Enter your answer in scientific notation. Be sure to answer all parts. Calculate the nuclear binding energy (in J) and the nuclear binding energy per nucleon of 107 Ag 47 (106.905093 amu).

Answer:

Distance covered by the car is 56.01 feet.

Explanation:

It is given that,

Initial velocity, u = 50 mi/h = 73.33 ft/s

Constant deceleration of the car, [tex]a=-48\ ft/s^2[/tex]

Final velocity, v = 0

Let s is the distance covered before the car comes to rest. It can be calculated using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0^2-(73.33\ ft/s^2)^2}{2\times -48\ ft/s^2}[/tex]

s = 56.01 ft

So, the distance covered before the car comes to a stop is 56.01 feet. Hence, this is the required solution.

The distance covered by the car as it comes to rest is 17.1 m.

To calculate the distance covered before the car come to stop, we use the formula below.

Formula:

Where:

make s the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, the distance covered by the car as it comes to rest is 17.1 m.

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Answer:

hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]

Explanation:

shift in wavelength due to compton effect is given by

[tex]\lambda ^{'}-\lambda =\frac{h}{m_{e}c}\times(1-cos\theta )[/tex]

λ' = the wavelength after scattering

λ= initial wave length

h= planks constant

m_{e}= electron rest mass

c= speed of light

θ= scattering angle = 180°

compton wavelength is

[tex]\frac{h}{m_{e}c}= 2.43\times10^{-12}m[/tex]

[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1-cos\theta )[/tex]

[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1+1 )[/tex]  ( put cos 180°=-1)

also given λ'=2λ

putting values and solving we get

[tex]\lambda =4.86\times10^{-12}m[/tex]

hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]

Answer:

Force, [tex]F=8.23\times 10^{-8}\ N[/tex]

Explanation:

It is given that,

Each ion in Na⁺ and Cl⁻ has a charge of, [tex]q=1.6\times 10^{-19}\ C[/tex]

Distance between two ions, [tex]d=5.29\times 10^{-11}\ m[/tex]

We need to find the electrostatic force. It is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{(1.6\times 10^{-19})^2}{(5.29\times 10^{-11})^2}[/tex]

[tex]F=8.23\times 10^{-8}\ N[/tex]

So, the magnitude of electrostatic force between them is [tex]F=8.23\times 10^{-8}\ N[/tex]. Hence, this is the required solution.

Answer:

Magnetic field, [tex]B=4.16\times 10^{-5}\ T[/tex]

Explanation:

It is given that,

Velocity of proton, [tex]v=3.5\times 10^7\ m/s[/tex]

Magnetic force, [tex]F=1.65\times 10^{-16}\ N[/tex]

Charge of proton, [tex]q=1.6\times 10^{-19}\ C[/tex]

We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]

[tex]B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}[/tex]

B = 0.0000416 T

[tex]B=4.16\times 10^{-5}\ T[/tex]

Hence, this is the required solution.

The strength of the magnetic field, given the force on the proton, its charge, and velocity, and the angle between the velocity and the magnetic field, is approximately 4.21 x 10^-5 T.

The strength of a magnetic field can be calculated using the formula for the force exerted on a moving charge in a magnetic field, which is F = q * v * B * sin(θ). Here, F is the magnetic force, q is the charge of the particle, v is the speed of the particle, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field direction.

In this problem, the magnetic force (F) is given as 1.65 x 10^-16 N, the charge of a proton (q) is +1.6 x 10^-19 C, the speed of the proton (v) is 3.5 x 10^7 m/s, and the angle between the velocity and the magnetic field direction (θ) is 45 degrees. Hence we can rearrange the formula to find the magnetic field (B), getting B = F / (q * v * sin(θ)).

Replacing the values into the equation gives: B = 1.65 x 10^-16 N / (+1.6 x 10^-19 C * 3.5 x 10^7 m/s * sin(45°)) which gives the strength of the magnetic field as approximately 4.21 x 10^-5 T.

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Final answer:

To determine the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper, use the formula L = (R·A)/ρ. The length of the copper wire, based on the provided resistance of 0.172 Ω, cross-sectional area of 7.85 × [tex]10^-^5[/tex] m², and resistivity of 1.72 × [tex]10^-^8[/tex]Ω·m, is calculated to be 7.85 meters.

