An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 15.0 kg, a radius of 0.400 m, and a length of 0.800 m. The mass of the end of the barrel equals a fourth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is =9.80 m/s2.
What is the translational speed f of the barrel at the bottom of the hill if released from rest at a height of 33.0 m above the bottom?

Hi there!

We can begin by using the work-energy theorem in regards to an oscillating spring system.

Total Mechanical Energy = Kinetic Energy + Potential Energy

For a spring:

[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]

A = amplitude (m)

k = Spring constant (N/m)

x = displacement from equilibrium (m)

m = mass (kg)

We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:

ME = KE + PE

ME - PE = KE

Thus:

[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]

Plug in the given values:

[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]

We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.

Thus, the object would have no kinetic energy since KE = 1/2mv².

Answer:

Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.

Explanation:

This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

A) Resistivity is given as,

[tex]p = \frac{RA}{l}[/tex]

where,

[tex]R = \frac{V}{I}[/tex]

Therefore,

[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]

B) Conductivity is given as,

[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]

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Answer: increases

Explanation:

Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.

Answer:

ma*XsinB

option 2 is correct

Explanation:

F = Icurrent×length×Bfieldstrength×sin(angle field to wire)

in our case

Icurrent = 10 A

length = 0.02km = 20 meters

B = 10^-6 T

angle = 30 degrees.

F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =

= 200 × 10^-6 = 2 × 10^‐4 N

Answer:

Explanation:

If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change

Answer: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".

Explanation:

Answer:

Explanation:

550 ft•lb/s / hp•(1.82 hp) / 500 lb = 2.00 ft/s

Answer:

Explanation:

Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.

In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.

That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.

Its relative speed compared to Earth is 0.921

The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.

Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.

So, v = 2πr/T

Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.

So, v' = 2πr'/T'

v'/v = 2πr'/T' ÷ 2πr/T

v'/v = r'/r × T/T'

From Kepler's law, T² ∝ r³

So, T'²/T² = r'³/r³

(T'/T)² = (r'/r)³

T'/T =  √[(r'/r)]³

T/T' = √[(r'/r)]⁻³

So, substituting this into the equation, we have

v'/v = r'/r × T/T'

v'/v = r'/r × √[(r'/r)]⁻³

v'/v = √[(r'/r)]⁻¹

Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18

So, v'/v = √[(r'/r)]⁻¹

v'/v = √[(1.18)]⁻¹

v'/v = [1.0863]⁻¹

v'/v = 0.921

So, its relative speed compared to Earth is 0.921

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Explanation:

1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2

Answer:

Explanation:

20 n is an unknown amount

If that is supposed to be 20 N(ewtons)

then W = Fd = 20(15) = 300 J

Answer: it will be 300 newton meters

Explanation:

The moral condition of the world today appears to be worse than it was in the antediluvian era.

The biblical account of the flood records that the world delved into apostasy in the days of Noah so much so that God regretted the fact that he created man. Some of the evils of the antediluvian world include; sodomy, drunkenness, lewdness and several forms of immorality.

We can see that these vices that led to the destruction of the world due to moral bankruptcy in the antediluvian era is still very much prevalent in our world today. The moral condition of the world today appears to be worse than it was in the antediluvian era.

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Answer:

all are harmful to the body

Answer:

the one in which the fluid has the lowest density

The force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.

Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.

Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.

Given:

The mass of the block, m = 2 kg,

The oscillation of spring, x  = 3 cos 4t,

Calculate the omega  by comparing the standard equation given below,

[tex]x = A cos \omega t[/tex]

ω = 4

Calculate the spring constant by the formula given below,

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

4² = k / 2

k = 32 N / m

Therefore, the force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.

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Answer:

Explanation:

1 Object C has an acceleration that is greater than the acceleration for D.

2 B

3 17M

4 The velocity is zero.

5 a straight line with negative slope

just took it

Answer:

Explanation:

s 0.12 + 1.20(1.80) = 2.28 m from the top.

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 20 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]

The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]

Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of  m·r² increases, the value of [tex]\omega _f[/tex] decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

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Answer:

Explanation:

F = GMm/d²

d = √(GMm/F)

d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)

d = 1.29398e8 = 1.29 x 10^8 m  center to center

Answer:

1.29 x 10^8 m  apart

Explanation:

Works in Acellus!

The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m

If we consider the potential energy (U) between the pions, then (U) can be expressed as:

[tex]\mathbf{U = \dfrac{kq^2}{r} ---- (1)}[/tex]

Given that at some instance, the potential energy becomes negligible compared to the final K.E.

As such the conservation of the total energy in the system can be given as:

Again, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:

[tex]\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}[/tex]

∴

Equating both equations (1) and (2) together, we have:

[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 K}[/tex]

[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }[/tex]

[tex]\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}[/tex]

where:

Replacing their values in the  above equation, the distance (r) between the pions is calculated as:

[tex]\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}[/tex]

distance (r) = 1.45 × 10⁻¹⁶ m

Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m

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Hi there!

We know that:

I = Δp = m(vf - vi)

Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.

I = 1.1(5 - (-23)) = 30.8 Ns

Answer:

Explanation:

Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.

v = at

v = 9.8(3.1)

v = 30.38

v = 30 m/s

max height can be found knowing the velocity is zero at the top of its flight.

v² = u² + 2as

s = (v² - u²) / 2a

s = (0.00² - 30.38²) / (2(-9.8))

s =  47.089

s = 47 m

A stomp rocket takes 3.1 seconds to reach its maximum height then the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.

Velocity is defined as rate of change of position with respect to time.

SI unit of velocity is m/sec. Velocity is a vector quantity.

Given that in the question time taken by rocket to reach maximum height is 3.1 sec. Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.

v = at

v = 9.8(3.1)

v = 30.38

v = 30 m/s

Max height can be found knowing the velocity is zero at the top of its flight.

v² = u² + 2as

s = (v² - u²) / 2a

s = (0.00² - 30.38²) / (2(-9.8))

s =  47.089

s = 47 m

So, the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.

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Answer:

3

Explanation:

state of matter are solid

liquid and

gases

Answer:

3

Explanation:

state of a matter are solid liquid and gas

We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is

From the question we are told

The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Generally the equation for the Block is mathematically given as

[tex]\sum Fy=0[/tex]

[tex]981cos21.80 = R_2cos53.6\\\\R_2=1535N[/tex]

the equation for the Wedge is mathematically given as

[tex]\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N[/tex]

the equation for the Screw is mathematically given as

[tex]\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1\\\\[/tex]

Therefore

[tex]M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m[/tex]

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The force of friction needed to overcome to move the box is 29.4N

According to Newton's second law;

[tex]\sum F_x = ma_x\\[/tex]

Taking the sum of force along the plane;

[tex]F_m -F_f = ma\\F_m -F_f = 0\\F_m=F_f = \mu R[/tex]

This shows that the moving force is equal to the frictional force

Given that

[tex]\mu = 0.6\\R = mg = 49N[/tex]

Get the frictional force;

Since

[tex]F_f = \mu R\\F_f = 0.6 \times 49\\F_f = 29.4N[/tex]

Hence the force of friction needed to overcome to move the box is 29.4N

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Answer:

Explanation:

The work of the brakes will equal the initial kinetic energy of the car

Fd = ½mv²

F = mv²/2d

F = 2457(18²) / (2(62))

F = 6,419.903...

F = 6.4 kN