A garden hose with an inside diameter of 0.75 inches is used to fill a round swimming pool 3.0 m in diameter. How many hours will it take to fill the pool to a depth of 1.0 m if water flows from the hose at a speed of 0.30 m/s? Enter your answer without units.

Explanation:

It is given that,

Mass of object, m = 2 kg

Initial velocity, u = 3 m/s (east)

Final velocity, v = - 7 m/s (west)

The Impulse can be calculated using the change in momentum of an object i.e.

J = m(v-u)

[tex]J=2\ kg(-7\ m/s-3\ m/s)[/tex]

J = -20 kg-m/s

So, the Impulse of this object is 20 kg-m/s but the direction is opposite. Hence, statement (1) is correct i.e. Student #1: "The impulse is equal to the change in momentum, which is (2 kg)(3 m/s + 7 m/s) = 20 kg m/s."

The correct student to agree with is Student #2. The impulse applied to an object is equal to the change in its momentum.

The change in momentum is calculated by multiplying the mass of the object by the change in its velocity.

First, let's calculate the change in velocity.

[tex]\[ \Delta v = v_f - v_i = (-7 \text{ m/s}) - (3 \text{ m/s}) = -10 \text{ m/s} \][/tex]

The negative sign indicates that the velocity has changed in the opposite direction. The magnitude of the change in velocity is 10 m/s.

Now, we multiply the mass of the object by the change in velocity to find the change in momentum:

[tex]\[ \Delta p = m \cdot \Delta v = (2 \text{ kg}) \cdot (-10 \text{ m/s}) = -20 \text{ kg m/s} \][/tex]

The impulse, which is equal to the change in momentum, is therefore:

[tex]\[ \text{Impulse} = \Delta p = -20 \text{ kg m/s} \][/tex]

The negative sign indicates that the impulse is in the opposite direction to the initial velocity. The magnitude of the impulse is 20 kg m/s.

Student #1 incorrectly adds the initial and final velocities without considering their directions. Student #3 incorrectly divides the change in momentum by the time interval, which is not necessary since impulse is the change in momentum, not the rate of change of momentum. Student #4 incorrectly states that we cannot find the impulse without knowing the force; however, we can find the impulse using the change in momentum, as we have done.

Answer:

part a)

k = 310 N/m

part b)

T = 0.51 s

Explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

[tex]mgx - \frac{1}{2}kx^2 = 0[/tex]

[tex](2.07)(9.8)(0.131) - \frac{1}{2}k(0.131)^2 = 0[/tex]

now we will have

[tex] k = 310 N/m[/tex]

Part B)

Time period of oscillation is given as

[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]

[tex]T = 2\pi\sqrt{\frac{2.07}{310}}[/tex]

[tex]T = 0.51 s[/tex]

Answer:

6.38 x 10^4 J

Explanation:

d = 0.33 cm = 0.33 x 10^-2 m, Area = 87 x 36 cm^2 = 0.87 x 0.36 m^2

ΔT = 14 degree C, t = 1 min = 60 second

K = 0.8 W / m K

Heat  = K A ΔT t / d

H = 0.8 x 0.87 x 0.36 x 14 x 60 / (0.33 x 10^-2)

H = 6.38 x 10^4 J

The question is related to calculating the rate of heat conduction through a glass window in Physics. For a complete answer, the thermal conductivity of glass is needed, which is not provided in the question. With that value, the formula for heat conduction can be used to find the heat flow per minute.

The subject of this question is Physics, and it applies to a High School grade level. To determine the rate of heat conduction through a window, we can use the formula for heat transfer through conduction:

Q = \(\frac{k \cdot A \cdot \Delta T \cdot t}{d}\)

where:
Q is the heat transferred,
k is the thermal conductivity,
A is the area,
\(\Delta T\) is the temperature difference,
t is the time, and
d is the thickness of the material.

However, the question does not provide the value of the thermal conductivity (k) for glass, which is necessary to calculate the rate of heat conduction. With the thermal conductivity value, we could then insert the given dimensions, temperature difference, and time into this formula to find the heat flow per minute. Without this value, the question cannot be fully answered as the data is incomplete. For a complete solution, thermal conductivity of the material is required.