Explanation:

The question involves determining the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper. The formula to calculate the length of the wire is derived from Ohm's law, which in terms of resistivity (ρ) is expressed as R = ρL/A, where R is the resistance, L is the length of the wire, A is the cross-sectional area, and ρ is the resistivity of the material.

We are provided with the resistance (R) as 0.172 Ω, the cross-sectional area (A) as 7.85 × [tex]10^-^5[/tex] m², and the resistivity (ρ) of copper as 1.72 × [tex]10^-^8[/tex] Ω·m. By rearranging the formula to solve for L, we get L = (R·A)/ρ. Substituting the given values into the formula, we find the length of the copper wire.

Let's calculate: L = (0.172 Ω × 7.85 × [tex]10^-^5[/tex] m^2) / (1.72 × [tex]10^-^8[/tex] Ω·m) = 7.85 m. Therefore, the length of the copper wire is 7.85 meters.

Answer:

[tex]Q = 8.61 \times 10^{-4} C[/tex]

Explanation:

Since in LC oscillation there is no energy loss

so here we can say that

initial total energy of capacitor = energy stored in capacitor + energy stored in inductor at any instant of time

so we can say

[tex]\frac{Q^2}{2C} = \frac{q^2}{2C} + \frac{1}{2}Li^2[/tex]

now we have

[tex]q = 3\mu C[/tex]

[tex]i = 75 \mu A[/tex]

now we have

[tex]Q^2 = q^2 + (LC) i^2[/tex]

we also know that

[tex]2\pi f = \frac{1}{\sqrt{LC}}[/tex]

[tex]2\pi(1.6) = \frac{1}{\sqrt{LC}}[/tex]

[tex]LC = 9.89 \times 10^{-3}[/tex]

now from above equation

[tex]Q^2 = (3\mu C)^2 + (9.89 \times 10^{-3})(75 \mu A)[/tex]

[tex]Q = 8.61 \times 10^{-4} C[/tex]

To find the maximum charge of the capacitor in the LC circuit, set up a cosine equation using the initial charge and current value.

To find the maximum charge of the capacitor in the LC circuit, we need to utilize the relationship between the charge on the capacitor and the current in the circuit. At time t = 0, the capacitor is fully charged, so the initial charge is Q = 3 µC. The current in the circuit is equal to 75 µA. Knowing these values, we can set up a cosine equation to find q(t). By solving the equation, we can find the maximum charge of the capacitor.

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Answer:

option (B) - true

option (C) - true

option (D) - true

option (E) - true

option (F) - true

Explanation:

Answer:

Initial velocity = [tex]29.43m/s[/tex] b) Height it reaches = 44.145 m

Explanation:

Using the first equation of motion we have

[tex]v=u+at[/tex]

here

v is the final velocity

u is the initial velocity

a is the acceleration of the object

t is time

When the ball reaches it's highest point it's velocity will become 0 as it will travel no further

Also the acceleration in our case is acceleration due to gravity ([tex]-9.8m/s^{2}[/tex]) as the ball moves in it's influence alone with '-' indicating downward direction

Thus applying the values we get

[tex]0=u-(9.81)m/s^{2}\times 3\\\\u=19.43m/s[/tex]

b)

By 3rd equation of motion we have

[tex]v^{2}-u^{2}=2gs[/tex]

here s is the distance covered

Applying the value of u that we calculated we get

[tex]s=\frac{-u^{2}}{2g}\\\\s=\frac{-29.43^{2}}{-2\times 9.81}\\\\s=44.145m[/tex]

Magnitude of force on each chain suspended with 100 ib load in a room, must be support is 314.54 N.

If the load, is hanging on chain or rope, and is in the equilibrium state, then the sum of all the horizontal component of it must is equal to zero.

Given information-

The load suspended by two chains in a room is 100-ib.

The angle between each chain and the horizontal ceiling is 45°.

The image attached below shows the given situation.

First convert the 100-ib into unit of N as,

[tex]F=100\rm ib=100\times4.4482\rm N\\F=444.82N[/tex]

As the load suspended by two chains in a room is 100-ib and the angle between each chain and the horizontal ceiling is 45°. Thus, the vertical component of the,

[tex]\sin (45)=\dfrac{F_y}{444.82}\\F_y=314.54\rm N[/tex]

Hence, the magnitude of the force each chain must be support is 314.54 N.

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Thus, the magnitude of the force each chain must support is approximately 70.7 lb.

