A2 kg object moves at a constant 5 m/s across a level surface. Determine its kinetic energy.

Answer:

3.44× 10⁻¹⁹Joules

Explanation:

Energy of the wavelength is expressed using the formula:

E = hc/λ

h is the Planck constant

c is the velocity of light

λ is the wavelength

Given

h = 6.63 × 10^-34 m² kg / s

c = 3×10⁸ m/s

λ = 579nm = 579 × 10⁻⁹m

λ = 5.79× 10⁻⁷m

Substitute the given values into the formula

E = hc/λ

E = (6.63 × 10⁻³⁴× 3×10⁸)/5.79× 10⁻⁷

E = 19.89× 10⁻³⁴⁺⁸/5.79× 10⁻⁷

E = 19.89× 10⁻²⁶/5.79× 10⁻⁷

E = 3.44× 10⁻²⁶⁺⁷

E = 3.44× 10⁻¹⁹Joules

Hence the energy of this wavelength of light is 3.44× 10⁻¹⁹Joules

A mechanical wave is a disturbance in matter that transfers energy through the matter. The matter through which a mechanical wave travels is called the medium

or media in plural.

Mechanical wave are waves that travels through a medium or matter.

A wave is a disturbance that travels or propagates from the place where it was created.

Waves transfer energy from one place to another, but they do not necessarily transfer any mass.

There are two types of waves;

mechanical wave and electromagnetic waves

electromagnetic waves are waves that that do not need any means of medium for propagation.

Examples of electromagnetic wave includes; light wave , radio wave e.t.c

Mechanical waves are waves that need medium for propagation. Example of mechanical waves includes ; sound wave , water wave e.t.c

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Answer:

First Blank. 30,000

Second Blank. 12,000

Third Blank. 20.0

Explanation:

Answer:

Distance and time.

Explanation:

Speed=Distance/time

The two factors which we need in order to calculate the speed of an object are the distance covered by the object and the time taken to cover that distance.

Speed is the rate of change of position of an object in any direction. Speed is a scalar quantity as it has only magnitude and no direction. It is measured as the ratio of the distance covered by an object to the time taken in which the distance was covered by that object.

Speed has the dimension of distance covered by the time taken. Thus, the SI unit of speed is the combination of the basic units of distance and the basic unit of Time. Thus, the SI unit of speed is meter per second (m/s).

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Answer:

3566.26 Watts

Explanation:

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

Mass (m) of crate = 151 kg

Coefficient of kinetic friction (μ) = 0.399

Velocity (v) = 6.04 m/s

Power (P) =?

Next, we shall determine the normal reaction. This can be obtained as follow:

Mass (m) of crate = 151 kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) = mg

Normal reaction (R) = 151 × 9.8

Normal reaction (R) = 1479.8 N

Next, we shall determine the force applied to move the crate. This can be obtained as follow:

Coefficient of kinetic friction (μ) = 0.399

Normal reaction (R) = 1479.8 N

Force (F) =?

F = μR

F = 0.399 × 1479.8

F = 590.44 N

Finally, we shall determine the power used to move the crate as follow:

Velocity (v) = 6.04 m/s

Force (F) applied = 590.44 N

Power (P) =?

P = F × v

P = 590.44 × 6.04

P = 3566.26 Watts

Therefore, the power used to move the crate is 3566.26 Watts.

momentum (m)=40

velocity (v)=2

now,

momentum (m)=m×v

40=m×2

m=40÷2

m=20kg

Answer:

P=mv

1x450,200=m40

m= 36,255kg

Answer:

11.25m

Explanation:

[tex]s = ut + \frac{1}{2} a {t}^{2} \\ = 0 \times 1.5 + \frac{1}{2} \times 10 \times {1.5}^{2} \\ = 5 \times 2.25 \\ = 11.25[/tex]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

[tex]E_{A}=E_{B}[/tex]

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

[tex]E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

[tex]E_{B}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]

Therefore we will have the following equation:

[tex](6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s][/tex]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

[tex]KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J][/tex]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

[tex]E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s][/tex]

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 141 × 13

We have the final answer as

Hope this helps you

The satellite is moved to a new circular orbit that is 2R from the center of the moon, then the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.

It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another.

As given in the problem A satellite of mass m orbits a moon of mass M in a uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of the moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon.

The gravitational field strength is inversely proportional to the square of the distance from the center of the planet.

Thus, the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.

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Answer:

Explanation:

F = m*a = (0.5)*15 = 7.5 N

Answer:

The answer is temperature lol

Explanation:

:)

Answer:

P = 1500 Watt

Explanation:

Mechanical Work and Power

Mechanical work is the amount of energy transferred by a force.

Being F the magnitude of the force vector and s the distance, the work is calculated as:

W=F.s

Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.

The power can be calculated as:

[tex]\displaystyle P=\frac {W}{t}[/tex]

Where W is the work and t is the time.