Answer:

[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]

Explanation:

Initial total mechanical energy is given as

[tex]ME = U + KE[/tex]

here we will have

[tex]U = mgh[/tex]

[tex]U = (0.066)(9.81)(1.05) = 0.68 J[/tex]

also we have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(0.066)(15.1)^2[/tex]

[tex]KE = 7.52 J[/tex]

[tex]ME_i = 0.68 + 7.52 = 8.2 J[/tex]

Now similarly final mechanical energy is given as

[tex]U = mgh[/tex]

[tex]U = (0.066)(9.81)(1.59) = 1.03 J[/tex]

also we have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(0.066)(10.9)^2[/tex]

[tex]KE = 3.92 J[/tex]

[tex]ME_f = 1.03 + 3.92 = 4.95 J[/tex]

Now change in mechanical energy is given as

[tex]\Delta E = ME_i - ME_f[/tex]

[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]

Answer:

21.85 C

Explanation:

mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C

mass of water = 20 kg, initial temperature of water, T2 = 18 C

let T be the equilibrium temperature.

Specific heat of iron = 449 J/kg C

specific heat of water = 4186 J/kg C

Use the principle of caloriemetry

heat lost by the hot body = heat gained by the cold body

mass of iron x specific heat of iron x decrease in temperature = mass of water  x specific heat of water x increase in temperature

1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)

336750 - 673.5 T = 83720 T - 1506960

1843710 = 84393.5 T

T = 21.85 C

Answer:

2.16 x 10^7 years

Explanation:

R = 6.9 x 10^20 m

ω = 9.2 x 10^-15 rad/s

T = 2π / ω

T = ( 2 x 3.14 ) / ( 9.2 x 10^-15)

T = 6.826 x 10^14 second

T = 2.16 x 10^7 years

By using the star's given angular velocity and the formula for a revolution's time period, it can be calculated that it takes the star approximately 2.17 x 10^8 years to complete one revolution around the center of the galaxy.

The timeframe in which a star completes one revolution around the center of a galaxy is referred to as a galactic year. In this scenario, the star's time period for one revolution, T, can be calculated using the formula T = 2π/ω, where ω represents the star's angular speed. So, T = 2π / (9.2 x 10^-15 rad/s). The resulting value is in seconds since the angular velocity was in rad/s. This can then be converted to years by using the relevant conversion factor: 1 year = 3.15 x 10^7 seconds.

By performing these calculations, we find that it takes the star about 2.17 x 10^8 years to complete one revolution around the center of the galaxy. It is important to note that this is a much larger timeframe than a human lifetime, meaning that humans have only seen a small fraction of a galactic year.

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Answer:

3.9 atm

Explanation:

Ideal gas law states:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The volume of the tank is constant, so we can say:

V₁ = V₂

Using ideal gas law to write in terms of P, n, R, and T:

n₁ R T₁ / P₁ = n₂ R T₂ / P₂

n₁ T₁ / P₁ = n₂ T₂ / P₂

Initially:

P₁ = 9.8 atm

T₁ = 29.0°C = 302.15 K

n₁ = n

Afterwards:

P₂ = P

T₂ = 83.0°C = 356.15 K

n₂ = n/3

Substituting:

n (302.15 K) / (9.8 atm) = (n/3) (356.15 K) / P

(302.15 K) / (9.8 atm) = (1/3) (356.15 K) / P

P = 3.85 atm

Rounding to 2 significant figures, P = 3.9 atm.

Answer:

f = 1.354*10^{20} Hz

Explanation:

By conservation of linear momentum, wavelength shift due to collision of photon to electron is given by following formula

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

where h is plank constant = 6.626*10^{-34}

c = speed of light = 3*10^{8} m/s

scattered angle  =  60 degree

m = rest mass of electron  = 9*10^{-31}

[tex]\lambda ^{'}=10^{-12} +\frac{6.626*10^{-34}}{9*10^{-31}*3*10^{8}}(1-cos60^{o} )[/tex]

[tex]\lambda ^{'}= 2.215 pm[/tex]

we know that 1 pm = 10^{-12}m

[tex]f = \frac{c}{\lambda ^{'}}[/tex]

[tex]f = \frac{3*10^{8}}{2.215 *10^{-12}} = 1.354*10^{20} Hz[/tex]

f = 1.354*10^{20} Hz

Answer:

a) 17 N

b) 21 N

Explanation:

At the 3 o'clock position, the sum of the forces towards the center is:

∑F = ma

T = m v² / r

19 = m v² / r

At the 12 o'clock position, the sum of the forces towards the center is:

∑F = ma

T + mg = m v² / r

T + (0.18)(9.8) = 19

T = 17.2 N

At the 6 o'clock position, the sum of the forces towards the center is:

∑F = ma

T − mg = m v² / r

T − (0.18)(9.8) = 19

T = 20.8 N

Rounding to two significant figures, the tensions are 17 N and 21 N.

To find the tension in the stick at the twelve o’clock and six o’clock positions, consider the forces acting on the ball at each position and calculate the tension using the equation T = mg + mv^2 / r for twelve o'clock and T = mg - mv^2 / r for six o'clock.

To find the tension in the stick when the ball is at the twelve o’clock and six o’clock positions, we need to consider the forces acting on the ball at each position. At the twelve o'clock position, the tension in the stick is equal to the weight of the ball plus the centripetal force required to keep it moving in a circle. The tension can be calculated using the equation T = mg + mv^2 / r, where T is the tension, m is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and r is the radius of the circle.

At the six o'clock position, the tension in the stick is equal to the weight of the ball minus the centripetal force. Using the same equation, the tension can be calculated by subtracting mv^2 / r from mg.

Substituting the given values into the equation for each position will give you the tension in the stick.

Answer:

(a) 0.204 Weber

(b) 0.22 Volt

Explanation:

N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.

(a) Magnetic flux, Ф = N x B x A

Ф = 100 x 0.0650 x 3.14 x 0.1 0.1

Ф = 0.204 Weber

(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s

dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s

induced emf, e = N dФ/dt

e = N x A x dB/dt

e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V

(a) The magnetic flux through a coil 0.204 Weber

(b) The emf induced in the coil during the time interval 0.22 Volt

It is given that

Number of the turns N = 100,  

Radius of the coil r = 10 cm = 0.1 m,

The magnetic field B = 0.0650 T

Since the angle is 90 degrees with the plane of the coil, so  [tex]\Theta[/tex]= 0 degrees with the normal coil.

(a) The Magnetic flux, will be given as

[tex]\rm\phi =N\times B\times A[/tex]

By putting the values

[tex]\phi =100\times0.0650\times3.14\times 0.1\times0.1[/tex]

[tex]\phi=0.204 \rm \ weber[/tex]

(b) The emf induced will to be given as

[tex]B_1=0.0650T\ \ B_2=0.1T\ \ dt=0.5s[/tex]  

[tex]\dfrac{dB}{dt} =\dfrac{B_2-B_1}{dt} =\dfrac{0.1-0.650}{0.5} =0.07\dfrac{T}{s}[/tex]

induced emf,

[tex]e=N\dfrac{d\phi}{dt}[/tex]

[tex]e=N\times A \times \dfrac{dB}{dt}[/tex]

[tex]e=100\times 3.14\times0.1\times0.1\times0.07=0.22V[/tex]

Thus

(a) The magnetic flux through a coil 0.204 Weber

(b) The emf induced in the coil during the time interval 0.22 Volt

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The power delivered to the solenoid is calculated by finding the current through the solenoid and its resistance, then applying the formula P=I²R. This involves applying known laws and formulas of electromagnetism.

The power delivered to the solenoid can be calculated given the field strength inside the solenoid and the physical dimensions and properties of the solenoid and wire. The field strength inside a solenoid is given by the formula B = µon, where n is the number of loops per unit length. We first find n, and then we need to find the current I because the power P delivered to the solenoid is given by P=I²R, where R is the resistance of the wire.

The resistance R of the wire can be found using the formula R = ρL/A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. After calculating R and I, we can compute P. Therefore, it's paramount to apply known formulas and laws of electromagnetism to solve this. It's important to note that B is the magnetic field inside the solenoid, µo is the permeability of free space and n is number of turns per unit length.

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To produce a magnetic field of 5.10 mT at the center of a solenoid, a power of 61.4 W must be delivered. This is calculated by determining the number of turns, the required current, the resistance of the copper wire, and then using the power formula. The key parameters are the number of turns (700), current (3.61 A), resistance (4.71 Ω), and power (61.4 W).

To solve this problem, we'll need to follow these steps:

Calculate the number of turns of wire on the solenoid.

Determine the current needed to produce the required magnetic field.

Find the resistance of the wire.

Calculate the power delivered to the solenoid.

1. Calculate the number of turns of wire

The diameter of the wire is 0.100 cm, so the circumference of each turn is approximately equal to the solenoid's length (70.0 cm). The number of turns (N) is:

N = 70.0 cm / 0.100 cm = 700 turns

2. Determine the current needed

The formula for the magnetic field (B) inside a solenoid is given by:

B = μ₀ * (N/L) * I

where:

B = 5.10 mT = 5.10 x 10⁻³ T

μ₀ = 4π * 10⁻⁷ T⋅m/A (permeability of free space)

N = 700 turns

L = 70.0 cm = 0.70 m

Solving for the current (I):

5.10 x 10⁻³ = (4π x 10⁻⁷) * (700 / 0.70) * I

5.10 x 10⁻³ = (4π x 10⁻⁷) * 1000 * I

5.10 x 10⁻³ = (4π x 10⁻⁴) * I

I = 3.61 A

3. Find the resistance of the wire

The resistance (R) of the wire is given by:

R = ρ * (L/A)

where:

ρ = 1.68 x 10⁻⁸ Ω⋅m (resistivity of copper)

L = N * circumference of each turn

A = * (diameter/2)²

Calculating the total length (L) of the wire:

L = 700 * 0.100 * m = 219.91 m (approximately)

Calculating the cross-sectional area (A):

A = * (0.001 m / 2)²

A = 3.14 x (0.0005)²

A = 7.85 x 10⁻⁷ m²

Thus, the resistance (R) is:

R = 1.68 x 10⁻⁸ * (219.91 / 7.85 x 10⁻⁷)

R = 1.68 x 10⁻⁸ * (28.014 x 10⁷)

R  = 4.71 Ω

4. Calculate the power delivered

Power (P) is given by:

P = I² * R

Substituting in the values:

P = (3.61)² * 4.71 = 13.032 * 4.71 =  61.4 W

Therefore, 61.4 W power must be delivered to the solenoid to produce the required magnetic field.

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

[tex]F = \mu mg[/tex]

so we have

[tex]F = ma = \mu mg[/tex]

now the acceleration is given as

[tex]a = \mu g[/tex]

[tex]a = (0.32)(9.8) = 3.14 m/s^2[/tex]

Final answer:

The force that accelerates the box when the truck moves forward is the force of static friction. The maximum acceleration the truck can have before the box slides can be found using the equation: max acceleration = coefficient of friction x acceleration due to gravity.

Explanation:

When the truck accelerates forward, the force that accelerates the box is the force of static friction between the box and the surface of the truck bed. This force opposes the motion of the box and allows it to stay in place on the truck.

The maximum acceleration the truck can have before the box slides can be found using the equation:

max acceleration = coefficient of friction x acceleration due to gravity

Using the given coefficient of friction between the box and the surface, you can substitute the value and solve for the maximum acceleration.

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

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

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

[tex]0-(17.7)^2=2a\times 33.1[/tex]

[tex]a=\dfrac{-(17.7)^2}{33.1}[/tex]

[tex]a=-9.46\ m/s^2[/tex]

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

[tex]F = ma[/tex]

[tex]F =1400\times(-9.46)[/tex]

[tex]F=-13244\ N[/tex]

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

[tex]|F| =13244\ N[/tex]

Hence,  The magnitude of the horizontal net force is 13244 N.

Answer:

The no of revolutions is 2.032 revolution.

Explanation:

Given that,

Moment of inertia = 0.85 Kgm²

Radius = 170 mm

Force = 32 N

Time =  2s

We need to calculate the angular acceleration

Using formula of torque

[tex]\tau=I\times\alpha[/tex]

[tex]\alpha=\dfrac{\tau}{I}[/tex]

[tex]\alpha=\dfrac{F\times r}{I}[/tex]

Where, F = force

r = radius

I = moment of inertia

Put the value into the formula

[tex]\alpha=\dfrac{32\times170\times10^{-3}}{0.85}[/tex]

[tex]\alpha=6.4\ m/s^2[/tex]

We need to calculate the rotational speed

Using equation of angular motion

[tex]\omega_{f}=\omega_{i}+\alpha t[/tex]

[tex]\omega_{f}=6.4\times2[/tex]

[tex]\omega=12.8\ rad/s[/tex]

We need to calculate the angular position

Using equation of angular motion

[tex]\theta=\omega_{i}+\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta=0+\dfrac{1}{2}\times6.4\times4[/tex]

[tex]\theta=12.8\ radian[/tex]

We need to calculate no of revolutions

[tex]n = \dfrac{\theta}{2\pi}[/tex]

[tex]n=\dfrac{12.8}{2\times3.15}[/tex]

[tex]n=2.032\ revolution[/tex]

Hence, The no of revolutions is 2.032 revolution.

The stress in the bungee cord due to the Max's weight is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].

Further Explanation:

The stress developed in the bungee cord is the amount of tensile force developed inside the bungee cord due to the Max's weight.

Given:  

The diameter d of the bungee cord is [tex]2.5\text{ cm}[/tex].  

The mass m of Max is [tex]15\text{ kg}[/tex].

Concept:

The stress developed in the bungee cord is given by:  

[tex]\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}[/tex]                                                                               ... (1)         

Here, [tex]\sigma[/tex] is the stress developed in the bungee cord, [tex]F[/tex]is the force due to Max's weight and [tex]A[/tex] is the area of cross-section of the bungee cord.  

The radius of the bungee cord will be the half of its diameter.  

[tex]r=\dfrac{d}{2}[/tex]

Substitute [tex]2.5\text{ cm}[/tex] for [tex]d[/tex].  

[tex]r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}[/tex] 

Convert the radius of the bungee cord in meter.  

[tex]r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}[/tex] 

The area of cross-section of the bungee cord is given by:  

[tex]A=\pi r^{2}[/tex]  

Substitute [tex]0.0125\text{ m}[/tex] for [tex]r[/tex].  

[tex]A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}[/tex]  

The force on the bungee cord due to Max's mass is the gravitational force acting on Max's body. The gravitational force acting on Max's body is given by:  

[tex]F=mg[/tex]  

Here, m is the mass of Max's body and g is acceleration due to gravity.  

Consider the value of acceleration due to gravity on Earth be [tex]9.80\text{ m}/\text{s}^{2}[/tex].  

Substitute [tex]15\text{ kg}[/tex] for m and [tex]9.80\text{ m}/\text{s}^{2}[/tex] for g in above expression.  

[tex]F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}[/tex]  

Substitute [tex]147\text{ N}[/tex] for F and [tex]4.980\times10^{-4}\text{ m}^{2}[/tex] for A in equation (1).  

[tex]\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}[/tex]

Thus, the stress developed in the bungee cord is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Stress and Strain

Keywords:

Max, Max's weight, force, stress, area, 299511 N/m2, 299511 N/m^2, 2.99x10^5 N/m2, gravitational, radius, bungee, cord, weight, developed, mass, 3x10^5 N/m^2.

The stress in the bungee cord due to Max’s weight is 294,000 N/m².

The given parameters;

The area of the bungee cord is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (0.0125)^2\\\\ A= 0.0005 \ m^2[/tex]

The weight of Max is calculated as follows;

[tex]F = mg\\\\F = 15 \times 9.8\\\\F = 147 \ N[/tex]

The stress in the bungee cord due to Max’s weight is calculated as;

[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{147}{0.0005} \\\\\sigma = 294,000 \ N/m^2[/tex]

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Answer:

213.18 volt

Explanation:

N = 500, A = 0.02 m^2, B = 0.30 T, f = 960 rpm = 960 / 60 = 16 rps

The maximum value of induced emf is given by

e0 = N x B x A x ω

e0 = N x B x A x 2 x π x f

e0 = 500 x 0.3 x 0.02 x 2 x 3.14 x 16

e0 = 301.44 Volt

The rms value of induced emf is given by

[tex]e_{rms} = \frac{e_{0}}{\sqrt{2}}[/tex]

[tex]e_{rms} = \frac{301.44}{\sqrt{2}}[/tex]

e rms = 213.18 volt

Answer:

All the balls hit the ground with same velocity.  

Explanation:

Let the speed of throwing be u and height of building be h.

Ball A is launched 45 degrees above the horizontal

Initial vertical speed = usin45

Vertical acceleration = -g

Height = -h

Substituting in v² = u² +2as

                       v² = (usin45)² -2 x g x (-h)

                       v² = 0.5u² +2 x g x h

Final vertical speed² = 0.5u² +2 x g x h

Initial horizontal speed = final horizontal speed =  ucos45

Final horizontal speed² = 0.5u²

Magnitude of final velocity

          [tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]  

Ball C is launched horizontally

Initial vertical speed = 0

Vertical acceleration = -g

Height = -h

Substituting in v² = 0² +2as

                       v² = 0² -2 x g x (-h)

                       v² = 2 x g x h

Final vertical speed² = 2 x g x h

Initial horizontal speed = final horizontal speed =  u

Final horizontal speed² =u²

Magnitude of final velocity

          [tex]v=\sqrt{2gh+u^2}=\sqrt{u^2+2gh}[/tex]  

Ball B is launched 45 degrees below the horizontal

Initial vertical speed = usin45

Vertical acceleration = g

Height = h

Substituting in v² = u² +2as

                       v² = (usin45)² +2 x g x h

                       v² = 0.5u² +2 x g x h

Final vertical speed² = 0.5u² +2 x g x h

Initial horizontal speed = final horizontal speed =  ucos45

Final horizontal speed² = 0.5u²

Magnitude of final velocity

          [tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]    

All the magnitudes are same so all the balls hit the ground with same velocity.  

The ball C which was launched horizontally will hit the ground fastest.

With respect to angle of projection above the horizontal, the time of motion each ball is calculated as follows;

[tex]T = \frac{u sin\theta }{g}[/tex]

where;

When θ is 45 degrees above the horizontal;

[tex]T = \frac{u \times sin(45)}{9.8}\\\\ T = 0.072u[/tex]

When θ is 45 degrees below the horizontal = 135 degrees above the horizontal

[tex]T = \frac{ u \times sin(135)}{9.8} \\\\T = 0.072 \ s[/tex]

When launched horizontally, θ = 90 degrees above the horizontal

[tex]T = \frac{u \times sin(90)}{9.8} \\\\T = 0.1 u[/tex]

Thus, the ball C which was launched horizontally will hit the ground fastest.

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Answer:

377.78 m/s

Explanation:

Let the speed of bullet is v.

Time taken by the sound is t' and time taken by the bullet is t.

Speed of sound = 340 m/s

Distance = 170 m

Time taken by the sound, t' = distance / speed of sound = 170 / 340 = 0.5 sec

time taken by the bullet, t = 0.95 - 0.5 = 0.45 sec

Speed of bullet = distance / time taken by bullet

Speed of bullet = 170 / 0.45 = 377.78 m/s

Answer:

3 x 10^-4 N/m

Attractive

Explanation:

r = 0.1 m, i1 = 10 A, i2 = 15 A

The force per unit length between two parallel wires is given by

[tex]F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}[/tex]

[tex]F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}[/tex]

F = 3 x 10^-4 N/m

Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.

Answer:

C) is at the equilibrium point

Explanation:

As we know that the equation of displacement of SHM executing object is given as

[tex]x = A sin(\omega t + \phi)[/tex]

now for velocity of the particle we will have

[tex]v = \frac{dx}{dt}[/tex]

now we will have

[tex]v = A\omega cos(\omega t + \phi)[/tex]

now if velocity is maximum then we will have

[tex]cos(\omega t + \phi) = \pm 1[/tex]

so at this situation we will have

[tex](\omega t + \phi) = N\pi[/tex]

now at this angle the value of

[tex]sin(\omega t + \phi) = 0[/tex]

so the position must be mean position at which the particle is at equilibrium position

In simple harmonic motion, an object oscillating vertically reaches its maximum speed at the equilibrium point. This is the position where the object would rest in the absence of force and is where the maximum velocity is attained during the oscillation.

In the context of simple harmonic motion, an object oscillating in the vertical direction attains its maximum speed at the equilibrium point (C). This equilibrium point represents the position where the object would naturally rest in the absence of force. An object attached to a spring and placed on a frictionless surface serves as a straightforward example of a simple harmonic oscillator. When displaced from its equilibrium, this object executes simple harmonic motion with an amplitude A and a period T. The maximum velocity or speed is achieved as the object moves through the equilibrium point, irrespective of the direction. The maximum displacement from the equilibrium, referred to as amplitude, does not correlate with the object's speed.

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Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

Answer:

Original speed of the bullet = 305.21 m/s

Explanation:

Here momentum is conserved.

Mass of bullet = 9.6 g = 0.0096 kg

Mass of wood = 4.95 kg

Let velocity of bullet be v.

Initial momentum = 0.0096 v

Final mass = 0.0096 + 4.95 = 4.9596 kg

Final velocity = 0.591 m/s

Final momentum = 4.9596 x 0.591 = 2.93 kgm/s

Equating both momentum

           0.0096 v = 2.93

                      v = 305.21 m/s

Original speed of the bullet = 305.21 m/s

Answer:

B.Exhaust Value

Explanation:

The waste products of combustion leave the internal combustion engine through the Exhuast Value.

Answer:

Exhaust Value

Explanation:

Answer:

10.2 m

Explanation:

A = 70 mm^2 = 70 x 10^-6 m^2

F = 7 N

Pressure sustained by ear drum

P = F / A = 7 / (70 x 10^-6) = 10^5 N/m^2

Let the depth of water is h

Pressure due to the depth of water should be same as the pressure sustained by the ear drum

P = h x d x g

where, d be the density of water.

10^5 = h x 1000 x 9.8

h = 10.2 m

The discomfort and pressure on the eardrums when diving underwater is due to increased pressure caused by depth. The maximum depth a person can dive without rupturing their eardrum can be calculated by dividing the maximum pressure the eardrum can sustain by the pressure increase per meter. In this case, the maximum depth is about 10 meters.

The discomfort and pressure on your eardrums when diving underwater is due to the increased pressure that comes with depth. The pressure exerted on your eardrums underwater is a result of the weight of the water above you and the air above the water's surface. To calculate the maximum depth you can dive without rupturing your eardrum, you need to determine the pressure that the eardrum can sustain and then divide it by the pressure difference caused by the depth.

The maximum force the eardrum can sustain is 7 N, and the area of the eardrum is 7x10-5 m2. The maximum pressure the eardrum can sustain is therefore 7 N / (7x10-5 m2) = 1x105 Pa.

The pressure underwater increases by 1 atmosphere for every 33 feet (10 meters) of saltwater depth. Using this information, you can calculate the maximum depth:

Maximum Depth = Maximum Pressure / Pressure Increase per Meter = 1x105 Pa / (1x105 Pa/m) = 10 meters

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Answer:

Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).

Explanation:

The forces acting on a rock thrown up are force due to gravity and air resistance. Air resistance is directly proportional to velocity of rock, when velocity is zero air resistance is zero. When it is at the top of its trajectory its velocity is zero. So air resistance is also zero. Hence only gravitational force acts on the rock.

Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).

At the peak of its trajectory, the only significant force acting on a rock is the force of gravity. Thus, the net external force on the system is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity. At this point, the rock begins its descent due to the gravitational pull.

The question is asking about the net external force acting on a rock when it is at the top of its trajectory. At the peak of its trajectory, the only significant force acting on the rock is the force of gravity (weight of the rock), assuming air resistance is negligible. Thus, the net external force on the system, if air resistance is neglected, is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity.

For example, if a rock is thrown straight upwards with an initial velocity, as it reaches the topmost point of its trajectory, the vertical velocity momentarily becomes zero before the rock starts descending again. At this point, the only major force acting on the rock is the gravitational pull exerted by the Earth, also known as the rock's weight.

It is important to note that this net external force (-mg) is what determines the acceleration of the object at the top of its trajectory. Simply put, even at the top of its path, the rock is still experiencing the force of gravity which pulls it back towards Earth, leading to its eventual fall back to the ground.

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Final answer:

The magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s. The magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

Explanation:

To find the magnitude of the impulse delivered by the bat to the ball, we need to use the formula for impulse:

Impulse = change in momentum

Momentum is given by the product of mass and velocity (momentum = mass × velocity). The change in momentum is the final momentum minus the initial momentum (change in momentum = final momentum - initial momentum).

Given that the mass of the baseball is 0.15 kg, the initial velocity is +17 m/s, and the final velocity is -17 m/s, we can calculate the magnitude of the impulse:

Impulse = (0.15 kg × -17 m/s) - (0.15 kg × 17 m/s) = -5.1 kg·m/s

So, the magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s.

To find the magnitude of the average force exerted by the bat on the ball, we can use the formula for average force:

Average force = impulse / time

Given that the time the ball is in contact with the bat is 1.5 ms (1.5 × 10^-3 s) and the magnitude of the impulse is 5.1 kg·m/s, we can calculate the magnitude of the average force:

Average force = 5.1 kg·m/s / (1.5 × 10^-3 s) = 3.4 × 10^3 N

Therefore, the magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.

That being said,

f=1/T=1/0.641=1.56 s

xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m

A=(x2-x1)/2=(0.395+0.285)/2=0.340 m

x(t)=Acos(wt+ϕ)

v(t)=-wAsen(wt+ϕ)

w=2πf=2*3.14*1.56=9.8 rad/s

vmax=-wA=9.8*0.340=-3.3 m/s

a(t)=-w^2Acos(wt+ϕ)

amax=-w^2A=96.04*0.340=-32.6 m/s^2

w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m

Etot=1/2kA^2

Etot=0.5*28.1*0.116=1.63 J

The characteristics of the simple harmonic movement allow to find the results for the questions about the movement of the mass with the spring are:

     a) The frequency is: f = 1.72 Hz

     b) The equilibrium position is: x₀ = 0.088 m

     c) The amplitude is: A = 0.686 m

     d) The maximum speed is: v = 7.42 m / s

     e) the maximum acceleration is: a = 80.16 m / s²

     f) The mechanical energy is: E = 3.44 J

Given parameters

To find.

    a) Frequency.

    b) the equilibrium position.

    c) The amplitude.

    d) Maximum speed.

    e) Maximum acceleration.

    f) The total mechanical energy.

Simple harmonic motion is a periodic motion where the restoring force is proportional to the elongation. This movement is fully described by the expression

       x = A cos (wt + Ф)

       w² = [tex]\sqrt{ \frac{k}{m} }[/tex]  

Where x is the displacement, A the amplitude of the movement, w the angular velocity, t the time, Ф a phase constant that meets the initial conditions, k the spring constant and m the mass attached to the spring.

a) Frequency and period are related.

          [tex]f= \frac{1}{T}[/tex]  

Let's calculate.

        [tex]f = \frac{1}{0.581 }[/tex]  

         f = 1.72 Hz

b) The equilibrium position occurs at the midpoint of the movement

       [tex]x_o = \frac{x_f + x_o}{2}[/tex]  

       [tex]x_o = \frac{0.387 + (-0.299)}{2}[/tex]  

       x₀ = 0.088 m

c) The amplitude is the maximum elongation of the spring

        [tex]A= x_f - x_o[/tex]  

        A = 0.387 - (-0.299)

        A = 0.686 m

d) The speed is defined by the variation of the position with respect to time.

          [tex]v = \frac{dx}{dt}[/tex]

Let's make the derivatives

          v = - A w cos (wt + Ф)

The angular velocity is related to the period.

        [tex]w= \frac{\pi }{T} \\w= \frac{2 \pi }{0.581}[/tex]

        w = 10.81 rad / s

The speed is maximum when the cosine  is maximum ±1

        v = A w

        v = 0.686 10.81

        v = 7.42 m / s

e) Acceleration is defined as the change in velocity over time.

           [tex]a = \frac{dv}{dt}[/tex]  

           a = -A w² sin (wt + Ф)

The acceleration is maximum when the sine function is maximum ±1

          a = A w²

          a = 0.686 10.81²

          a = 80.16 m / s²

f) The mechanical energy is calculate for the point of maximum elongation.

          E = ½ k A²  

          k = w² m

Let's replace.

          E = ½ w² m A²

Let's calculate.

          E = ½ 10.81² 0.125 0.686²

          E = 3.44 J

In conclusion, using the characteristics of simple harmonic motion we can find the results for the questions about the motion of the mass with the spring are:

     a) The frequency is: f = 1.72 Hz

     b) The equilibrium position is: x₀ = 0.088 m

     c) The amplitude is: A = 0.686 m

     d) The maximum speed is: v = 7.42 m / s

     e) the maximum acceleration is: a = 80.16 m / s²

     f) The mechanical energy is: E = 3.44 J

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Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

[tex]W = F\cdot d[/tex]

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

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

[tex]W=120\times3\cos0^{\circ}[/tex]

[tex]W=360\ J[/tex]

Hence, The work done is 360 J.

Answer:

33.33 m/s

Explanation:

m = 450 kg. T = 5000 N, t = 3 seconds,

let the net acceleration is a.

T = m a

a = 5000 / 450 = 11.11 m/s^2

u = 0 , v = ?

Let v be the velocity after 3 seconds.

Use first equation of motion

v = u + a t

v = 0 + 11.11 x 3 = 33.33 m/s