With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

Answer:

a) 2πR

b) greater

Explanation:

Radius of the cylinder = R

a) after one complete revolution, the center of mass would have moved a distance equal to the circumference of the radius formed by the cylinder.

distance moved = 2πR

b) If slipping occurred, there will be some distance lost during rotation, which will increase the distance that the center of mass will have to move.

Answer:

18.83m/s

Explanation:

In the first 11meters, we can calculate the Kinectic energy since we know that

the floor is frictionless then work done by the horizontal force= Kinectic energy gain by the block in the first 11m.

Then Kinetic Energy = wordone

But work done= Force × distance

Kinetic Energy=(225×11)= 2475J

In the next 10.0 m the coefficient of friction is 0.20, the Kinectic energy has equal value to the difference in workdone by both the horizontal and frictional force

K.E= (Force× distance) - ( mass × gravity× coefficient of friction × distance at 10 m)

K.E= [(225×10)-(.20× 9.8×10× 24)]

K.E= 2250-470.4

= 1779.6J

The addition of the Kinectic energy above give us the total Kinectic energy experience by the crate.

Total Kinectic energy= 1779.6+2475

= 4254.6J

the final speed of the crate after being pulled these 21.0 m?

Total distance= 11m + 10m = 21 m

Then the final speed can be calculated from the total Kinectic energy, since we know that

K.E= 0.5mv^2

V= √(2K.E/m)

= √(2×4254.6)/24

Final speed v = √354.55

Final speed v= 18.83m/s

Therefore, the final speed of the crate after being pulled these 21.0 m is 18.83m/s

Explanation:

The pressure of hurricane which is high decreases gradually as we move higher . The pressure is maximum at the surface . Hence the relative air pressure is higher at the surface and low at the top . Like wise the pressure is low at the center and keeps on increasing when we move outside . Hence the answer is pressure is high at surface and low aloft.

Speed, weight, and time between impact and stopping all are the factors that affect force.

A force is an influence in physics that can change the motion of an object. A force can cause a mass object to change its velocity, or accelerate.

Intuitively, force can be described as a push or a pull. A force is a vector quantity because it has both magnitude and direction.

The magnitude of a force expresses its strength. The magnitude of a force is expressed in Newtons, the SI unit of force.

One Newton is the force that can cause an object weighing one kilogram to move one meter per second.

The mass of the objects and the coefficient of friction between them are the two factors. The angle between them is also important.

Thus, these are some factors affecting strength of force.

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Answer:

D. The object splitting into two parts to form two organisms.

Explanation:

The splitting of cell of an organism actually shoes if it's Alive or not .

The splitting can be either mitosis or meosis for either plant or animal.

But for the fact that it's has cell membrane shoes it's either a plant or an animal.

So the splitting will confirm it's alive.

Answer:

D the correct answer is D

Explanation:

I took the test and was going to pick c but changed my mind

Answer:

aaawwwwwwwsssaaaasasss

Answer:

e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

Explanation:

This is an ejercise in special relativity, where the speed of light is constant.

Let's carefully analyze the approach, we see the two events at the same time.

The closest event time is

       c = (x₁-300) / t

       t = (x₁-300) / c

The time for the other event is

       t = (x₂- 600) / c

since they tell us that we see the events simultaneously, we can equalize

        (x₁ -300) / c = (x₂ -600) / c

         x₁ = x₂ - 300

We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

Answer:

form a hypothosis

Explanation:

Explanation:

Solidity means the quality or state of being firm or strong in structure. Density means the degree of compactness in a structure.

Complete question:

A certain organ pipe, open at both ends, produces a fundamental frequency of 293 Hz in air. If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.97 g/mol and the molar mass of helium to be 4.00 g/mol. Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 293 Hz in air. How does your answer to first question, fHe, change?

Answer:

The fundamental frequency that helium produced is 788.6 Hz.

Since the fundamental frequency of air in closed pipe is still the same value as open pipe, the fundamental frequency of helium will not change.

Explanation:

The speed of sound in pipe when filled with air is given by;

[tex]v = \sqrt{\frac{\gamma RT}{M} }[/tex]

Where;

γ = 1.4

R = 8.31 J/mol.K

M = 0.02897 kg/mol

T = 20°C = 293 K

[tex]v_{air} = \sqrt{\frac{1.4* 8.31*293}{0.02897} } = 343 \ m/s[/tex]

The speed of sound in pipe when filled with helium is given by

[tex]v_{He} = \sqrt{\frac{1.4* 8.31*293}{0.004} } =923.14 \ m/s[/tex]

Now, determine the fundamental frequency Helium will it produce

v = fλ

[tex]\frac{V_{air}}{F_o{air}} = \frac{V_{He}}{F_o{He}}\\\\F_o{He} = \frac{F_o{air}*V_{He}}{V_{air}} \\\\F_o{He} = \frac{(293)*(923.14)}{343}\\\\F_o{He} = 788.6 \ Hz[/tex]

Since the fundamental frequency of air in closed pipe is still the same value as open pipe, the fundamental frequency of helium will not change.

Answer:

True

Explanation:

Without proper development during the critical period, the growth of some life skills may be severely delayed or stunted.

Answer:

The  value is  [tex]h = 11930 \ m[/tex]

Explanation:

From the question we are told that

    The  temperature is  [tex]T = 300 \ K[/tex]

Generally the root mean square speed of the  oxygen molecules is mathematically represented as

        [tex]v = \sqrt{\frac{3 * R * T }{M} } = \sqrt{ 2 * g * h }[/tex]

Here  R is the gas constant with a value  [tex]R = 8.314 \ J\cdot K^{-1} \cdot \ mol^{-1}[/tex]

    M  is the molar mass of oxygen molecule with value [tex]M = 0.032 \ kg /mol[/tex]

So  

     [tex]\frac{3 * 8.314 * 300 }{0.032} = 2 * 9.8 * h[/tex]

=>    [tex]h = 11930 \ m[/tex]

Answer:

The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]

The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]

Explanation:

Given that,

Charge = q

Acceleration = a

The rate at which energy is emitted from an accelerating charge

[tex]\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon c^3}[/tex]....(I)

We know that,

Acceleration for circular motion is

[tex]a=\dfrac{v^2}{r}[/tex]....(II)

The kinetic energy is

[tex]K.E=\dfrac{1}{2}mv^2[/tex]

[tex]v^2=\dfrac{2(K.E)}{m}[/tex]

Put the value of v in equation (II)

[tex]a=\dfrac{2(K.E)}{mr}[/tex]

Put the value of a in equation (I)

[tex]\dfrac{dE}{dt}=\dfrac{q^2(\dfrac{2(K.E)}{mr})^2}{6\pi\epsilon c^3}[/tex]

[tex]\dfrac{dE}{dt}=\dfrac{q^24(K.E)^2}{6\pi\epsilon c^3\times m^2 r^2}[/tex]

[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=\dfrac{q^24(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]

Suppose that,

[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=R[/tex]

So,

[tex]R=\dfrac{q^2\times4(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]....(III)

(a). For proton,

We need to calculate the fraction of its energy does it radiate per second

Using equation (III)

[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(1.67\times10^{-27})^2\times(0.540)^2}[/tex]

[tex]R=2.23\times10^{-11}[/tex]

(b). For electron,

We need to calculate the fraction of its energy does it radiate per second

Using equation (III)

[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(9.1\times10^{-31})^2\times(0.540)^2}[/tex]

[tex]R=0.0000753[/tex]

Hence, The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]

The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

                v² = v₀² - 2 g (y -y₀)

                v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

              v² = 2g y₀

              v = Ra (2g i)

let's calculate

              v =√ (2 9.8 16)

              v = 17.71 m / s

Explanation:

Pressure = force / area

P = 55 N / (0.55 m × 0.45 m)

P = 220 Pa

Answer:

The average velocity of an object is its total displacement divided by the total time taken. In other words, it is the rate at which an object changes its position from one place to another. Average velocity is a vector quantity. The SI unit is meters per second.

hope this helps you

Answer:

displacement divided by time

Explanation:

Explanation:

We need to add 28.97 and 45.876.

Here, the first no is 28.97 and other no is 45.876

Here, 28.97 is the least precise value and 45.876 is most precise.

28.97 + 45.876 = 74.846

We need to write the final answer such that there is two digit after the decimal. So, 28.97 + 45.876 = 74.85 because 6 is more than 5 i.e. we add 1 to 4.

Answer:

 28.97

+ 45.876

 74.846

Explanation:

Because 28.97 has only two decimal places, round off the answer to two decimal places: 74.85.

Answer:

the correct amount

Explanation:

The pan balance has two pans . When the elevator is accelerating upward , the apparent weight of both objects placed on either pan increase . The net effect is that they cancel out the effect of accelerating elevator . Hence the apparent weight of salt remains same as and equal to its real weight.

In case of spring balance,  its apparent weight would have increased . Its apparent weight would have measured more than its true weight.

Answer: -1   or -4/4

Explanation:

4/4  simplifies to 1   and the opposite of 1 is -1.

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / [tex]n_{f}^{2}[/tex] - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

Answer:2

Explanation:

Suppose that the clay balls model the growth of a planetesimal at various stages during its accretion, the planetesimal that is most likely to pull in debris particle A would be option B.

Science is the methodical, empirically-based pursuit and application of knowledge and understanding of the natural and social worlds.

Science is a way of learning about the world.

People may contribute to the development of new knowledge through science and utilize it to promote their objectives.

It is a method, a thing, and an organization all at once.

If the clay balls represent the formation of a planetesimal at different phases of its accretion, then option B represents the planetesimal that is most likely to draw in debris particle A.

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Answer:

4.167m/sec

Explanation:

1km=1000m

1.5km=1500m

1min=60sec

6min=360sec

In 360sec they travel 1500m

In 1 sec they travel=1500m/360

1sec=4.167m

Answer:

a

  When [tex]r \ge R[/tex]

      [tex]B = \frac{ \mu_o * I}{ 2 \pi r }[/tex]

b

 When [tex]r< R[/tex]

   [tex]B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r[/tex]

Explanation:

From the question we are told that

   The  radius is  R  

   The  current is  I

    The  distance from the center

Ampere's law is mathematically represented as

       [tex]B[2 \pi r] = \mu_o * \frac{I r^2 }{R^2 }[/tex]

      [tex]B = \frac{ \mu_o}{2 \pi } * \frac{r}{R^2}[/tex]

When [tex]r \ge R[/tex]

=>     [tex]B = \frac{ \mu_o * I}{ 2 \pi r }[/tex]

But when [tex]r< R[/tex]

   [tex]B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r[/tex]

Answer:

7.92 × 10^8 M

A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time between the generation of the wave and the detection of the reflected wave is 2.6444s. then the distance from the surface of the earth to the surface of the moon would be 3.9632×10⁸ meters.

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

As given in the problem A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time between the generation of the wave and the detection of the reflected wave is 2.6444s

the given speed of the light is 2.9979 ×10⁸ m/s

As we know that

Distance =speed ×time

Distance = 2.9979 ×10⁸× 2.6444

               =7.92644× 10⁸

This is the total distance covered by both ways from the moon to earth

The actual distance of the moon from earth would be half of this distance

Thus, the distance from the surface of the earth to the surface of the moon would be 3.9632×10⁸ meters.

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Answer:

T' = 2T

Explanation:

The time period of a simple pendulum is given by the relation as follows :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of the pendulum

g is acceleration due to gravity

If the length is increased four time, new length is l' = 4l

So,

New time period is :

[tex]T'=2\pi \sqrt{\dfrac{l'}{g}}\\\\T'=2\pi \sqrt{\dfrac{4l}{g}}\\\\T'=2\times 2\pi \sqrt{\dfrac{l}{g}}\\\\T'=2\times T[/tex]

So, the new time period is 2 times of the initial time period.

Answer:

The  value is  [tex]d = 0.000125 \ m[/tex]

Explanation:

From the question we are told that

   The  wavelength is  [tex]\lambda = 500 \ nm = 500 *10^{-9 } \ m[/tex]

    The  distance is of  the screen is  [tex]D = 2.0 \ m[/tex]

    The  width of the fifth dark spot is [tex]y = 4.00 \ cm = 0.04 \ m[/tex]

Generally the width of a fringe is mathematically represented as

           [tex]y = \frac{ n * \lambda * D }{d }[/tex]

=>        [tex]0.04 = \frac{ 5 * 500 *10^{-9} * 2 }{d}[/tex]

=>        [tex]d = \frac{ 5 * 500 *10^{-9} * 2 }{0.04}[/tex]

=>        [tex]d = 0.000125 \ m[/tex]

Answer:

option B is correct

Explanation:

in geocentric model earth is at the center of the universe and sun,moon star etc orbited the earth  thats why option B is correct

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + [tex]a_{c}[/tex]²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure =  P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa