Suppose a man's scalp hair grows at a rate of 0.49 mm per day. What is this growth rate in feet per century?

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

[tex]s=ut+\frac{gt^2}{2}[/tex]

[tex]1.45=0+\frac{9.81\times t^2}{2}[/tex]

[tex]t^2=frac{2\times 1.45}{9.81}[/tex]

t=0.54 s

Motion of ball will be straight line

Answer:

A) The maximum height of the blue ball is 29.7 m above the ground.

B) The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) The balls are at the same height 1.41 s after the red ball is thrown

Explanation:

A) At maximum height, the velocity of the blue ball is 0 because for that instant, the ball does not go up nor down.

The equation for velocity for an accelerated object moving in a straight line is:

v = v0 + a*t

where

v = velocity.

v0 = initial velocity.

a = acceleration, in this case, it is the acceleration of gravity, g, 9.81 m/s².

t = time.

Then:

0 = v0 + g * t  (if the origin of the reference system is the ground, then g is negative)

0 = 23.8 m/s - 9.81 m/s² * t

-23.8 m/s / -9.81 m/s² = t

t = 2.43 s

With this time, we can calculate the position of the blue ball. The equation for position is:

y = y0 + v0 * t + 1/2 * g * t²

y = 0.8 m + 23.8 m/s * 2.43 s - 1/2 * 9.81 m/s² * (2.43 s)²

y = 29.7 m

the maximum height of the blue ball is 29.7 m above the ground.

B) 1.8 s after throwing the red ball, the blue ball was in the air for (1.8 s - 0.6) 1.2 s. Then, using the equation for the position of the blue ball:

y = 0.8 m + 23.8 m/s * 1.2 s - 1/2 * 9.81 m/s² * (1.2 s)² = 22.3 m

The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) Now, we have to find the time at which both positions are equal. Notice that the time of the blue ball is not the same as the time for the red ball. The time for the blue ball is the time of the red ball minus 0.6 s:

t blue = t red - 0.6 s

Then:

position red ball = position blue ball

 y0 + v0 * t + 1/2 * g * t² = y0 + v0 * (t- 0.6) + 1/2 * g * (t-0.6s)²

25 m + 1.2 m/s * t -1/2 * 9.81 m/s² * t² = 0.8 m + 23.8 m/s * (t-0.6 s) - 1/2 * 9.81 m/s² * (t-0.6 s)²

24.2 m + 1.2 m/s * t -4.91 m/s² * t² = 23.8 m/s * t - 14.28 m - 4.91 m/s² * (t-0.6 s)²

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² (t² - 1.2 s * t + 0.36 s²)

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² * t² + 5.89 m/s * t - 1.77 m

40.3 m - 28.5 m/s * t = 0

t = -40.3 m / -28.5 m/s

t = 1.41 s

The balls are at the same height 1.41 s after the red ball is thrown and 0.81 s after the blue ball is thrown.

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.

We have to use Coulomb's Law and the principle of charge conservation.

1. Conservation of Charge:

  The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:

[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]

2. Coulomb's Law:

  The force between two charges is given by Coulomb's Law:

[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]

Let's set up the equation:

[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]

Rearrange to solve for [tex]\( q_A q_B \)[/tex]:

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]

Now, we have two equations:

1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]

2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]

To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:

[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]

[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]

[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]

This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:

[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]

Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]

This gives us two solutions for [tex]\( q_A \)[/tex]:

1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]

2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]

Since bead A has the greater charge, we take:

[tex]\[q_A = 21.907 \, \text{nC}\][/tex]

And the charge on bead B:

[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]

So, the charges on the beads are:

[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]

Answer:

Explanation:

Moles of helium ( n ) = 156.7 / 4 = 39.175

Temperature T₁ = 35.73 +273 = 308.73 K

Volume V₁ = V

Pressure P₁ = 3.55 atm

V₂ =1.75 V

a ) For isothermal change

P₁ V₁ = P₂V₂

P₂ = P₁ V₁ / V₂

= 3.55 X V / 1.75 V

= 2.03 atm.

b ) Work done by the gas = nRT ln(V₂/V₁)

= 39.175 X 8.321 X 308.73 X ln 1.75

= 56318.8 J

Work done on the gas = - 56318.8 J

c ) Since there is no change in temperature , internal energy of gas is constant

Q = ΔE + W

ΔE  = 0

Q = W

Work done by gas = heat absorbed

heat absorbed = 56318.8 J

d ) Change in the internal energy of gas is zero because temperature is constant.