To find the magnitude of the force each chain must support, you can use the principles of equilibrium.

Given:

Since the system is in equilibrium, the vertical forces must balance the weight of the load, and the horizontal components of the forces must cancel each other out.

1. Determine the vertical component of the force in each chain:

Each chain supports part of the total load. Let [tex]\( F \)[/tex] be the force in each chain.

Since the chains are symmetric, each chain supports half the load.

The vertical component of the force in each chain is given by:

[tex]\[ F_{\text{vertical}} = F \sin(45^\circ) \][/tex]

where [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex].

2. Calculate the vertical component for each chain:

For equilibrium, the sum of the vertical components of the forces in the two chains must equal the total load:

[tex]\[ 2 \times F \sin(45^\circ) = 100 \text{ lb} \][/tex]

Substitute [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex] :

[tex]\[ 2 \times F \times \frac{\sqrt{2}}{2} = 100 \][/tex]

Simplify:

[tex]\[ F \sqrt{2} = 100 \][/tex]

Solve for [tex]\( F \)[/tex]:

[tex]\[ F = \frac{100}{\sqrt{2}} \][/tex]

Rationalize the denominator:

[tex]\[ F = \frac{100 \sqrt{2}}{2} = 50 \sqrt{2} \][/tex]

3. Approximate the force:

Using [tex]\( \sqrt{2} \approx 1.414 \)[/tex]:

[tex]\[ F \approx 50 \times 1.414 = 70.7 \text{ lb} \][/tex]

This question requires Newton's Second Law and the formula for kinetic energy to solve. Unfortunately, we lack the man's mass in the problem as posed, making it impossible to determine the velocity. If it is a class question, the student should consult with the professor considering possible typographical errors.

This question is about Newton's Second Law of Motion, which states that the acceleration of an object is equal to the net force acting on it divided by the object's mass. In this case, the forces acting on the man are a forward force of 1.200×10³ N and a backwards air resistance force of 615 N. The net force the man feels is therefore (1.200×10³ - 615) N = 585 N.

We also know that the formula for work done (Work = Force x Distance) and that this work is transferred into kinetic energy (Work = 1/2 x Mass x Velocity²). He moved 25.0 m, but we are not given the man's mass to find the velocity directly, making it impossible to solve unless we have the mass. However, if you received this question in class and believe that there is a typo, please consult with your professor for clarification.

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The man will be going approximately 17.1 m/s after 25.0 m.

Sure, here is the solution to the problem:

Given:

Force applied (F_a) = 1.200 × 10³ N

Force resisting (F_r) = 615 N

Distance (d) = 25.0 m

To find:

Final velocity (v)

Solution:

Step 1: Calculate the net force (F_net)

The net force is the difference between the applied force and the resisting force.

F_net = F_a - F_r

F_net = 1.200 × 10³ N - 615 N

F_net = 585 N

Step 2: Calculate the work done (W)

The work done is equal to the net force times the distance.

W = F_net × d

W = 585 N × 25.0 m

W = 14,625 J

Step 3: Calculate the final velocity (v)

The work done is also equal to the change in kinetic energy (ΔKE).

ΔKE = W

(1/2)mv² = 14,625 J

where:

m is the mass of the man and the bicycle

Since the man and the bicycle are initially at rest, their initial velocity (u) is 0. Therefore, the final velocity (v) can be calculated as follows:

v² = 2W / m

v = √(2W / m)

Assuming a mass of 100 kg for the man and the bicycle, we can calculate the final velocity as follows:

v = √((2 × 14,625 J) / (100 kg))

v = √(292.5 J/kg)

v ≈ 17.1 m/s

Answer:

The angular speed and tangential speed are 58.69 rad/s and 7.92 m/s.

Explanation:

Given that,

Radius = 0.344 m

Speed v= 20.1 m/s

(I). We need to calculate the angular speed

Firstly we will calculate the time

Using formula of time

[tex]t = \dfrac{d}{v}[/tex]

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

[tex]t =\dfrac{2\times3.14\times0.344}{20.1}[/tex]

[tex]t=0.107[/tex]

The angular velocity of the tire

[tex]\omega=\dfrac{2\pi}{t}[/tex]

[tex]\omega=\dfrac{2\times3.14}{0.107}[/tex]

[tex]\omega=58.69\ rad/s[/tex]

Now, using formula of angular velocity

(II). We need to calculate the tangential speed of a point located 0.135 m from the axle

The tangential speed

[tex]v = r\omega[/tex]

Where,

r = distance

[tex]\omega[/tex]= angular velocity

Put the value into the formula

[tex]v= 0.135\times58.69[/tex]

[tex]v=7.92\ m/s[/tex]

Hence, The angular speed and tangential speed are 58.69 rad/s and 7.92 m/s.

Answer:

Frequency, f = 481.8 Hz

Explanation:

Given that,

Length of windpipe, l = 4.6 ft = 0.182 m

We need to find the lowest resonant frequency of this pipe at 33 degrees Celcius. Firstly, we will find the speed of sound at 33 degrees Celcius as :

[tex]v=331+0.6T[/tex]

[tex]v=331+0.6\times 33[/tex]

v = 350.8 m/s

At resonance, wavelength is equal to 4 times length of pipe i.e.

λ = 4 l

We need that, [tex]f=\dfrac{v}{\lambda}[/tex]

[tex]f=\dfrac{350.8\ m/s}{4\times 0.182\ m}[/tex]

f = 481.8 Hz

So, the resonant frequency of the windpipe is 481.8 Hz. Hence, this is the required solution.

The lowest resonant frequency of the whooping crane's windpipe, assuming it is closed at one end and a temperature of 33°C, is approximately 62.82 Hz.

The lowest resonant frequency of the whooping crane's windpipe, which can be thought of as a pipe that is closed at one end, can be found using the formula for the fundamental frequency of such a pipe: f = v/λ, where f is the frequency, v is the speed of sound, and λ is the wavelength. In this case, the speed of sound is dependent on the temperature and can be approximated as v = 331.4 + 0.6*T m/s, where T is the temperature in degrees Celsius.

At a temperature of 33°C, the speed of sound, v, is approximately 351.8 m/s. Note that we need to convert the length of the windpipe from feet to meters, with 4.6 ft being approximately 1.4 m. The fundamental wavelength for a pipe closed at one end is four times the length of the pipe (λ=4L), so in this case, λ=4*1.4 m = 5.6 m.

Substituting these values into the frequency formula gives us f = v/λ = 351.8 m/s / 5.6 m = 62.82 Hz. So, the lowest resonant frequency of the whooping crane's windpipe is approximately 62.82 Hz.

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Answer:

The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]

Explanation:

Given that,

Upward current = 24 A

Force per unit length[tex]\dfrac{F}{l} =88\times10^{4}\ N/m [/tex]

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force

[tex]F=ILB[/tex]

[tex]\dfrac{F}{l}=\dfrac{\mu I_{1}I_{2}}{2\pi r}[/tex]

Where,

[tex]\dfrac{F}{l}[/tex]=force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

[tex]88\times10^{4}=\dfrac{4\pi\times10^{-7}\times24\times I_{2}}{2\pi \times7\times10^{-2}}[/tex]

[tex]I_{2}=\dfrac{88\times10^{4}\times7\times10^{-2}}{2\times\times10^{-7}\times24}[/tex]

[tex]I_{2}=1.3\times10^{10}\ A[/tex]

Hence, The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]

Answer:

(a) 142.5 rad/s

Explanation:

Torque = 45 Nm

Moment of inertia = 30 kgm^2

w0 = 0, t = 95 second

(a) Let it is rotating with angular speed w.

Torque = moment of inertia x angular acceleration

45 = 30 x α

α = 1.5 rad/s^2

Use first equation of motion for rotational motion

w = w0 + α t

w = 0 + 1.5 x 95

w = 142.5 rad/s

(b) if he hold rocks, then the moment of inertia change and then angular acceleration change and then final angular velocity change.

Explanation:

It is given that,

Speed of proton, [tex]v=5\times 10^7\ m/s[/tex]

Magnetic force, [tex]F=1.7\times 10^{-16}\ N[/tex]

(1) The strength of the magnetic field if there is a 45º angle between it and the proton’s velocity. The magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]

[tex]B=\dfrac{1.7\times 10^{-16}\ N}{1.6\times 10^{-19}\ C\times 5\times 10^7\ m/s\ sin(45)}[/tex]

B = 0.00003 T

or

B = 0.03 mT

The magnitude of Earth's magnitude of 25 to 65 Tesla. The value obtained in part (a) is not consistent with the known strength of the Earth’s magnetic field on its surface.

