How fast should a rocket ship move for its mass to be dilated to 167% of its rest
mass?​

Answers

Answer 1

Answer:

The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.

Hopefully, this answer helps you! :)


Related Questions

explain and derive the equation for capillary action in the phenomenon of surface tension​

Answers

Answer:

Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.

What is the frequency of a wave with a Wavelength of 200 cm?

Answers

Explanation:

wavelength =200cm=2m

frequency is reciprocal of wavelength

f=1/2

f=0.5Hz or 1/s

The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.

Calculation of the frequency:

Since the wavelength is 200 cm or 2m

Also we know that

frequency is reciprocal of wavelength

So,

f=1/2

f=0.5Hz

Hence, The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.

Learn more about frequency here: https://brainly.com/question/24470698

A rifle is aimed at a target 40m away. The bullet hits the target 2.2cm below the horizontal plane (ignore wind and rotational effects).
(a) What is the time of flight for bullet?
(b) What is the muzzle velocity (velocity of bullet when it leaves the rifle)?

Answers

Answer:

The value is [tex]u_x = 597 \ m/s[/tex]

Explanation:

From the question we are told that

    The distance of the target from the riffle is  [tex]d = 40 \ m[/tex]

    The height at which the bullet hit the target is  [tex]y = 2.2 \ cm = 0.022 \ m[/tex]

Considering the vertical motion  

Generally from kinematic equations we have that

        [tex]y = u_y t + \frac{1}{2} * gt^2[/tex]

At the initial stage the velocity is  zero  i.e [tex]u_y = 0 \ m/s[/tex]

=>     [tex]0.022 = 0 * t + \frac{1}{2} * 9.8 t^2[/tex]

=>    [tex]t = 0.067 \ s[/tex]

Generally the velocity of the bullet when it leaves the riffle is mathematically represented as

         [tex]u_x = \frac{ d}{t}[/tex]

=>     [tex]u_x = \frac{40 }{ 0.067 }[/tex]

=>     [tex]u_x = 597 \ m/s[/tex]

2. Fossil fuel has been used more in the

existing world​

Answers

Answer: fossil fuels are using more in the existing world because fossil fuels are of great importance because they can burned producing significant amount of energy prr unit mass the use of coal as a fuel prefates recorded history

Explanation:

Answer:

A fossil fuel is a fuel formed by natural processes, such as anaerobic decomposition of buried dead organisms, containing organic molecules originating in ancient photosynthesis[1] that release energy in combustion.[2] Such organisms and their resulting fossil fuels typically have an age of millions of years, and sometimes more than 650 million years.[3] Fossil fuels contain high percentages of carbon and include petroleum, coal, and natural gas.[4] Peat is also sometimes considered a fossil fuel.[5] Commonly used derivatives of fossil fuels include kerosene and propane. Fossil fuels range from volatile materials with low carbon-to-hydrogen ratios (like methane), to liquids (like petroleum), to nonvolatile materials composed of almost pure carbon, like anthracite coal. Methane can be found in hydrocarbon fields alone, associated with oil, or in the form of methane clathrates.

Coal, a fossil fuel

As of 2018, the world's main primary energy sources consisted of petroleum (34%), coal (27%), and natural gas (24%), amounting to an 85% share for fossil fuels in primary energy consumption in the world. Non-fossil sources included nuclear (4.4%), hydroelectric (6.8%), and other renewables (4.0%, including geothermal, solar, tidal, wind, wood, and waste).[6] The share of renewables (including traditional biomass) in the world's total final energy consumption was 18% in 2018.[7] Compared with 2017, world energy-consumption grew at a rate of 2.9%, almost double its 10-year average of 1.5% per year, and the fastest since 2010.[8]

Although fossil fuels are continually formed by natural processes, they are generally classified as non-renewable resources because they take millions of years to form and known viable reserves are being depleted much faster than new ones are generated.[9][10]

Most air pollution deaths are due to fossil fuel combustion products, it is estimated to cost over 3% of global GDP,[11] and fossil fuel phase-out would save 3.6 million lives each year.[12]

The use of fossil fuels raises serious environmental concerns. The burning of fossil fuels produces around 35 billion tonnes (35 gigatonnes) of carbon dioxide (CO2) per year.[13] It is estimated that natural processes can only absorb a small part of that amount, so there is a net increase of many billion tonnes of atmospheric carbon dioxide per year.[14] CO2 is a greenhouse gas that increases radiative forcing and contributes to global warming and ocean acidification. A global movement towards the generation of low-carbon renewable energy is underway to help reduce global greenhouse-gas emissions.

