An archer puts a 0.285 kg arrow in a bow and uses an average force of 182 N to draw the string back 1.32 m. Assume the energy stored in the bow is transferred to the arrow when it is shot. (a) What is the speed of the arrow as it leaves the bow

Answer:

the total trajectory length is 349.39 km

Explanation:

for the first trajectory

Time taken in first trajectory = First trajectory length /velocity  = 180 km/95 km/h = 1.894 hours

therefore since the total time is 4.5 hours

Time taken in second trajectory =  Second trajectory length /velocity  

4.5 hours- 1.894 hours = Second trajectory length / 65 km/h

Second trajectory length = 169.39 km

therefore the total trajectory length is 180 km + 169.39 km = 349.39 km

Explanation:

Below is an attachment containing the solution.

Answer:

a. the work done by the gravitational force on Block A is less than the work done by the gravitational force on Block B.

b. the speed of Block A is equal to the speed of Block B.

c. the momentum of Block A is less than the momentum of Block B.

Explanation:

a. The  work done by the gravitational force is equal to:

w = m*g*h

where m is mass, g is the standard gravitational acceleration and h is height. Given that both blocks are released from rest at the same height, then, the bigger the mass, the bigger the work done.

b. With ramps frictionless, the final speed of the blocs is:

v = √(2*g*h)

which is independent of the mass of the blocks.

c. The momentum is calculated as follows:

momentum = m*v

Given that both bocks has the same speed, then, the bigger the mass, the bigger the momentum.

The work done by gravitational force on blocks A and B is equal as the work is independent of the path. Both blocks have the same speed when they reach the final height due to the conversion of potential energy into kinetic energy. However, the momentum of Block B is greater due to its larger mass.

This question is about the principles of work, energy and momentum in physics. Let's address each part of it:

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Answer:

The acceleration of the refrigerator is [tex]a= 0.056 ms^{-2}[/tex]

Explanation:

The expression of the equation of the net force acting on the refrigerator is as follows;

F-f= ma

Here, F is the applied force, f is the force of friction, m is the mass and a is the acceleration.

It is given in the problem that you're having a hard time pushing a refrigerator having mass 355 kg across the kitchen floor. The force of your own push is 993 N. The force of friction opposing your own push is 973 N.

Put F= 993, f= 973 N and m = 355 kg in the above expression of the equation to calculate the acceleration of the refrigerator.

993 - 973 = (355)a

20 = 355 a

[tex]a= 0.056 ms^{-2}[/tex]

Therefore, the acceleration of the refrigerator is [tex]a= 0.056 ms^{-2}[/tex].

The refrigerator's acceleration is 0.0563 m/s², calculated by subtracting the force of friction from the pushing force and then dividing by the refrigerator's mass.

To find the acceleration of the refrigerator, we first need to calculate the net force acting on it. The net force is the difference between the force of the push and the force of friction.

Step 1: Calculate the Net Force

Net Force = Force of the push - Force of friction
Net Force = 993 N - 973 N
Net Force = 20 N

Step 2: Apply Newton's Second Law of Motion

According to Newton's Second Law of Motion, the acceleration (a) can be calculated using the formula:

a = Net Force / Mass

Substituting the known values:

a = 20 N / 355 kg
a ≈ 0.0563 m/s²

The refrigerator's acceleration when being pushed across the kitchen floor is approximately 0.0563 meters per second squared.

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Answer:

D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Explanation:

D)When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

The gold nucleus and alpha particle are both definitely charged therefore there is a repulsive pressure between the (gold) nucleus and the alpha particle. This causes the alpha particle to be deflected through a massive angle.

Maximum alpha debris surpassed instantly through the gold foil, which implied that atoms are ordinarily composed of open space. a few alpha particles had been deflected barely, suggesting interactions with different definitely charged particles in the atom.

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The potential difference between the bird's feet is 0.71 mV

Explanation:

Given-

Distance between the feet, x = 5 cm = 0.05 m

Diameter of the wire = 2 cm

Radius, r = 2/2 = 1 cm = 0.01 m

Current, I = 170 A

Potential difference, ΔV = ?

