A x h x 1000 x g = A x H x 500 x g

1000 h = 500 x 0.4

h = 0.2 m

Answer:

The magnitude of the force per unit of lenght between the wires are of  F/L= 3.17 * 10⁻⁵ N/m.

Explanation:

d=0.0665m

I1=I2= 3.25A

μo= 4π * 10⁻⁷ N/A²

F/L= (μo * I1 * I2) / (2π * d)

F/L= 3.17 * 10⁻⁵ N/m

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]