The force to be considered is the weight of the mass of m=100 kg, [tex]g= 10\ m/s^2[/tex]:

F = 100 * 10 = 1000 N

The distance is s=3 m, thus the work done by the weight lifter is:

W = 1000 N * 3 m

W = 3000 J

Finally, the power is:

[tex]\displaystyle P=\frac {3000}{2}[/tex]

P = 1500 Watt

For every complete revolution the boy makes around the center of the carousel, he travels a distance of 2π (5.0 m) = 10π m, which gives a linear speed of

v = (10π m) / (8.0 s) ≈ 3.927 m/s

Then his centripetal acceleration would be

a = v ² / (5.0 m) ≈ 3.084 m/s²

so that the centripetal force exerted on him has magnitude

F = (45 kg) a ≈ 138.791 N ≈ 140 N

(rounded to 2 significant digits)

Answer

I feel the answer is C because it could cause mental and physical trauma

Explanation:

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = [tex]H_{0}[/tex] x d

 [tex]H_{0}[/tex] = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x [tex]10^{19}[/tex] km      

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( [tex]H_{0}[/tex] xd)

t = 1/[tex]H_{0}[/tex]

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from [tex]H_{0}[/tex] = 73 km/sec/Mpc to [tex]H_{0}[/tex] = 700 km/sec/Mpc

then the age of universe will be:

t = 1/[tex]H_{0}[/tex]

first convert the units of new [tex]H_{0}[/tex] into 1/s

[tex]H_{0}[/tex] = (700) x (/3.08 x [tex]10^{19}[/tex] )

[tex]H_{0}[/tex] = 227.27 x[tex]10^{-19}[/tex]  = 2.27 x [tex]10^{-21}[/tex] 1/s

So,

Age of universe will be:

t = 1/[tex]H_{0}[/tex] = 1/2.27x[tex]10^{-21}[/tex] 1/s

t = 2.27 x [tex]10^{21}[/tex] s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Given value:

Hubble Constant,

We know,

By substituting the value of "Mpc" in Hubble constant, we get

→ [tex]H_o = \frac{700}{3.086\times 10^{19}}[/tex]

        [tex]= 227\times 10^{-19} \ 1/s[/tex]

        [tex]= 2.27\times 10^{-21} \ 1/s[/tex]

The Hubble's time will be:

→ [tex]H_o = \frac{1}{t}[/tex]

or,

→ [tex]t = \frac{1}{H_o}[/tex]

     [tex]= \frac{1}{2.27\times 10^{-21}}[/tex]

     [tex]= 4.4\times 10^{20} \ seconds[/tex]

Thus the above approach is right.

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Answer: 1 How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-.                                                                                                                                     2  34 °C? 4. How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of ... A saturated solution of KClO3 was made with 300 g of H2O at. 34 °C.

Explanation:

Answer:

a is the correct answer hope it will help you

Answer:

Answer:

9.23m

Explanation:

Max height = u²sin²theta/2g

u is the speed = 34.7m/s

theta is the angle of elevation = 33°

g is the acceleration due to gravity = 9.8m/s²

Substitute into the formula

Max height = 24.7²sin²33/2(9.8)

Max height = 610.09sin²33/2(9.8)

Max height = 610.09(0.29663)/19.6

Max height = 180.97/19.6

Max Height = 9.23m

Hence the max height the football reaches during flight is 9.23m

Answer:

The weight of the combined system is 1989.8 N.

Explanation:

The weight, W, of an object can be expressed as;

W = mg

where m is the mass of the object, and g is the acceleration due to gravity.

Weight of the combined system = weight of tank + weight of water

mass of the plastic tank = 3 kg

weight of the plastic tank = m x g

                               = 3 x 9.8

                               = 29.4 N

Weight of the plastic tank is 29.4 N

But,

density = [tex]\frac{mass}{volume}[/tex]

mass = density x volume

volume of water = 0.2 [tex]m^{3}[/tex], density of water = 1000 kg/[tex]m^{3}[/tex].

mass  = 1000 x 0.2

         = 200

mass of water = 200 kg

weight of water = 200 x 9.8

                   = 1960 N

Thus,

combined weight of the system = 29.4 + 1960

                        = 1989.8 N

The weight of the combined system is 1989.8 N.

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

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

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

[tex]v=\sqrt{2\times 9.8\times 15}[/tex]

[tex]v=\sqrt{294}[/tex]

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

Answer:

Explanation:

An object following a circular path can be covering the same distance along the circle's circumference with every passing minute, giving it a constant speed. Since a change in either speed OR direction means a change in velocity, the object's velocity is not constant.

velocity is a vector so therefore direction affects it being constant.

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.  

Velocity is a vector so therefore direction affects it being constant.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time

An object following a circular path can be covering the same distance along the circle's circumference with every passing minute, giving it a constant speed. Since a change in either speed OR direction means a change in velocity, the object's velocity is not constant. An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.  

Velocity is a vector so therefore direction affects it being constant.

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Answer:

The potential at the center of square due to four identical charges each at the corner is 5.09 q x [tex]10^1^0[/tex] Volts, where q is the charge.

Explanation:

From the question it is given that

side of square = a = 1 m

let the charge of each square is q

potential at center due to 1 charge is V = [tex]\frac{q}{4\pi \epsilon r}[/tex] , where

[tex]\epsilon[/tex] is electrical permittivity of space

r is the distance between charge and point.

since the charge is present at at the corner so the distance between charge and point is the half the length of diagonal of square.

⇒ distance between charge and point = r =[tex]\frac{a}{\sqrt{2}}[/tex] = [tex]\frac{1}{\sqrt{2} }[/tex] = 0.707 m

thus, V= [tex]\frac{q}{4\pi \epsilon r}[/tex]  , on substituting the respected values of r = 0.707m and [tex]\frac{1}{4\pi \epsilon}[/tex] = 9 x [tex]10^9[/tex] we get,

V = 1.272 q x [tex]10^1^0[/tex] V

thus potential due to all 4 charges is

V = 1.272 q x [tex]10^1^0[/tex] x 4 = 5.09 q x [tex]10^1^0[/tex] Volts