We know,

ΔV = IR

Where,

ΔV is the potential difference

I is the current

R is the resistance of the wire

And also

R = ρ L/ A

where,

ρ is the resistivity of aluminium wire (ρ = 2.65 X 10⁻⁸ Ωm)

L is the length

A is the area

Equating both the equations,

ΔV = I * ρL/A               (The value of R is replaced in the first equation)

ΔV =

[tex]170 * \frac{(2.65 X 10^-^8) * 0.05}{\pi (0.01)^2} \\\\\frac{22.525 X 10^-^8}{0.000314} \\\\\frac{22.525 X 10^6}{314 X 10^8} \\\\0.0717 X 10^-^2\\\\7.17 X 10^-^4[/tex]

Therefore, the potential difference between the bird's feet is 0.71 mV

Answer: Heat Energy

Explanation:

Heat is energy in its most disordered form. heat energy is the random jostling of molecules and is therefore not organized. As cells perform the chemical reactions that generate order within, some energy is inevitably lost in the form of heat. Because the cell is not an isolated system, the heat energy produced by the cell is quickly dispersed into the cell's surroundings where it increases the intensity of the thermal motions of nearby molecules. This increases the entropy of the cell's environment and keeps the cell from violating the second law of thermodynamics.

Question:

a). Use the de Broglie relation λ=h/p to find the wavelength of a raindrop with mass m=1 mg and speed 1cm/s.

ii). Does it seem likely that the wave properties of a raindrop could be easily detected?

b). Find the wavelength of electrons with KE = 500 eV.

c). If a neutron has the same wavelength as blue light (λ=450 nm) what is it's KE?

ii). What if it's an electron?

Answer:

The answers to the question are

a). The wavelength of the raindrop is 6.626*10⁻²⁶ m

The properties of the rain drop will be hardly detected

b). The wavelength of the electrons is 5.491×10⁻¹¹ m

c). The KE of the neutron is 5.242510⁻²⁸ J

ii). For an electron it will increase to be KE (electron) = 9.6392639×10⁻²⁵ J

Explanation:

Using de Broglie relations, we have

p = h/λ and E = h·f also E = 1/2·m·v²

a). λ= h/p, E= p²/2·m, p = √(2·m·E), λ = h/√(2·m·E)

Where

λ=wavelength

E = energy

p = momentum

m = mass

The kinetic energy of the rain drop  is [tex]\frac{1}{2}[/tex]×m×v² = 0.5×(‪1×10⁻⁶)(0.01)2

=  5× 10⁻¹¹ J

λ = h/√(2·m·E) = 6.626*10-34 Js/√(2×‪‪1×10⁻⁶×5× 10⁻¹¹)

= 6.626*10⁻²⁶ m

The properties of the rain drop will not be easily detected

b). The electron energy  is equivalent to 500 eV ⇒500 eV × 1.6×10⁻¹⁹ J/eV

= 8×10⁻¹⁷ J

λ = h/√(2·m·E) = 6.626×10⁻³⁴ Js/√(2*×9.1×10⁻³¹×8×10⁻¹⁷)

= 5.491×10⁻¹¹ m

c). λ = h/√(2·m·E) then √(2·m·E) = h/ λ or E = (h/λ)²/(2·m)  

= (6.626×10⁻³⁴/‪5.0×10⁻⁷‬)²/(2×1.674927471×10⁻²⁷)

E = 5.242510⁻²⁸ J

ii). For an electron, we have m = 9.10938356 × 10⁻³¹ kg

λ  = (6.626×10⁻³⁴/‪5.0×10⁻⁷‬)²/(2×9.10938356×10⁻³¹)  = 9.6392639×10⁻²⁵ J

Answer:

Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.

Explanation:

Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.  

Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:

[tex]\eta_{engine} \cdot U_{engine} = \frac{1}{2} \cdot m_{car} \cdot v^{2} + \mu_{r} \cdot m_{car} \cdot g \cdot \Delta s[/tex]

Augmented reality has the potential to superimpose digital data over real photos so that GPS maps can be combined with real pictures of stores and streets to help people locate their position.

Explanation:

An engaging perception of an original globe atmosphere, where by computer-generated perceptual knowledge the transformation of real-world entities take place and also by multiple sensory modalities, involving somatosensory, visual, auditory, haptic and olfactory forms, thus known as augmented reality.

AR app uses GPS and camera from a smartphone to deploy an augmented reality-enabled GPS navigation system. As in the web, AR tool termed as Real View Navigation is accessible to all Android and iOS clients. Google is brought its first virtual reality walking directions, now recognized as Live Experience on Google Maps.

Answer:

Total angle through which the wheel has turned 58.9s after it begins rotating is 1709.52 rad

Explanation:

The image attached would offer a better explanation

Answer:

The answer to the question is;

The total angle through which the wheel has turned 58.9 s after it begins rotating is approximately 1709.67 rad.

Explanation:

To solve the question we note the equation for the motion of the flywheel as

ω₂ = ω₁ + α·t

Where:

ω₁ = Initial angular velocity = 0 rad/s as the body is initially at rest

ω₂ = Final angular velocity

α = angular acceleration = 1.35 rad/s²

t = Time = 28.3 s

Plugging in the values, we find ω₂

ω₂ = 0 + 1.35 rad/s²× 28.3 s = 38.205 rad/s

Since the acceleration is constant, only the mean velocity is required to determine the angle traveled during the first 28.3 seconds thus

Average velocity

ω[tex]_{average}[/tex]= [tex]\frac{Final .Velocity +Initial . Velocity}{2} = \frac{\omega_2+\omega_1}{2} = \frac{38.205 rad/s+0 rad/s}{2}[/tex]

= 19.1025 rad/s

The total angle traveled in 28.3 s is ω[tex]_{average}[/tex] × time

= 19.1025 rad/s × 28.3 s = 540.60075 rad

After this the remaining time left is

58.9 s - 28.3 s = 30.6 s

Since the flywheel is moving at a constant velocity of 38.205 rad/s during the last 30.6 s we have

Angle traveled in 30.6 s at an angular velocity of 38.205 rad/s is given by

Angle traveled = Time × Angular velocity = 30.6 s × 38.205 rad/s

= 1169.073 rad

Therefore, the total angle traveled by the flywheel in 58.9 s is given by

540.60075 rad + 1169.073 rad = 1709.67375 rad ≈1709.67 rad.

the total angle traveled by the flywheel in 58.9 s ≈ 1709.67 rad.

Answer:

38 m

Explanation:

Number of turns=N=100

Magnetic field=B=0.50 T

Frequency of the generator=f=60 Hz

Rms value of emf=[tex]E_{rms}=120 V[/tex]

We have to find the length of the wire from which the coil is made.

Peak value of emf=[tex]E_0=E_{rms}\sqrt 2=120\times \sqrt 2=169.7 V[/tex]

Length of wire=[tex]4\sqrt{\frac{NE_0}{2\pi fB}}[/tex]

Substitute the values

Length of wire=[tex]4\times \sqrt{\frac{169.7\times 100}{0.50\times 2\pi\times 60}}[/tex]

Length of wire=38 m

Hence, the length of wire from which the coil is made=38 m

To determine the length of wire in a generator coil with 100 turns in a 0.50-T field and 120 V rms at 60 Hz, one can use the formula for rms emf of a generator and solve for the area of one turn to find the length per turn and multiply by the number of turns.

The question is asking to determine the length of wire used to make a coil in a generator. The generator has 100 turns of wire, operates with a 0.50-T magnetic field, and has an rms value of the emf of 120 V with a 60.0 Hz frequency. Assuming the turns are squares, we can use the formula for the rms value of the emf (Erms) for a generator, which is Erms = NABωrms, where N is the number of turns, A is the area of the turn, B is the magnetic field, and ωrms is the rms angular velocity. The rms angular velocity ωrms is related to the frequency (f) by the equation ωrms = 2πf/√2.