Answer:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

Explanation:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

(2) Navier Stokes equation is used to model weather and ocean current and water flow in the pipe and air flow around wing

( 3) equation is

[tex]\nabla .\overrightarrow{v} = 0[/tex]   momentum equation

[tex]\rho \frac{d\overrightarrow{v}}{dt} = \nabla p + \rho \overrightarrow{g} + \mu \nabla ^2 v^2[/tex]

here [tex]\nabla p[/tex] is pressure gradient and [tex]\rho \overrightarrow{g}[/tex] is body force and [tex]\mu \nabla ^2 v^2[/tex] is diffusion term

and

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

(2) it is drive in 1886 from Navier Stokes law

(3) equation is attach

here

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = [tex]5.0\times 10^{- 5} T[/tex]

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

[tex]B = \farc{\mu_{o}I}{2\pi d}[/tex]

where

d = distance from current carrying wire

Now,

[tex]d = \frac{\mu_{o}I}{2\pi B}[/tex]

[tex]d = \frac{4\pi\times 10^{- 7}\times 5.0}{2\pi\times 5.0\times 10^{- 5}}[/tex]

d = 0.02 m 2 cm

The distance from the wire where the magnitude of the magnetic field equals the Earth's magnetic field is approximately 0.4 meters.

To find the distance from the wire where the magnitude of the magnetic field is equal to the strength of the Earth's magnetic field, we can use the equation:

B = μ0 * I / (2π * r)

Where B is the magnetic field, μ0 is the permeability of free space (4π x [tex]10^-7[/tex]A), I is the current, and r is the distance from the wire.

Plugging in the given values, we have:

B_wire = μ0 * 5.0A / (2π * r) and B_earth = 5.0 x [tex]10^-5 T[/tex]

Setting B_wire equal to B_earth and solving for r:

5.0 x[tex]10^-5[/tex]* 5.0A / (2π * r)

Solving for r, we find that the distance from the wire where the magnitude of the magnetic field is equal to the Earth's is approximately 0.4 meters.

To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. The distance of the earthquake center can be calculated by multiplying the time difference by the speed difference between the two waves.

To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. We know that the speed of the P-wave is 9000 m/s and the speed of the S-wave is 5000 m/s.

We can calculate the distance using the formula: distance = speed × time.

So, the distance of the earthquake center can be calculated as follows:

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The speed of the ball when it leaves Sarah's hand is 8.2 m/s.

Given that;

The ball leaves Sarah's hand at a distance of 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head.

For the speed of the ball when it leaves Sarah's hand, use the equations of motion and consider the vertical motion of the ball.

Since the ball is thrown vertically upward and then comes back down, the time taken to reach the maximum height is half of the total time of flight.

Therefore, the time to reach the maximum height is,

t/2 = 1.619 s / 2.

So, the time to reach the maximum height is 0.8095 s.

Now, let's find the initial vertical velocity of the ball using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

At the maximum height, the final vertical velocity is 0 m/s because the ball momentarily stops.

The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downward).

Using the equation, we have:

0 m/s = u + (-9.8 m/s²) × 0.8095 s

Simplifying the equation:

u = 7.93 m/s

So, the initial vertical velocity of the ball is 7.93 m/s.

Since the ball travels in a parabolic path, the time taken to reach Julie's horizontal position is the same as the time taken to reach the maximum height, which is 0.8095 s.