During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approximately 103 cm/s. The blood volume traverses the aortic arch, exiting with the same speed but opposite direction. Assume the mass density of the blood is 1060 kg/m3 blood, the aortic arch remains stationary, and that the heart rate is 59 bpm. What is the average force exerted by the blood on the wall of the aorta.

Answers

Answer:

The value  [tex]F  =  0.1396 \ N [/tex]

Explanation:

From the question we are told that

   The volume blood  ejected is  [tex]V = 65 \ cm^3 = 65*10^{-6} \ m^3[/tex]

    The velocity of the blood ejected is  [tex]v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s[/tex]

    The density of blood is  [tex]\rho = 1060 \ kg/m^3[/tex]

     The heart beat is [tex]R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps[/tex]

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      [tex]F = 2 * \rho * V * R * v[/tex]

=>    [tex]F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03[/tex]

=>    [tex]F  =  0.1396 \ N [/tex]

The index of refraction of quartz is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation. Enter your answer in accordance to the question statement

Answers

Answer:

1.04*10⁻⁵

Explanation:

light wave do showcase some behaviors whenever there is encounters with the end of the medium, some of the behaviors are - reflection, refraction, as well as diffraction. When visible light wave strikes a boundary that exist two different media, a portion of the energy will be transmitted into the new medium and some reflected.

Reflection of a light wave can be regarded as bouncing off of light wave from boundary. refraction on other hand is bending of the path of a light wave.

We were to calculate the reflectivity at the boundary,

reflectivity at the boundary can be calculated using the expression below

Reflectivity= (n₂ - n₁)² /(n₂ + n₁ ) ²

where

n₁= Indices of refraction at first grain= 1.545

n₂= Indices of refraction at second grain=

1.555

(1.555 - 1.545)² / (1.555 - 1.545)²

=(0.01)²/ (3.1)²

= 0.0001/ 9.61

= 1.04*10⁻⁵

Hence, the reflectivity at the boundary if the indices of refraction for the two grains are 1.545 and 1.555 in the direction of light propagation is 1.04*10⁻⁵

A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
that you must drop into the bucket. Where should the bucket be when you drop the mass (so it is moving
purely vertically when it lands in the bucket) if your goal is to:
(a) Maximize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.
(b) Minimize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.

Answers

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = [tex]\sqrt{2y_{o}/g}[/tex]

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = [tex]\sqrt{k/m}[/tex]

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - [tex]\sqrt{k/3}[/tex] [tex]\sqrt{2y_{o}/g}[/tex]

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

A stiuden goes for a bike ride ayt 20 meters

Answers

What did you need help with? If you needed help correcting the sentence it is, A student goes for a bike ride at 20 meters.

what are the importance of informal education?​

Answers

Answer:

Informal education is important because it can help individual to learn how to react and control situations.

It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.

Explanation:

Informal education is a type of education that is learned from different life experiences, happenings outside a structured curriculum.

Informal education is important because it can help individual to learn how to react and control situations.

It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.

This 200-kg horse ran the track at a speed of 5 m/s. What was the average kinetic energy?

Answers

Answer:

2500 J

Explanation:

The average kinetic energy can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 200 \times {5}^{2} \\ = 100 \times 25[/tex]

We have the final answer is

2500 J

Hope this helps you

A certain electrical circuit contains a battery with three cells, wires, and a light bulb. Which of the following would cause the bulb to shine less brightly?


A. decrease the resistance of the circuit


B. increase the resistance of the circuit

Answers

The answer is B increase the resistance of the circuit

Physical science deals with the ... of matter

Answers

Physical science, by definition....is the sciences concerned with the study of inanimate natural objects, including physics, chemistry, astronomy, and related subjects.