To find the side length (L) of the square turns, we rearrange the formula to solve for A and then take the square root. Once we have L, we multiply by 4 to get the perimeter of one turn and then by 100 to find the total length of wire needed for all turns.

Answer:

Where the height of the cap is less than or equal to the radius of the bore of the plastic pipe then the ⇵⇵shear stress  is less than o equal to 60 psi.

Explanation:

Shear stress = F/A

where F = Applied force

A = Cross sectional area of the member experiencing the force

If the cap covers a section of the pipe of height, h then the area = 2πrh

Where the 60 psi pressure is acting on the pipe bore, we have P₀ = 60 psi  = F₀/A₀ where A₀ = area of pipe bore ≈ πr². Therefore if F₀ = F then we have

F₀/A₀ = F₀/πr² and F/A = F₀/2πrh = where h greater than or equal to 0.5×r then the shear stress is less than or equal to 60 psi

To calculate the shear stress in the adhesive, use the formula for shear stress and the given pressure and area.

To calculate the shear stress present in the adhesive, we can use the formula for shear stress:

Shear stress = Force / Area

In this case, the force exerted by the water pressure on the capped end of the pipe is equal to the pressure multiplied by the area of the capped end. Since the water pressure is given as 60 psi, we need to convert it to a consistent unit, such as Pascals. 1 psi is equal to approximately 6895 Pascals. The area of the capped end can be calculated using the formula for the area of a circle: Area = pi * radius^2.

Once we have the force and area, we can substitute them into the formula for shear stress to find the value.

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Answer:

a) Q = 80,000 cal

b) Q = 100,000 cal

c) Q = 540,000 cal

d) Q = 720,000 cal

Explanation:

a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion

[tex]Q_{l} = ML_{f}[/tex] = 1 * 80

[tex]Q_{l}[/tex] = 80 kCal = 80,000 cal

b) 1 kg of O°C ice water to 1 kg of 100°C boiling water

Specific heat capacity, c =  1000cal/kg.C

[tex]Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal[/tex]

c) 1 kg of 100°C boiling water to 1 kg of 100°C steam

Latent heat of vaporization is needed for this conversion

[tex]Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal[/tex]

d)  1 kg of O°C ice to 1 kg of 100°C steam.

Q = [tex]Q_{L} + Q_{c} + Q_{v}[/tex]

Q = 80,000 + 100,000 + 540,000

Q = 720,000 cal

Final answer:

To change 1 kg of ice at 0°C to water at 0°C, 80 kcal of energy is required. To change 1 kg of water at 0°C to boiling water at 100°C, 100 cal of energy is required. To change 1 kg of boiling water at 100°C to steam, 540 kcal of energy is required.

Explanation:

The quantity of heat required to change the phase of a substance is given by the equation Q = mL, where m is the mass of the substance and L is the heat of fusion or heat of vaporization. For example, to change 1 kg of ice at 0°C to water at 0°C, the energy required is Q = (1 kg)(80 kcal/kg) = 80 kcal. To change 1 kg of water at 0°C to boiling water at 100°C, the energy required is Q = (1 kg)(1 cal/g °C)(100°C) = 100 cal. To change 1 kg of boiling water at 100°C to steam, the energy required is Q = (1 kg)(540 kcal/kg) = 540 kcal.

Answer:

a. 1/2 b. 1/2 c, 20 cm d. 40 cm

Explanation:

Here is the complete question

A proton ( = +, = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron ( = +, = 2.0 u) and an alpha particle ( = +2, = 4.0 u) are accelerated from rest through the same potential difference , and then enter the same region of uniform magnetic field ⃗⃗ , moving perpendicularly to the direction of the magnetic field.

A) What is the ratio of the proton’s kinetic energy to the alpha particle’s kinetic energy ?

B) What is the ratio of the deuteron’s kinetic energy to the alpha particle’s kinetic energy ?

C) If the radius of the proton’s circular orbit = 10 cm, what is the radius of the deuteron’s orbit ?

D) What is the radius of the alpha particle’s orbit ?

Solution

a. For both particles, kinetic energy = electric potential energy

For proton K.E= K₁ = 1/2m₁v₁² = +eV , for alpha particle K.E = K₂ = 1/2m₂v₂²= +2eV

where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the proton and alpha particle. So, the ratio of their kinetic energies is

1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV

m₁v₁²/m₂v₂² = 1/2.

So the ratio K₁/K₂ = 1/2

b. For both particles, kinetic energy = electric potential energy

For deuteron  K₁ = 1/2m₁v₁² = +eV , for alpha particle K₂ = 1/2m₂v₂²= +2eV

where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the deuteron and alpha particle. So, the ratio of their kinetic energies is

1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV

m₁v₁²/m₂v₂² = 1/2.

So the ratio K₁/K₂ = 1/2

c. The radius of the proton's circular is gotten from the centripetal force which equal the magnetic force. So,

mv²/r = Bev

r₁ = mv/Be

Since mass of deuteron m₂ equals twice mass of proton m₁, m₂ = 2m₁

So, radius of deuteron's circular orbit equals

r₂ = m₂v/Be = 2m₁v/Be = 2r₁ = 2 × 10 cm = 20 cm

d. The radius of the alpha particle is given by r₃ = m₃v/Be. Since mass of alpha particle equal four times mass of proton, m₃ = 4m₁.

So, radius of alpha particle's circular orbit equals

r₃ = m₃v/Be = 4m₁v/Be = 4r₁ = 4 × 10 cm = 40 cm

Answer:

He will need to support the joint above and below the deformity, this will allow the xray to be taken without generating more damage to the child.

I hope you find this information useful and interesting! Good luck!

Answer:

349 m

Explanation:

Parameters given:

Mass of climber, m = 92.6 kg

Amount of food calories = 735

1 food calorie = 103 calories

735 food calories = 75705 calories

1 joule is equal to 0.239 calories. Therefore, 75705 calories will be 316749.72 joules.

Hence, this is the amount of work the climber must do work off the food he ate.

Work done is given as:

W = Force * distance

W = m * g * h

h = W/(m * g)

h = 316749.72/(92.6 * 9.8)

h = 349 m

Answer:

Explanation:

When it was approaching, the wavelength is blue-shifted and the rest wavelength when it was receding the wavelength is redshifted.

Any planet is an extremely faint light source compared to its parent star. Comets are considered as the most primitive objects in the Solar System.

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

Explanation:

Below is an attachment containing the solution.

The distance covered by a freely falling object in a certain time is calculated by the formula d = 1/2 * g * t^2 where d is the distance, g is acceleration due to gravity and t is time. When plugging in the values g = 32 ft/sec^2 and t = 5 sec, we find that the object would have fallen 400 ft in the first 5 seconds.

In physics, the distance that a free-falling object covers is given by the formula d = 1/2 * g * t^2, where g is the acceleration due to gravity. In typical physics problems, g is approximated as 32 feet/second^2 on Earth. If we plug t = 5 sec into the equation, we get:

d = 1/2 * 32 ft/sec^2 * (5 sec)^2 = 400 ft.

Therefore, the object would have fallen 400 ft in the first 5 seconds of free fall.

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Answer:

Explanation:

Given that

Beam current (i)=23.3µA

And the time to strike(t)=28s

Also, a fundamental charge e=1.602×10^-19C

Then, the charge quantity is given as,

q=it

Then, q=23.3×10^-6×28

q=6.524×10^-4C

Also, the number of electron N is given as

q=Ne

Therefore, N=q/e

So, N=6.524×10-4/1.602×10^-19

N=4.072×10^15

There are 4.072×10^15 electrons strike the tube screen every 28 s.

When The fundamental charge is 1.602 × 10−19 C So, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.