Now, let's calculate the initial horizontal velocity of the ball, using the equation:

s = ut

where s is the horizontal distance travelled, u is the initial horizontal velocity, and t is the time.

The horizontal distance travelled is equal to the horizontal distance between Sarah and Julie, which we don't have.

However, since we are only interested in the initial horizontal velocity, we can assume that the horizontal distance travelled is equal to the distance between Sarah and Julie.

Therefore, s = 1.5 m.

Using the equation, we have:

1.5 m = u × 0.8095 s

u = 1.5 m / 0.8095 s

Calculating u, we find:

u ≈ 1.853 m/s

So, the initial horizontal velocity of the ball is 1.853 m/s.

Finally, the speed of the ball when it leaves Sarah's hand by combining the horizontal and vertical components of velocity using the Pythagorean theorem:

speed = √(horizontal velocity² + vertical velocity²)

speed = √(1.853 m/s)² + (7.951 m/s)²

Calculating the speed, we find:

speed ≈ 8.2 m/s

Therefore, the speed of the ball when it leaves Sarah's hand is 8.2 m/s.

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Final answer:

Using the kinematic equations for vertical motion, the initial speed of the ball Sarah threw is calculated to be approximately 7.94 m/s, based on the given maximum height and time to reach Julie's horizontal position.

Explanation:

To find the initial speed of the ball when it leaves Sarah's hand, we need to use the information about the ball's motion under gravity. We know it reaches a maximum height of 8 m and takes 1.619 s to get directly over Julie's head, starting from a height of 1.5 m.

The motion of the ball can be divided into two segments: ascending and descending. During the ascending part, the ball slows down due to gravity until it reaches its maximum height. In the descending part, it accelerates back down. Since the motion is symmetrical, the time to reach the maximum height is half of the total time, which is 1.619 s / 2 = 0.8095 s.

To find the initial velocity (v_i), we can use the kinematic equation for vertical motion:
v_i = v_f - g*t
where v_f is the final velocity (0 m/s at the maximum height), g is the acceleration due to gravity (9.81 m/[tex]s^2[/tex]), and t is the time to reach maximum height.

Plugging in the values, we get:
v_i = 0 m/s - (-9.81 m/[tex]s^2[/tex] * 0.8095 s) = 7.94 m/s
Therefore, the initial speed of the ball when it leaves Sarah's hand is about 7.94 m/s.

Answer:

mechanical energy per unit mass is 887.4 J/kg

power generated is 443.7 MW

Explanation:

given data

average velocity = 3 m/s

rate = 500 m³/s

height h = 90 m

to find out

total mechanical energy and power generation potential

solution

we know that mechanical energy is sum of potential energy and kinetic energy

so

E = [tex]\frac{1}{2}[/tex]×m×v² + m×g×h    .............1

and energy per mass unit is

E/m =  [tex]\frac{1}{2}[/tex]×v² + g×h

put here value

E/m =  [tex]\frac{1}{2}[/tex]×3² + 9.81×90

E/m = 887.4 J/kg

so mechanical energy per unit mass is 887.4 J/kg

and

power generated is express as

power generated = energy per unit mass ×rate×density

power generated = 887.4× 500× 1000

power generated = 443700000

so power generated is 443.7 MW

The total mechanical energy per unit mass of the river water is the sum of the potential and kinetic energy. The potential energy is calculated using the height and the gravitational constant, while kinetic energy is calculated using velocity. The power generation potential of the river is the total mechanical energy multiplied by the volume flow rate.

The total mechanical energy per unit mass of the river water at this location can be determined using the principle of mechanical energy conservation and the knowledge of potential and kinetic energy. The total mechanical energy (E) equals the potential energy (PE) plus the kinetic energy (KE) per unit mass.

The potential energy is given by: PE = mgh, where m is the mass, g is the acceleration due to gravity (~9.81 m/s^2), and h is the height (90 m above lake surface). But since we are looking for energy per unit mass, m cancels out and we get PE = gh.