Answer:

The speed of her at the bottom is 24.035 m/s.

Explanation:

Given that,

Speed of cyclist = 12 m/s

Height = 30 m

Distance d = 450 m

Mass of cyclist and bicycle =70 kg

Drag force = 12 N

We need to calculate the speed at the bottom

Using conservation of energy

K.E+P.E=drag force+K.E+P.E

Potential energy is zero at the bottom.

K.E+P.E=drag force+K.E

[tex]\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}[/tex]

[tex]\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2[/tex]

[tex]25620=5400+35v^2[/tex]

[tex]35v^2=25620-5400[/tex]

[tex]35v^2=20220[/tex]

[tex]v^2=\dfrac{20220}{35}[/tex]

[tex]v=\sqrt{\dfrac{20220}{35}}[/tex]

[tex]v=24.035\ m/s[/tex]

Hence, The speed of her at the bottom is 24.035 m/s.

Answer:

Speed at the bottom of the slope is 24.03 m /sec

Explanation:

We have given speed of the cyclist v = 12 m/sec

She starts down a 450 m long that is 30 m high

Si distance = 450 m and height h = 30 m

Combined mass of cyclist and bicycle m = 70 kg

Drag force = 12 N

We have to find the speed at the bottom

According to conservation theory

KE +PE = work done by drag force + KE + PE

As we know that at the bottom there will be no potential energy

So [tex]\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2[/tex]

[tex]\frac{1}{2}\times 70\times v^2=20220[/tex]

[tex]v=24.03m/sec[/tex]

Answer:

1.28 x 10^4 N

Explanation:

m = 1500 kg, h = 450 km, radius of earth, R = 6400 km

Let the acceleration due to gravity at this height is g'

g' / g = {R / (R + h)}^2

g' / g = {6400 /  (6850)}^2

g' = 8.55 m/s^2

The force between the spacecraft and teh earth is teh weight of teh spacecraft

W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N

Answer:

2.7 J

Explanation:

The energy of one photon is given by

[tex]E=hf[/tex]

where

h is the Planck constant

f is the frequency

For the photons in this problem,

[tex]f=6.8\cdot 10^9 Hz[/tex]

So the energy of one photon is

[tex]E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J[/tex]

The number of photons contained in 1.0 mol is

[tex]N_A = 6.022\cdot 10^{23} mol^{-1}[/tex] (Avogadro number)

So the total energy of [tex]N_A[/tex] photons contained in 1.0 mol is

[tex]E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J[/tex]

Final answer:

The Physics question involves calculating the total momentum before and after a traffic collision. Total momentum before the collision is 76800 kg·m/s northward. Assuming a perfectly inelastic collision where the vehicles stick together, the total momentum remains unchanged after the collision.

Explanation:

To calculate the total momentum before the collision, we use the formula: momentum = mass × velocity. For the 1200 kg car traveling north at 14 m/s, its momentum is 1200 kg × 14 m/s = 16800 kg·m/s northward.

For the 2000 kg truck traveling at 30 m/s, its momentum is 2000 kg × 30 m/s = 60000 kg·m/s northward. The total momentum before the collision is the sum of these two momenta, so 76800 kg·m/s northward.

Assuming the collision is perfectly inelastic and the two vehicles stick together, the law of conservation of momentum states that the total momentum before the collision must be equal to the total momentum after the collision.

Therefore, the total momentum after the collision will also be 76800 kg·m/s northward.

Answer: a) 7,00 centimeters

(b) 259. 19 feet

(c) 3110.28 inches

(d) 0.049 miles

Explanation:

(a) We know that 1 meter = 100 centimeters

Therefore,

[tex]79\ m= 7,900\text{ centimeters}[/tex]

(b)Since 1 meter = 3.28084 feet

Then, [tex]79\ m= 79\times3.28084=259.18636\approx 259.19\text{ feet}[/tex]

(c) Since, 1 feet = 12 inches.

[tex]79\ m=259.19\text{ feet}=259.19\times12=3110.28\text{ inches}[/tex]

(d) [tex]\text{Since 1 feet= }\dfrac{1}{5280}\text{ mile}[/tex]

[tex]79\ m=259.19\text{ feet}=\dfrac{259.19}{5280}= 0.0490890151515\approx0.049\text{ miles}[/tex]

Answer:

vf = (m1*v1i)/(m1 + m2)

Explanation:

take * as multiplication

from the conservation of linear momentum, we know that:

sum of pi = sum of pf,   where sum of pi = m1*v1i + m2*v2i

                                                              pf = vf(m1 + m2) since the masses stick                        

                                                                                           together.

v2i = o, since mass m2 is initially stationary.

m1*v1i + m2*(0) = vf(m1 + m2)

m1*v1i = vf(m1 + m2)

therefore

vf = m1*v1i/(m1 + m2)

Answer:

It takes 0.43 seconds before the pendulum attains the maxium speed.

Explanation:

L= 0.75m

g= 9.8 m/s²

T= 2π * √(L/g)

T=1.73 sec

T(vmax) = T/4

T(vmax) = 0.43 sec

The time elapsed before a simple pendulum attains its greatest speed is given by t = T/4 = π√(L/g)/2, where T is the period and L is the length of the pendulum.

When a simple pendulum is released from rest, it oscillates back and forth. The time it takes for the pendulum to reach its greatest speed depends on the length of the pendulum. The formula for the period of a simple pendulum is given by:

T = 2π√(L/g)

Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. To find the time it takes for the pendulum to reach its greatest speed, we can divide the period by 4. This is because the pendulum reaches its greatest speed when it is at the bottom of its swing, which is halfway through one period.

So, the time elapsed before the pendulum attains its greatest speed is:

t = T/4 = (2π√(L/g))/4 = π√(L/g)/2

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The magnetic force acting on a charged particle moving perpendicular to the field is:

[tex]F_{b}[/tex] = qvB

[tex]F_{b}[/tex] is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

[tex]F_{c}[/tex] = mv²/r

[tex]F_{c}[/tex] is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set [tex]F_{b}[/tex] equal to [tex]F_{c}[/tex] and solve for B:

qvB = mv²/r

B = mv/(qr)

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

v = 7.50×10⁷m/s

q = 1.60×10⁻¹⁹C (proton charge)

r = 0.800m

Plug these values in and solve for B:

B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)

B = 0.979T

The magnetic field strength of the proton is 0.979Tesla

The magnetic force acting on a charged particle moving perpendicular to the field is expressed using the equation.

Fm = qvB

The centripetal force traveled by the proton in a circular path is expressed as:

Fp = mv²/r

To get the field strength, we will equate both the magnetic force and the centripetal force as shown:

Fm = Fp

qvB = mv²/r

qB = mv/r

m is the mass of a proton

v is the velocity = 7.50×10⁷ m/s

m is the mass on the proton = 1.67 × 10⁻²⁷kg

q is the charge on the proton =  1.60×10⁻¹⁹C

r is the radius = 0.800m

Substitute the given parameters into the formula as shown:

[tex]B=\frac{mv}{qr}\\B = \frac{1.67 \times 10^{-27} \times 7.5 \times 10^7}{1.60 \times 10^{-19} \times 0.8}[/tex]

[tex]B=0.979Tesla[/tex]

On solving, the magnetic field strength of the proton is 0.979Tesla

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Answer:

The pressure is 2.167 psi.

Explanation:

Given that,

Diameter = 1.5 feet

Height = 10 feet

We need to calculate the psi at 5 feet

Using formula of pressure at a depth in a fluid

Suppose the fluid is water.

Then, the pressure is

[tex]P=\rho g h[/tex]

Where, P = pressure

[tex]\rho[/tex] = density

h = height

Put the value into the formula

[tex]P=1000\times9.8\times1.524[/tex]

[tex]P=14935.2\ N/m^2[/tex]

Pressure in psi is

[tex]P=2.166167621\ psi[/tex]

[tex]P=2.167\ psi[/tex]

Hence, The pressure is 2.167 psi.  

Answer:

Distance of closest approach, [tex]r=1.91\times 10^{-14}\ m[/tex]

Explanation:

It is given that,

Charge on proton, [tex]q_p=e[/tex]

Charge on alpha particle, [tex]q_a=2e[/tex]

Mass of proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]

Mass of alpha particle, [tex]m_a=4m_p=6.68\times 10^{-27}\ kg[/tex]

The distance of closest approach for two charged particle is given by :

[tex]r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}[/tex]

[tex]r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}[/tex]

[tex]r=1.91\times 10^{-14}\ m[/tex]

So, their distance of closest approach, as measured between their centers [tex]1.91\times 10^{-14}\ m[/tex]. Hence, this is the required solution.

Answer:

The velocity of the boat after Batman lands in it is 2.08 m/s.

Explanation:

Given that,

Mass of batman [tex]m_{1}= 88.8\ kg[/tex]

Mass of boat [tex]m_{2}=530\ kg[/tex]

Initial velocity = 14.5 m/s

We need to calculate the velocity of boat

Using conservation of momentum

[tex]m_{1}u=(m_{1}+m_{2})v[/tex]...(I)

Where, [tex]m_{1}[/tex]=mass of batman

[tex]m_{2}[/tex] =mass of boat

u=initial velocity

v = velocity of boat

Put the value in the equation

[tex]88.8\times14.5=530+88.8\times v[/tex]

[tex]v=\dfrac{88.8\times14.5}{530+88.8}[/tex]

[tex]v=2.08\ m/s[/tex]

Hence, The velocity of the boat after Batman lands in it is 2.08 m/s.

The ball thrown by a major-league pitcher to a catcher 17.0m away, with a speed of 41.0m/s, would drop approximately 0.84 meters due to the gravitational acceleration.

The subject of this question involves physics, specifically the concept of projectile motion. When a pitcher throws a ball with a horizontal speed, the ball also experiences a vertical motion due to gravity. Therefore, even though the ball is thrown horizontally, it will gradually drop as it travels towards the catcher. To figure out how far it drops, we can use the physics equation for motion under constant acceleration: distance = 0.5 * acceleration * time2.

Here, the acceleration is due to gravity, which is approximately 9.8m/s2. The time can be calculated by dividing the horizontal distance (17.0 m) by the horizontal speed (41.0 m/s), giving us approximately 0.4146 s. Substituting these into the equation: 0.5 * 9.8m/s2 * (0.4146s)2 = 0.84 meters, the drop of the ball is approximately 0.84 meters.

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Answer:

Mass of the skater, m = 72.75 kg

Explanation:

It is given that,

Radius of the circular path, r = 31 m

Speed of the skater, v = 14 m/s

Centripetal force, F = 460 N

We need to find the mass of the skater. It can be determined using the formula of centripetal force. It is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]m=\dfrac{Fr}{v^2}[/tex]

[tex]m=\dfrac{460\ N\times 31\ m}{(14\ m/s)^2}[/tex]

m = 72.75 kg

So, the mass of the skater is 72.75 kg. Hence, this is the required solution.

The centripetal acceleration is found to be approximately 6.32 m/s², which helps determine the mass of the skater to be around 72.78 kg.

To find the mass of the speed skater, we will use the formula for centripetal force:

F = m * ac, where:

F is the centripetal force (460 N)

m is the mass of the skater (unknown)

ac is the centripetal acceleration

Centripetal acceleration can be calculated using:

ac = v² / r, where:

v is the speed of the skater (14 m/s)

r is the radius of the turn (31 m)

First, calculate the centripetal acceleration:

ac = (14 m/s)² / 31 m = 196 m²/s² / 31 m ≈ 6.32 m/s²

Now, substitute into the centripetal force formula to solve for the mass (m):

460 N = m * 6.32 m/s²

Solving for m:

m = 460 N / 6.32 m/s² ≈ 72.78 kg

Therefore, the mass of the skater is approximately 72.78 kg.

Answer:

Tangential Speed equals 5.57m/s

Explanation:

In the figure shown for equilibrium along y- axis we have

[tex]\sum F_{y}=0[/tex]

Resolving Forces along y axis we have

[tex]Tcos(\theta )=mg............(i)[/tex]

Similarly along x axis

[tex]\sum F_{x}=ma_{x}[/tex]

[tex]Tsin(\theta )=m[tex]\frac{v^{2} }{r}[/tex]............(ii)[/tex]

Dividing ii by i we have

[tex]tan(\theta )=\frac{v^{2}}{rg}[/tex]

In the figure below we have [tex]r=lsin(\theta )[/tex]

Thus solving for v we have

[tex]v=\sqrt{lgsin(\theta) tan(\theta )}[/tex]

Applying values we get

v=5.576m/s

Answer:

≈1,39 and 4.56 kg.

Explanation:

for more details see the attachment. Note, the ρ(Cu) means 'Density of copper', ρ(Al) means 'Density of aluminium'.