Given that information as per the question

Beam current (i) is =23.3µA

And also the time to strike(t)=28s

Also, a fundamental charge e=[tex]1.602\times10\wedge -19C[/tex]

Then, the charge quantity is given as,

q is =it

Then, q=[tex]23.3\times 10\wedge-6×28[/tex]

q is =[tex]6.524\times10\wedge -4C[/tex]

Also, When the number of electron N is given as

Now, q=Ne

Therefore, N is =q/e

So, N is = [tex]6.524\times10-4/1.602\times10\wedge-19[/tex]

N is = [tex]4.072\times 10\wedge 15[/tex]

Therefore, There are [tex]4.072\times 10\wedge 15[/tex] electrons that strike the tube screen every 28 s.

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The electric field inside the hollowed insulator can be calculated considering the missing charge in the cylindrical hole. This electric field is uniform and its magnitude is E = ρ.b/ϵ0; where ρ is the uniform charge density and ϵ0 is the permittivity constant. The direction of the electric field is from the axis of the cylinder toward the hole for positive ρ and opposite for negative ρ.

The question refers to the electric field inside a hollowed cylindrical insulator. This is a physics problem related to the study of electrostatics. The concept of electric field refers to the influence exerted in the space around a charged object, which can result in forces on other charged objects.

We know the electric field inside a uniformly charged insulator is zero because charges move to the surface in an insulator. In this case, though, the insulator is not uniform; it has a cylindrical hole. So, we must calculate the contribution from the missing charge in the hole.

The electric field generated by the missing volume, dV, with charge density ρ is given by Coulomb's law, where the electric field corresponds to the charge divided by the square of the distance: E = k.Q/R^2.

Because the volume dV, at radius r from the axis of the cylinder has a charge equal to the volume of the cylindrical disk multiplied by ρ, we have dQ = ρ.dV and thus E = k.ρ.dV/R^2. Integrating over the total volume of the cylindrical hole, we find that the electric field is uniform and its magnitude, E, is given by E = ρ.b/ϵ0, where ϵ0 is the permittivity constant.

The direction of the E is from the axis of the cylinder towards the hole if ρ is positive, because positive charges repel, and vice versa if ρ is negative.

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To find the electric field inside the hole of the insulating cylinder, we use the superposition principle and Gauss's law, resulting in a uniform electric field due to the subtracted cylindrical charge distribution.

The question is asking for the magnitude and direction of the electric field inside a cylindrical hole bored inside another charged cylindrical insulator. To determine the electric field (ES) inside the hole, we apply the principle of superposition, which involves considering the charge distribution that would have been present had there been no hole and subtracting the charge distribution of a cylinder of radius a with charge density -ρ, placed such that it would carve out the hole.

Using Gauss's Law, the electric field due to the entire charged cylinder without the hole would be zero inside the cylinder. Therefore, the electric field inside the hole is solely due to the cylindrical charge distribution we subtracted. Since the cylindrical charge distribution is symmetric, the electric field ES within the hole will also be uniform and directed radially outward from the axis of the imaginary cylinder we subtracted. The magnitude of ES can be calculated using Gauss's Law and considering the symmetry of the problem.

Answer:

F = 0.1153 N

Explanation:

Given:

- Rain fall rate = 5 cm/hr

- Velocity before impact vi = 8.3 m/s

- The Area of the pan A = 1.0 m^2

- The density of water ρ = 1000 kg/m^3

Find:

Estimate the magnitude of the force

Solution:

- Consider rain drops impacting a surface  looses all its momentum. So the change in its momentum is just the momentum with which it impacted.

- We do not know the size of each drop. But the rain fall rate allows us to calculate the rate of change of momentum.

- Total Force: The total force experience by the surface due to the momentum transfer from the impacting rain drops is:

                         F = m(Δv) / t = ρ*v*A*Δh/Δt

Where  

                         Δh/Δt = rain-fall rate.