So it depends on what subject you are in.

Consider a rigid 3-mass system ( with origin at the leftmost mass 1 kg) which can rotate about an axis perpendicular to the system. The mass are separated by rods of length 5m, so that the entire length is 10m. Find the x-coordinate of the center of the mass for the three-mass system with respect to the origin. Treat mass as particles. Answer in unit of m.
now consider a rotation axis perpendicular to the system and passing through the point Xo at distance 3.8 m from the leftmost mass 1kg. find the moment of inertia of the 3-mass system about the new axis. Answer in unit of kg.m^2

Answers

Answer:

1)    x_{cm} = 5 m ,  2)  I = 168.32 kg m²

Explanation:

1) An important concept of center of mass is

          [tex]x_{cm} = \frac{1}{M} \sum x_{i} m_{i}[/tex]

where M is the total mass of the system

Let's apply this equation to our case, suppose that all masses are equal and are worth 1 kg

         [tex]x_{cm}[/tex] = ⅓ (1 0 + 1 5 + 1 10)

         x_{cm} = 5 m

2) In this para indicates that there is an axis of rotation at the point xo = 3.8 m and they ask to calculate the moment inertia.

Let's use the parallel axes theorem

          I =    I_{cm} + M D

where    I_{cm}  is the moment of inertia with respect to the center of mass, D the distance between the two axes of rotation and M the total mass of the system

Let's look for the moment of inertia of the center of mass

         [tex]I_{cm}[/tex] = 1  0 + 1  5² + 1  10²

         I_{cm} = 125 kg m²

the total moment of inertia is

        I = 125 + 3 3.8²

        I = 168.32 kg m²

The moment of inertia of the 3-mass system about the new axis is 54.32 kgm/s².

We have three masses each of mass = 1kg such that they are in line with mass m at origin, m at 5m and m at 10m

(a) The center of mass:

[tex]X=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X =\frac{ 1*0+1*5+1*10}{1+1+1}\\ \\ X = 5m[/tex]

Hence the center of mass of the system is at x = 5m.

(b) The moment of inertia about the axis passing through x = 3.8m

from the parallel axis theorem:

[tex]I = I_{cm} + Md^2[/tex]

where, [tex]I_{cm}[/tex] is the moment of inertial along an axis passing through the center of mass of the system, M is the total mass of the system and d is the distance of the given axis from center of mass.

M = 3kg

d = 5 - 3.8 = 1.2m

[tex]I_{cm}=1*5^2+1*0+1*5^2\\\\ I_{cm}=50 kgm/s^2[/tex]

Md² = 3×(1.2)²

Md² = 4.32 kgm/s²

I = 50 + 4.32

I = 54.32 kgm/s² is the moment of inertial about the given axis.  

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help please Derive an equation
Ta=1.44T1/2​​

Answers

Explanation:

To derive an equation you must indicate the variable you want to solve for.

Here we have tension of an object A and Tension 1.

Two variables or unknown are given hence we cannot derive any other equations.

Which lists three organic biological molecules?
O carbohydrates, salts, metals
O salts, proteins, minerals,
O proteins, lipids, carbohydrates
O lipids, metals, minerals

Answers

Answer:

B

Explanation:

I'm learning it in science.

Answer:

its not b i just took the test and b was wrong

Explanation:

Pelican hunts fish in the water as th pelican the waters surface why must it aim for the fish in a slightly different place than where the fish appears to be located​

Answers

The Pelican  must aim differently than where the fish appears to be because light is refracted from air to water.

Refraction is the change in the direction of a wave at the interface between two media. The effects of refraction are easily seen when light is travelling from air to water hence objects are not really where they appear to be when viewed above the water surface.

Hence, the Pelican  must aim differently than where the fish appears to be because light is refracted from air to water.

Learn more about refraction: https://brainly.com/question/11155928

The motion of a nightingale's wingtips can be modeled as simple harmonic motion. In one study, the tips of a bird's wings were found to move up and down with an amplitude of 8.0 cm and a period of 0.80 s.
Part A: What is the wingtips' maximum speed?
Part B: What is the wingtips' maximum acceleration?

Answers

Answer:

PART A: Maximum speed = 0.314 m/s

PART B: Maximum acceleration = 1.23 m/s²

Explanation:

A simple Harmonic motion is a repetitive motion through an equilibrium point.

Amplitude = 8.0cm = 8/100 = 0.08m (highest displacement)

period (T) = 0.80s

A) maximum speed [tex](V_{max)[/tex]

[tex]V_{max} = 2\pi fA\\where:\\A = Amplitude = 0.08m\\f = frequency = \frac{1}{period(T)} = \frac{1}{0.8} = 1.25 Hz\\\therefore 2\pi fA = 2\pi \times 1.25 \times 0.08\\= 0.314\ m/s[/tex]

B) maximum acceleration [tex](a_{max})[/tex]

[tex]a_{max} = (2\pi f)^2A\\where:\\f = 1.25Hz\\A = 0.08m\\a_{max} = (2\pi \times 1.25)^2 \times 0.08\\= 1.23\ m/s^2[/tex]

Which would best help a student determine the net force acting on a rollercoaster car as it moves from one point on its track to another?

Answers

Answer:

you can show them a vid

Explanation:

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 59.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.852 m/s2. Calculate her mass.

Answers

Answer:

The value is  [tex]m = 69.24 \ kg[/tex]

Explanation:

From the question we are told that

   The value of the external force is  [tex]F = 59.0 \ N[/tex]

    The magnitude of the astronaut's acceleration is  [tex]a = 0.852 \ m/s[/tex]

Generally Newton's Second Law of Motion from the mass of the astronauts is mathematically represented as

            [tex]m = \frac{F}{a}[/tex]

=>         [tex]m = \frac{59 }{0.852 }[/tex]

=>         [tex]m = 69.24 \ kg[/tex]  

g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket, and the drive

Answers

Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

Resonance frequency ​

Answers

Answer:

resonance frequency is the frequency when capacitive reactance and inductive reactance become equal and opposite to each other and all impedence is given by resistance.

Explanation:

              f=1/2[tex]\pi[/tex][tex]\sqrt{LC}[/tex]

A convex mirror has a focal length of -12 cm. A lightbulb with
a diameter of 6.0 cm is placed 60.0 cm in front of the mirror.
Locate the image of the lightbulb. What is its diameter?​

Answers

Answer:

PLEASE MARK AS BRAINLIEST!!

Explanation:

ANSWER IS IN THE IMG BELOW

It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway (D-Hall), a distance of 47.0 meters. The second hallway (C-Hall) is filled with students, and she covers its 63.0 m length quickly. The final hallway (B- Hall) is empty, and Susie sprints its 76.0 m length. How fast does Susie need to go to make it to class on time (Hint: Calculate the total distance. Then calculate her total average speed rounded to the nearest tenths in meters/seconds.)?

Answers

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m and 0.6 m, respectively. The wavelength is 630 nm. Suppose that the radius of the aperture can be increased from 0.5 mm, determine: (a) The first two radii when the center intensity at the reception screen is maximum. (b) The first two radii when the center intensity is minimum.

Answers

Explanation:

The width of the central maximum is given by

W = 2 λ L / a

where W is the width of the central maximum

λ is the wavelength of the light used.

L is the distance between the aperture and screen

a is the width of the slit or aperture

So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.



At a sports event, the car starts from rest. in 5.0 s its acceleration is 5.0 m/s2.
Calculate the distance travelled by car.

Answers

Answer:

62.5m

Explanation:

Given parameters:

Initial velocity  = 0m/s

Time  = 5s

Acceleration  = 5m/s²

Unknown:

Distance traveled  = ?

Solution:

To solve this problem, we use the motion equation given below:

S = ut + [tex]\frac{1}{2}[/tex] at²

S is the distance traveled

u is the initial velocity

a is the acceleration

t is the time taken

Now, insert the parameters and solve;

 S =( 0 x 5) +( [tex]\frac{1}{2}[/tex] x 5 x 5²) = 62.5m

An automobile travels to the right, with the center Aof the wheel moving at a constant speed of 48 mi/h. If the diameter of a wheel is 22in.,determine the velocities of points B, C, D, and E on the rim of the wheelassuming there is no slipping. [Hint: at any given instant the velocity of point C, which is in contact with the ground must be zero]. Please introduce axes and give your answers in vector form.

Answers

Answer:

Explanation:

From the question we are told that

Constant speed =48mile/hr

Diameter of a wheel = 22inch therefore [tex]r=\frac{22}{2} =>11[/tex]

Generally Convert from mile/hr to inches/sec

The length in inches is equal to the miles multiplied by 63,360.

an hour is 3600seconds

[tex]\frac{48*63360}{3600}[/tex]

48miles/h = 849.8 inch/sec

[tex]V_a =844.8 inch/sec[/tex]

therefore

[tex]\omega= \frac{v}{r}[/tex]

[tex]\omega= \frac{844,8}{11}[/tex] =>[tex]76.8 sec ^-^1[/tex]

a)Considering the velocity of Vb  in inches per seconds

Generally the formula is stated as

[tex]V_b= V_a + V_b_/_a[/tex]

[tex]V_b = 844.8 + \omega r[/tex]

[tex]V_b= 1639.6in/s[/tex]

b)Considering the velocity of Vc  in inches per seconds

Since the tire doesn't slip as earlier stated in the question

Therefore [tex]V_c = 844.8 -w(r) =0[/tex]

c)Considering the velocity of Ve  in inches per seconds

Generally the formula is stated as

[tex]V_e=V_a + V_e_/_a[/tex]

[tex]V_e = 844.8 \uparrow \theta -844.8 \uparrow = 844.8\sqrt{2}[/tex]

Expressing result with vector

[tex]V_e =844.54 + 20.85j[/tex]

d)Considering the velocity of Vd  in inches per seconds

Generally the formula is stated as

[tex]V_d= V_a + V_d_/_a[/tex]

Mathematically

[tex]V_d =844.8 \uparrow +( 844.8\frac{\sqrt{3} }{2} \uparrow + \frac{844.8}{2} \uparrow)[/tex]

[tex]V_d= (1576.4 \uparrow + 422.4\uparrow)[/tex]

[tex]V_d= 1632.028in/s[/tex]

Numerical Problems
A bus covers a distance of 600 metres in 30 seconds. Calculate the speed of
the bus.​

Answers

Answer:

20 m/s.

Explanation:

From the question given above, the following data were obtained:

Distance travelled = 600 m

Time taken = 30 s

Speed =?

Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can easily calculate the speed of the bus as follow:

Distance travelled = 600 m

Time taken = 30 s

Speed =?

Speed = Distance /time

Speed = 600 / 30

Speed = 20 m/s

Therefore, the speed of the bus is 20 m/s.

A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of the circle so that the cord remains taut and the ball’s path remains circular.

Answers

Answer:

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

Explanation:

By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:

[tex]K =U_{g}[/tex] (1)

Where:

[tex]K[/tex] - Translational kinetic energy, measured in joules.

[tex]U_{g}[/tex] - Gravitational potential energy, measured in joules.

By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h[/tex]

[tex]v = \sqrt{2\cdot g\cdot h}[/tex] (2)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Initial speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]h[/tex] - Maximum height of the ball, measured in meters.

If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]h = 5\,m[/tex], then the initial speed of the ball is:

[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]

[tex]v\approx 9.903\,\frac{m}{s}[/tex]

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

A phosphodiester bond is used to:

A. Join glycerol to fatty acids
B. Join two nucleotides into a polynucleotide
C. Join two glucose molecules
D. Join two amino acids into a polypeptide

Answers

Answer:

A. Join glycerol to fatty acids

Explanation:

I majored in Physics.

I think it’s A I believe sorry if I get it wrong

I dont understand physics.

Answers

Answer:

same but from what i know is the newtons thingy

Explanation:

hope that helps!

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