The kinetic energy is given by: KE = 0.5mv^2, where v is the velocity (3 m/s). In per unit mass terms, this simplifies to KE = 0.5v^2.

Therefore, E = PE + KE = gh + 0.5v^2.

The power generation potential can be calculated using the equation: P = E * volume flow rate (in our case, 500 m^3/s).

So,​​ P = (gh + 0.5v^2) * 500.

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Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

[tex]F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}[/tex]

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

[tex]acceleration=\frac{30}{5}=6m/s^{2}[/tex]

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

[tex]v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds[/tex]

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

Answer:

The speed of the train is 7.75 m/s towards station.

The speed of the train is 8.12 m/s away from the station.

Explanation:

Given that,

Frequency of the whistles f= 175 Hz

Beat frequency [tex]\Delta f= 4.05 Hz[/tex]

Speed of observer = 0

We need to calculate the frequency

Using formula of beat frequency

[tex]\Delta f=f'-f[/tex]

[tex]f'=\Delta f+f[/tex]

[tex]f'=4.05+175[/tex]

[tex]f'=179.05\ Hz[/tex]

When the train moving towards station, then the frequency heard is more than the actual

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v-v_{s}})[/tex]

[tex]v=v-\dfrac{vf}{f'}[/tex]

Put the value into the formula

[tex]v=343-\dfrac{343\times175}{179.05}[/tex]

[tex]v=7.75\ m/s[/tex]

The speed of the train is 7.75 m/s towards station.

When the train moving away form the station

Again beat frequency

[tex]\Delta f=f-f'[/tex]

[tex]f'=f-\Delta [/tex]

[tex]f'=175-4.05[/tex]

[tex]f'=170.95\ Hz[/tex]

We need to calculate the speed

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v+v_{s}})[/tex]

[tex]v=\dfrac{vf}{f'}-v[/tex]

Put the value into the formula

[tex]v=\dfrac{343\times175}{170.95}-343[/tex]

[tex]v=8.12\ m/s[/tex]

The speed of the train is 8.12 m/s away from the station.

Hence, This is the required solution.

Answer:

force is 96 N

Explanation:

given data

mass = 40 kg

acceleration = 2.4 m/s²

to find out

force

solution

we know force is mass time acceleration so

we will apply here force formula that is express as

force = m × a   ..............1

here m is mass and a is acceleration so

put here value in equation 1 we get force

force = 40 × 2.4

force = 96

so force will be 96 N

Answer:

1.93 m

Explanation:

Initial velocity of fish u = 11 m /s

Angle of jump = 34 degree.

Vertical component of its velocity = u sin34 = 11 x.5592 = 6.15 m /s

Considering its motion in vertical direction ,

u = 6.15 m/s

g = - 9.8 m /s

maximum height attained = h

From the formula

v² = u² - 2gh

0 = 6.15x 6.15 - 2 x 9.8 h

h = (6.15 x 6.15 )/ 2 x 9.8

= 1.93 m .

Answer:

(a) d = 20 m

(b) d' = 80 m

(c) x = 28 m

(d) x' = 96 m

Solution:

As per the question:

Initial velocity of the object, v = 0

Constant acceleration of the object, [tex]a_{c} = 10 m/s^{2}[/tex]

(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:

[tex]d = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]d = 0.t + \frac{1}{2}\times 10\times 2^{2} = 20 m[/tex]

(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:

[tex]d' = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]d' = 0.t + \frac{1}{2}\times 10\times 4^{2} = 80 m[/tex]

Now, when initial velocity, v = 4 m/s, then

(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:

[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]x = 4\times 2.0 + \frac{1}{2}\times 10\times 2^{2} = 28 m[/tex]

(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:

[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]x = 4\times 4.0 + \frac{1}{2}\times 10\times 4^{2} = 96 m[/tex]

Answer:

acceleration, a = [tex]20 m/s^{2}[/tex]

distance, d = 80 m

Given:

Initial velocity of object, v = 20 m/s

Final velocity of object, v' = 60 m/s

Time interval, [tex]\Delta t = 2.0 s[/tex]

Solution:

(a) Acceleration is the rate at which velocity changes and constant acceleration is when the velocity changes by equal amount in equal intervals of time.

Thus

acceleration, a = [tex]\frac{v' - v}{\Delta t}[/tex]

a = [tex]\frac{60 - 20}{2.0} = 20 m/s^{2}[/tex]

(b) Now, the distance covered is  given by:

[tex]d = vt + \frac{1}{2}at^{2}[/tex]

[tex]d = 20\times 2.0 + \frac{1}{2}20\times 2^{2} = 80 m[/tex]

Final answer:

To find the time the football is in the air before hitting the ground, we can analyze the vertical motion using the given initial velocity and launch angle.

Explanation:

To find the time it takes for the football to hit the ground, we need to analyze the vertical motion of the football. We can use the formula:
t = (2 * vy) / g
where t is the time, vy is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Given that the initial velocity is 22.0 m/s and the launch angle is 58.5°, we can find the vertical component of the velocity using the formula:
vy = v * sin(θ)
where v is the initial velocity and θ is the launch angle.

Using this information, we can calculate the time it takes for the football to hit the ground.

Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

[tex]V = \dfrac{4}{3}\pi r^3\ ft^3[/tex]

The density of shot is

Using formula of density

[tex]\rho = \dfrac{m}{V}[/tex]

Put the value into the formula

[tex]\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}[/tex]

We need to calculate the radius

Using formula of specific gravity

[tex]specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}[/tex]

Put the value into the formula

[tex]6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}[/tex]

[tex]r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}[/tex]

[tex]r^3=0.0067514[/tex]

[tex]r =(0.0067514)^{\frac{1}{3}}[/tex]

[tex]r=0.1890\ ft[/tex]

The diameter will be

[tex]d = 2\times r[/tex]

[tex]d =2\times0.1890[/tex]

[tex]d =0.378\ ft[/tex]

Hence, The diameter is 0.378 ft.

Answer:

option (C)

Explanation:

Speed of car = 30 m/s

Let the time taken by the police car to catch the speeding car is t

The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t

Distance traveled by the police car in time t

[tex]s=ut + 0.5 at^{2}[/tex]    .... (1)

Distance traveled by the speeding car in t + 1 second

s = 30 (t + 1) = 300

t + 1 = 10

t = 9 s

Put the value of t in equation (1), we get

300 = 0 + 0.5 x a x 9 x 9

a = 7.41 m/s^2

C. 7.41 meters per square second.

In this question, the car is travelling at constant speed, whereas the police car accelerates uniformly after some delay to catch the car, the respective kinematic formulas are shown below:

Car

[tex]x_{C} = x_{o} + v_{C}\cdot t[/tex] (1)

Police car

[tex]x_{P} = x_{o} + \frac{1}{2}\cdot a_{P}\cdot (t-t')^{2}[/tex] (2)

Where:

If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{C} = x_{P} = 300\,m[/tex], [tex]t' = 1\,s[/tex] and [tex]v_{C} = 30\,\frac{m}{s}[/tex], then we have the following system of equations:

[tex]300 = 30\cdot t[/tex] (1)

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (t-1)^{2}[/tex] (2)

By (1):

[tex]t = 10[/tex]

Then we find that acceleration of the police car must be:

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (10-1)^{2}[/tex]

[tex]a_{P} = 7.407\,\frac{m}{s^{2}}[/tex]

Therefore, the correct choice is C.

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Answer:

W = 289.70 kg

Explanation:

Given data:

Pressure in tank = 23 atm

Altitude 1000 ft

Air temperature  in tank T = 700 F

Volume of tank = 800 ft^3 = 22.654 m^3

from ideal gas equation we have

PV =n RT

Therefore number of mole inside the tank is

[tex]\frac{1}{n} = \frac{RT}{PV}[/tex]

              [tex] = \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}[/tex]

               [tex]= 1.02\times 10^{-4}[/tex]

               [tex]n = 10^4 mole[/tex]

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is [tex]W = 10^4\times 28.91 g = 289700 g[/tex]

               W = 289.70 kg

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

[tex]Q=mL[/tex], here m is mass of ice and L is latent heat of fusion

So heat [tex]Q=mL=0.4535\times 334=151.469kj[/tex]

So heat required to melt 1 lb of ice is equal to 151.469 KJ

Answer:

The force on X Fx=0 N

The force on Y Fy=-2.18 N

Explanation:

We have an array of charges, we will use the coulomb's formula to solve this:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

but we first have to find the distance and the angule of the charge respect the charges on the Y axis:

[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]

we can notice that it is the same distance from both charges on Y axis.

we can find the angle with:

[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]

for the charge of 5µC [tex]\alpha =-66.80^o[/tex]

for the charge of -5µC [tex]\alpha =66.80^o[/tex]

the net force on the X axis will be:

[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]

and

[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]

So the net force on X will be Zero.

for the force on Y we have:

[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]

and

[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]

Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]

So the net force on Y is Fy=-2.18N

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. The force is calculated to be 5.12 N.

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k * |q1 * q2| / r^2

Where F is the force, k is the electrostatic constant (k ≈ 8.99 * 10^9 N * m² / C²), q1 and q2 are the magnitudes of the charges, and r is the distance between them.

In this case, one charge is 5 µC and the other charge is -5 µC, with a distance of 14 cm between them (7 cm on the positive y-axis and 7 cm on the negative y-axis). The charge we're interested in is 2 µC at x = 3 cm. Plugging these values into the formula:

F = (8.99 * 10^9 N * m² / C²) * |(5 µC) * (2 µC)| / (0.14 m)²

F = 5.12 N

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Answer:

a) [tex]x=26m[/tex]

b) [tex]v_{y}=-21.5m/s[/tex]

Explanation:

From the exercise we know initial velocity, initial height

[tex]y_{o}=20m[/tex]

[tex]v_{o} =12m/s[/tex] [tex]\beta =45[/tex]º

a) The range of the stone is defined by how far does it goes. From the theory of free falling objects, we have:

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

The stone strike the water at y=0

[tex]0=20+12sin(45)t-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t, using the quadratic formula

[tex]t=-1.33s[/tex] or [tex]t=3.06s[/tex]

Since time can't be negative, the answer is t=3.06s

Now, we can calculate the range of the stone

[tex]x=v_{o}t=(12cos(45))m/s(3.06s)=26m[/tex]

b) We can calculate the velocity were the stone strike the water using the following formula  

[tex]v_{y}=v_{oy}+gt=12sin(45)m/s-(9.8m/s^{2})(3.06s)=-21.5m/s[/tex]

The negative sign indicates that the stone is going down

Answer:

(a) x= 26m

(b) vf= 23.15m/s  

Explanation:

Given data

h=20m

Ɵ=45°

to find

(a) range of stone=x=?

(b) velocity=vf=?

Solution

For part (a)

You need to solve for time first using

yf = yi + visinƟt + 1/2gt^2

0 = 20m + 12sin45t + 1/2(-9.8)t^2

and use the quadratic equation to solve for t

t = 3.064 sec

To solve for the distance traveled use

x = xi + vicosƟt + 1/2at^2 there is no acceleration in the x direction so that cancels

x = 12cos(45)(3.064)

x= 26m

For part(b)

For b I'm not sure if you what direction you want the final velocity in the x, y, or the direction its traveling so I'll just give all 3.

Theres no change in the velocity in the x direction so its just vfx = vixcosƟ = 12cos45 = 8.49m/s

For the y direction its vfy^2 = viy^2 + 2g(Δy)

vfy = sqrt((12sin(45))^2 + 2(-9.8)(0-20m)) = 21.54m/s

The velocity the direction the stone is traveling is vf = sqrt(vx^2 + vy^2) = sqrt(8.49^2 + 21.54^2)

vf= 23.15m/s

Final answer:

By applying the conservation of energy principle, accounting for the work done by the electric field as change in potential energy, we calculate the change in kinetic energy to find the speed of the charged particle at a different electric potential.

Explanation:

To calculate the speed of the particle at point B, we need to consider the change in electric potential energy and convert it to kinetic energy. The work done on the particle by the electric field is equal to the change in potential energy, which is given by W = q(VB - VA), where q is the charge of the particle, VA and VB are the electric potentials at points A and B respectively. Since no other forces act on the particle, the work done by the electric field causes a change in kinetic energy, which is defined by ½mv22 - ½mv12 = q(VB - VA), where m is the mass of the particle and v1 and v2 are the speeds at points A and B.

Rearranging the equation to solve for v2 gives us v2 = √(2q(VB - VA)/m + v12). Substituting in the given values, we can find the speed at point B.

Answer:

Three linear operators A,B, and C will satisfy the condition [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].

Explanation:

According to the question we have to prove.

[tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex]

Now taking Left hand side of the equation and solve.

[tex][[A, B],C] + [[B,C), A] + [[C, A], B][/tex]

Now use commutator property on it as,

[tex]=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0[/tex]

Therefore, it is proved that [tex][[A, B],C] + [[B,C), A] + [[C, A], B] = 0[/tex].

Answer:

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

[tex]f=1.363684118Hz[/tex]

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

[tex]\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}[/tex]

Replacing the data provided:

[tex]\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}[/tex]

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

Finally, to calculate the frequency of oscillation we use this:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }[/tex]

Replacing m and k:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz[/tex]

Answer:

28.5 m/s

Explanation:

The speed has two orthogonal components, horizontal and vertical. To decompose the speed into these components we can use these trigonometric equations:

Vh = V * cos(a)

Vv = V * sin(a)

The forwards velocity is the horizontal component, so we use

Vh = V * cos(a)

Vh = 50 * cos(55)

Vh = 50 * 0.57

Vh = 28.5 m/s

Answer:

a)[tex]v=6.19m/s[/tex]

b)[tex]v=5.51m/s[/tex]

c)[tex]a=3.3*10^{3}m/s^{2}[/tex]

d)[tex]x=5.78*10^{-3}m[/tex]

Explanation:

h1=195m

h2=1.55

a) Velocity just before the ball strikes the floor:

Conservation of the energy law

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=mgh_{1}[/tex]

[tex]E_{f}=1/2*mv^{2}[/tex]

so:

[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]

b) Velocity just after the ball leaves the floor:

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=1/2*mv^{2}[/tex]

[tex]E_{f}=mgh_{2}[/tex]

so:

[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]

c) Relation between Impulse, I, and momentum, p:

[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]

d) The compression of the ball:

The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.

[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]

Kinematics equation:

[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]

Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.

a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3

So:

[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]

Answer:

Explanation:

a) Hardness is a measure of the resistance of a material to permanent deformation (plastic) on its surface,

Hardness tests play an important role in material testing, quality control and component acceptance.

Hardness test are needed to be perform as a quality assurance procedure, to validate materials are according to the specific hardness required,

We depend on the data to verify the quality of the components to determine if a material has the necessary properties for its intended use.

Through the years, the establishment of increasingly productive and effective means of testing, has given way to new cutting-edge methods that perform and interpret hardness tests more effectively than ever. The result is a greater capacity and dependence on "letting the instrument do the work", contributing to substantial increases in performance and consistency and continuing to make hardness tests very useful in industrial and R&D applications.

b)

"The hardness test is a critical aspect of engineering practice for several reasons:

1. Material Performance: Hardness is a measure of a material's resistance to deformation, which is directly related to its wear resistance, strength, and durability. By performing hardness tests, engineers can predict how a material will perform under service conditions.

2. Quality Control: Hardness testing is used to ensure that materials meet the required specifications. It is a non-destructive (in some cases) or semi-destructive method to verify the quality of the material without compromising its integrity.

3. Material Selection: The results of hardness tests help engineers to select appropriate materials for different applications. A material that is too soft or too hard for a particular application can lead to premature failure or unnecessary costs.

4. Heat Treatment Verification: Heat treatment processes such as quenching, annealing, and tempering are used to alter the mechanical properties of materials. Hardness testing is used to confirm that the desired properties have been achieved after such treatments.

5. Failure Analysis: In the event of a material failure, hardness testing can provide valuable information about the cause of the failure. It can indicate if the material was too brittle or too soft for the intended application.

6. Comparative Analysis: Hardness tests provide a simple way to compare the properties of different materials or different batches of the same material.

7. Research and Development: In the development of new materials, hardness testing is an essential tool for characterizing materials and understanding their properties.

B) Two possible sources of error in hardness testing, other than parallax error, and their effects on the results are:

1. Indenter Geometry Errors: The geometry of the indenter (e.g., ball, cone, or pyramid) must be precise according to the test standard being used. Wear, damage, or incorrect shape of the indenter can lead to inaccurate measurements. For instance, a worn or blunt indenter will give lower hardness values than the actual hardness of the material.

2. Loading Rate and Duration Errors: The rate at which the load is applied and the duration for which it is held can significantly affect the hardness measurement. Some materials exhibit time-dependent plasticity (creep), which can affect the size of the indentation. If the loading rate or duration is not consistent with the standards or previous tests, the results may not be comparable or may not accurately represent the material's hardness. For example, applying the load too quickly may result in an underestimation of hardness due to the material's inability to fully resist the indenter.

To minimize these errors, it is essential to follow standardized testing procedures, maintain and calibrate equipment regularly, and ensure that the test conditions are consistent. Additionally, operators should be well-trained and aware of the potential sources of error to ensure the accuracy and reliability of the hardness test results."

Answer:

(a) 0.9 s

(b) 7.78 m/s

Explanation:

height, h = 4 m

Horizontal distance, d = 7 m

Let it takes time t to reach the ground and u be the initial velocity of the jet.

(a) Use second equation of motion in vertical direction

[tex]s = ut + \frac{1}{2}at^{2}[/tex]

In vertical direction, uy = 0 m/s, a = g = - 9.8 m/s^2, h = - 4 m

By substituting the values, we get

[tex]-4 = 0 - \frac{1}{2}\times 9.8\times t^{2}[/tex]

t = 0.9 second

Thus, the time taken by water jet in air is 0.9 second.

(b) Use

Horizontal distance = horizontal velocity x time

d = u t

7 = u x 0.9

u = 7.78 m/s

Thus, the initial velocity of water jet is 7.78 m/s.

The duration the water is in the air is found using the formula for the motion under gravity, which depends on the vertical distance and the acceleration due to gravity. After finding the time, the initial velocity of the water is calculated using the horizontal distance and the fraction of time the water was in motion.

To determine how long the water is in the air (a), and the initial velocity of the water (b) when a water gun is fired horizontally from a hill, we can use the principles of projectile motion. The time a projectile is in the air is solely determined by its vertical motion. Since the water gun is fired horizontally, it has an initial vertical velocity of 0 m/s.

For part (a), the time (t) it takes for the water to reach the ground can be calculated using the formula for the motion under gravity, which is y = 0.5 * g * t2, where y is the vertical distance (4 meters in this case) and g is the acceleration due to gravity (approximated to 9.81 m/s2). Solving for t gives us the time the water is in the air.

For part (b), once we have the time, we can use the horizontal distance (7 meters) to find the initial velocity (v0) using the formula x = v0 * t. This provides the initial horizontal velocity of the water gun's jet. The overall process involves solving for time first and then using that time to find the initial velocity.