                         F = 1000*8.3*1*0.05 / 3600

                         F = 0.1153 N

The magnitude of the force is 0.1153N

Given-

The velocity of the raindrop=8.3m/sec

Let when the raindrop hits the surface it loses its momentum

Now the magnitude of the force on the bottom of the given area can be calculated as

[tex]F=\dfrac{m\bigtriangleup v}{t}[/tex]

[tex]F= \rho vA \frac{\bigtriangleup h}{\bigtriangleup t}[/tex]

here rainfall rate

=[tex]\dfrac{\bigtriangleup h}{\bigtriangleup t}[/tex]

therefore,

[tex]F=1000\times 8.3\times 1 \times \dfrac{0.05}{3600}[/tex]

[tex]F=0.1153N[/tex]

Hence the magnitude of the force is 0.1153N

For more detail about the momentum, follo9w the link

https://brainly.com/question/4956182

Answer:

C. It decreases

Explanation:

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualizing the particles of gas in the container moving with a greater energy when the temperature is increased.

A common example is cooking gas when refilled, there is a perceptible change in the temperature of the cylinder.

Explanation:

Below is an attachment containing the solution.

Answer:

The answer to the question is

Momentum

Explanation:

Momentum in physics refer to the attribute that a body has by virtue of its mass and velocity. It is found by multiplying the mass of the moving object and the velocity, hence

Momentum = Mass × Velocity = m·v

Newton first law of motion states that the force acting on an object is proportional to the rate of change of momentum produced

Therefore when we find the momentum of the two cars, the one that has the greater momentum will require the most force to stop it.

Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum

Momentum

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

[tex]d=30,000 pc =9.26\cdot 10^{20} m[/tex]

The speed of the cosmic ray is

[tex]v=0.98 c[/tex]

where

[tex]c=3.0 \cdot 10^8 m/s[/tex] is the speed of light. Substituting,

[tex]v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s[/tex]

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

[tex]T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s[/tex]

Converting into years,

[tex]T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years[/tex]

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

[tex]T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}[/tex]

And substituting v = 0.98c, we find:

[tex]T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years[/tex]

Answer:

a. [tex]F_f = \mu mg cos\theta[/tex]

b. h = Lsinθ

c. 22.78 m

Explanation:

a. The kinetic friction is the product of kinetic coefficient and normal force N, which is the gravity force in the direction normal to the incline

[tex]F_f = \mu N = \mu mg cos\theta[/tex]

b. As the car travels a distance L down the incline of θ degrees, vertically speaking it would have traveled a distance of:

h = Lsinθ

As we can treat L and h in a right triangle where L is the hypotenuse and h is a side length in opposite of incline angle θ

c. Let g = 9.81 m/s2. the acceleration caused by kinetic friction according to Newton's 2nd law is

[tex]a = F_f/m = \mu g cos\theta = 0.45*9.81*cos13^o = 4.3 m/s^2[/tex]

We can use the following equation of motion to find out the distance traveled by the car:

[tex]v^2 - v_0^2 = 2a\Delta s[/tex]

where v = 0 m/s is the final velocity of the car when it stops, [tex]v_0[/tex] = 14m/s is the initial velocity of the car when it starts braking, a = -4.3 m/s2 is the deceleration of the car, and [tex]\Delta s[/tex] is the distance traveled, which we care looking for:

[tex]0^2 - 14^2 = 2(-4.3)\Delta s[/tex]

[tex]\Delta s = 14^2 / (2*4.3) = 22.78 m[/tex]

a. An expression for the magnitude of the force of kinetic friction is [tex]\(f_{\text{friction}} = \mu_k \cdot N\).[/tex]

b. An expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline is: [tex]\(h = L \cdot \sin(\theta)\).[/tex]

c. The car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.

The detailed explanation is as follows:

a. The magnitude of the force of kinetic friction can be calculated using the formula:

[tex]\(f_{\text{friction}} = \mu_k \cdot N\),[/tex]

Where:

[tex]\(f_{\text{friction}}\)[/tex] is the force of kinetic friction,

[tex]\(\mu_k\)[/tex] is the coefficient of kinetic friction (given as 0.45),

[tex]\(N\)[/tex] is the normal force.

The normal force can be calculated using the equation:

[tex]\(N = m \cdot g \cdot \cos(\theta)\),[/tex]

Where:

[tex]\(m\)[/tex] is the mass of the car (1030 kg),

[tex]\(g\)[/tex] is the acceleration due to gravity (approximately 9.81 m/s²),

[tex]\(\theta\)[/tex] is the angle of the incline (13 degrees converted to radians).

b. The magnitude of the change in the car's height [tex](\(h\))[/tex] along the y-direction can be found using trigonometry. When the car travels a distance [tex]\(L\)[/tex] down the incline, the vertical displacement [tex](\(h\))[/tex] can be calculated as:

[tex]\(h = L \cdot \sin(\theta)\).[/tex]

c. To calculate the distance the car travels down the hill until it comes to a stop, you can use the work-energy theorem. The work done by the force of kinetic friction will be equal to the initial kinetic energy of the car. The work-energy theorem is given as:

[tex]\(W = \Delta KE\),[/tex]

Where:

[tex]\(W\)[/tex] is the work done by friction (negative, as it opposes motion),

[tex]\(\Delta KE\)[/tex] is the change in kinetic energy.

The initial kinetic energy is:

[tex]\(KE_0 = \frac{1}{2} m v_0^2\).[/tex]

The final kinetic energy is zero because the car comes to a stop.

So, the work done by friction is:

[tex]\(W = -\frac{1}{2} m v_0^2\).[/tex]

Now, you can use the work-energy theorem to find the distance \(L\) down the incline:

[tex]\(W = -\frac{1}{2} m v_0^2 = \Delta KE = KE_f - KE_0\),[/tex]

Where [tex]\(KE_f = 0\)[/tex] (final kinetic energy).

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

[tex]\(-\frac{1}{2} m v_0^2 = -\mu_k m g L \cos(\theta)\).[/tex]

Now, solve for [tex]\(L\):[/tex]

[tex]\[L = \frac{v_0^2}{2 \mu_k g \cos(\theta)}.\][/tex]

Substitute the known values:

[tex]\[L = \frac{(14 m/s)^2}{2 \cdot 0.45 \cdot 9.81 m/s^2 \cdot \cos(13^\circ)} \approx 94.69 \, \text{meters}.\][/tex]

So, the car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.

For more such questions on kinetic friction:

https://brainly.com/question/14111192

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Answer:

Farther,

Because the stars are far from one another

Explanation:

The star look dim because a star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.

The sun appears very bright to us because it is closer to us, the sun distance from the earth is one light year which is around 92,955,807 miles. Now the closest star to the earth is 4.22 light-years, which is four times that of the sun and so it slowly spread out over time.

Therefore, if two objects have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!

The same applies to star.

Answer:

W = 3.4x0³ J.

Explanation:

The work done by the man is given by the following equation:

[tex] W = F_{t}\cdot d [/tex]     (1)

where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.  

We need to find first the total force Ft, which is:

[tex] Ft = Fm + W [/tex]

where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°.  

[tex] F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N [/tex]

Hence, the work is:

[tex] W = 802.0 N \cdot 4.25 m = 3.4 \cdot 10 ^{3} J [/tex]  

Therefore, the work done by the man is 3.4x10³ J.  

I hope it helps you!      

Answer:

Definitely Spinning permanent magnets within an array of fixed permanent magnets

Explanation:

Any relative motion between magnets (be they permanent or electromagnetic) and a coil of wire will induce an electric current in the coil.

What will not induce an electric current is the relative motion between the two coils of wire (because there is no change in magnetic field), or the relative motion between two magnets (there are no coils of wire to induce the current into).

Therefore, spinning permanent magnets within an array of fixed permanent magnets does not induce an electric current.

Answer:

Spinning permanent magnets within an array of fixed permanent magnets

Explanation: