An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by the force on the box?

Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor

The impulse given to the ball by the floor is 0.2865 kg.m/s.

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.

The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s

The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s

Substitute the values into the expression, we get

Impulse = m(v- u)

Impulse=0.5 x (4.852- 5.425 )

Impulse = - 0.2865 kg.m/s

Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.

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The proper connection between kinetic energy and speed is depicted in Graph D.

The kinetic energy (KE) is defined as one-half of the mass times multiplied by the square of velocity.

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

where,

KE is the kinetic energy

m is the mass of each molecule

(V) is the  velocity

As the square of the velocity is exactly proportional to kinetic energy. Consequently, the relationship between velocity and kinetic energy must be parabolic.

The proper connection between kinetic energy and speed is depicted in Graph D. Graph showing the link between kinetic energy on the y-axis and speed on the x-axis.

On the y-axis, a semi-curve line begins above the origin and ascends as it accelerates.

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Answer:

i think the answer is 20.......

The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.

Three equation of motion are:-

Where v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.

In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.

Applying equation 1 to find the initial speed of plane

v = u + at

0 = u + (-6.34 × 5400)   {v=0 as plane will stop after 5400 sec}

u =  6.34 × 5400

u = 34236 m/s

Initial velocity of plane is 34236 m/s

Applying equation 2 to find the displacement of plane in that time period

s = ut + (1/2)at²

s = ( 34236 × 5400 )  - ( (1/2) × 6.34 × 5400² )

s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )

s = 5400 × ( 34236 - 17118 )

s = 5400 × 17118 metres

s = 5.4 × 17118 Km

s = 92437.2 Km

Distance travelled by plane is 92437.2 Km

So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.

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The merits of and problems with the manager’s statement are explained below.

Extension of the social identity of the individual in groups such as working in company or studying in college is called as the stereotyping in organizations.

Merits of Stereotypes in organizations:

Demerits of Stereotypes:

Thus, the merits and demerits of stereotypes in organizations are explained above.

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Answer:

Explanation:

Hope I give all correct answer please mark as brainlest answer

The rod is stretched due to the force of gravitational attraction.

The magnitude of the deformation force on the rod is  -GMm×((1/r² + 1/(L+r)²)

The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.

The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.

F = GMm/R²

Given are two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M.  Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects.

Let the mass closer to the planet is A and the other is B

FA = Force on planet A = -GMm/r²

FB = Force on planet B = -GMm/(L+r)²

Net Force on the rod is given by the addition of the individual forces.

Fnet= FA + FB = -GMm/r² -GMm/(L+r)²

Fnet = -GMm×((1/r² + 1/(L+r)²)

Thus,  the magnitude of the deformation force on the rod is derived above.

A experience more force than B, so A will stretch out from B. Hence the force is stretching the rod.

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Answer:

the 1 gram wasp

Explanation:

To start off with this problem, write down every piece of information and do neccessary conversions.

mass of bee = 2 grams = 0.002 kg

speed of bee = 1 m/s

mass of wasp = 1 gram = 0.001 kg

speed of wasp = 2 m/s

now, we will use the kinetic energy formula and compare the answers

KE BEE = 0.5 (0.002 kg)(1 m/s)^2 = 0.001 Joules

KE WASP = 0.5(0.001 kg)(2 m/s)^2 = 0.002 Joules

0.002 J > 0.001 J

The estimated density of the golf ball is  700 kg/m³

Density is defined as Mass per unit Volume.

In displacement method,  

First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.

Then calculate the density by dividing the mass by the volume.

i.e.  D = m/V

Given,  Density of water is 1.0 g/cm³

Using displacement method  , The estimated volume of golf ball is 100 cm³  and estimated mass is 7g

Then ,

Density =  10 cm³ / 7 g= 0.07 g/ cm³ = 700 kg/m³

So the estimated density of golf ball is 700 kg/m³

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C. The Densities are equal.

Density is mass per unit volume or mass of a unit volume of a material substance.

If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]

Therefore,

[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]

Hence, D1 = D2

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Wave I stands for a wave with an equal frequency as wave II. Option c is correct.

Frequency is defined as the number of repititions of a wave occurring waves in 1 second. Its unit is Hz.

Frequency is given by the formula as,

[tex]\rm f = \frac{1}{t}[/tex]

Where,

f is the frequency

t is the period of the wave

From the digrame it is observed that both the wave has the same period. So that they will have the same frequency.

Hence option c is correct.

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The total time required is 1 minute and 0.3 seconds.

Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.

Here, the time taken by the person is 0.1 seconds.

Total distance covered by the wave = Circumference of the circle

So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m

As the distance between each of the people is 1 m.

So, the number of personal interactions is 618.

Time taken by each person is 0.1 seconds.

So, total time, t = 0.1 × 618 = 61.8 seconds.

So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.

Hence, the total time required is 1 minute and 0.3 seconds.

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Answer:

He has a hangover.

Explanation:

Just something I know.

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

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Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by

[tex]x = v_i \cos(\theta_i) t[/tex]

[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]

and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are

[tex]v_x = v_i \cos(\theta_i)[/tex]

[tex]v_y = v_i \sin(\theta_i) - gt[/tex]

The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when

[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]

which means

[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]

At the same time, the ball will have traveled half its horizontal range, so

[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]

Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :

[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]

Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)

[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]

Now,

[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]

so it follows that (d)

[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]

Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is

[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]

Finally, in the work leading up to part (e), we showed the time to maximum height is

[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]

but this is just half the total time the ball spends in the air. The total airtime is then

[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]

and the ball is in the air over the interval (a)

[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

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The distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.

The intensity of light is given as power emitted by the light by unit area.

I = P/A

I = P/L²

I₁L₁² = I₂L₂²

L₂² = I₁L₁²/I₂

L₂² = (3 x 1.5²)/(9)

L₂² = 0.75

L₂ = 0.87 cm

Thus, the distance x equal to 0.87 cm when the intensity of the light decreases by 9 times.

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Option C. The measurement with the most significant number is 6.600 cm.

Significant numbers are numbers that have significance or meaning and give more precise details about the value of the entire numbers.

Thus, the measurement with the most significant number is 6.600 cm.

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Answer:

6800

Explanation:

100 x 0.68=6800

Answer: 68

Explanation:

Based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.

From the standpoint of  brain lateralization, differences do exist such as based on experience such as handedness or specific skills such as playing a guitar.

Note that Left-handers are said to have reduced or little lateralized brains, which tells us that the two halves of the brain are little different than as seen in the right-handers.

Therefore, I can say that based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.

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The gauge pressure at bottom of vaccine solution will be 16 kPa

Positive pressure is another name for gauge pressure. When a system's internal pressure exceeds that of its surroundings, it is said to be under positive pressure. Any leak that develops in the positively pressured system will therefore escape into the outside world. In contrast, a negative pressure chamber draws air into it.

Given As seen in the illustration, a syringe is held vertically. The container carries a 3 cm tall column of vaccine solution and has an open inner diameter of 1 cm. The needle contains a 2 cm column of vaccine solution and has an open inner diameter of 0.5 mm. At the needle's open end, the solution is exposed to the air. The vaccination solution has a density of 1200 kg/m3.

We have to find the gauge pressure at bottom of vaccine solution

Since the 5N force is applied to vaccine solution the pressure exerted will be much more

Hence the gauge pressure at bottom of vaccine solution will be 16 kPa

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Answer:

The scientific method

Explanation:

Answer:

Discrimination is the worst thing in the world you can't even do a thing so you will do such a physical thing or do a mathematics problem ok done it's ok

According to Piaget, that means that she has made the transition from the sensorimotor to preoperational stage

The first of Piaget's four stages of cognitive development is the sensorimotor stage.

A child's understanding that the outside world exists apart from them is what distinguishes it.

Within Piaget's stages of development, the kid will advance to the following stage after they have completely grasped this.

"Charlene has just begun to be able to form a mental representation of an object that is not visibly present.

According to Piaget, that means that she has made the transition from the sensorimotor to preoperational stage

Only recently has Charlene been able to create a mental image of something that is not physically present.

That indicates that she has moved from the sensorimotor to the preoperational level, in accordance with Piaget.

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Here is your answer mate,

Question,

[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]

Answer,

[tex][/tex]

Solution,

[tex][/tex]

Given,

[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]

[tex][/tex]

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm

[tex][/tex]

Now

POWER =

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]

POWER

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]

[tex]Therfore\: your \: answer\: is\: 6.25[/tex]

[tex][/tex]

Check this,

[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]

[tex][/tex]

Have a good day

Answer:

λ = c / f     or    f = c / λ

f = 3.0E8 / 4.0E-7 = .75E15 / sec = 7.5E14 / sec = 7.5 X 10^14 /sec

The distance of the Image will be -33.75 cm

A concave mirror has an inward-curving reflecting surface that faces away from the light source. Unlike convex mirrors, a concave mirror's image forms a variety of images based on the object's proximity to the mirror.

Given that, an object placed 27 cm from a concave mirror having the focal length of 15 cm

We have to find distance of the Image

Using Mirror Formula:

1/f = 1/v + 1/u

Where,

f = focal length

v =  Image distance from the mirror

u = object distance from the mirror (concave)

Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.

1/(-15) = 1/v + 1/(-27)

1/(-15) = 1/v - (1/27)

1/v = -0.029

v = -33.75 cm

Therefore the distance of the Image will be -33.75 cm

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Answer:

r= 1.09×10^-4

Explanation:

Given

V=speed=4.0×10^5 m/s

B= magnetic field= 0.040 T

©=angle= 35°

m= mass of electron= 9.11×10^-31

q= charge of electron= 1.60×10^-19

solution

qv×B= mv²/r

qvBsin©=mv²/r

qBsin©=mv/r

r=mv/qBsin©

r=9.11×10^-31× 4.0×10^5/1.064×10^-19×0.04T(sin35°)

r= 1.09×10^-4 m

a) r = 1.09 * [tex]10^{-4}[/tex] m

b) Distance travelled : 6.845 *  [tex]10^{-4}[/tex] m    

An electron is a  stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids.

given

charge of electron : 1.6 * [tex]10^{-19}[/tex] C

mass of electron = 9.11 * [tex]10^{-31}[/tex] kg

v = 4.0 x 10 m/s

B =  0.040 T

theta = 35 degrees

since ,

force in magnetic field on electron = centripetal force

a) q(v*B) = m [tex]v^{2}[/tex] / r

q v B sin(theta) =  m [tex]v^{2}[/tex] / r

r =m v /q B sin(theta)

r =  9.11 * [tex]10^{-31}[/tex] * 4.0 x [tex]10^{5}[/tex]/ 1.6 * [tex]10^{-19}[/tex] sin (35)

r = 1.09 * [tex]10^{-4}[/tex] m

b)  far forward will the electron have moved after completing one circle will be equal to circumference of the circle = 2πr

 = 2 * 3.14 * 1.09 * [tex]10^{-4}[/tex] m  = 6.845 *  [tex]10^{-4}[/tex] m  

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10¹¹ is the approximate number of atoms in a bacterium.

The atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu).

Given,

Mass of a bacterium atom = 10⁻¹⁵ kg.

Mass of a hydrogen atom  = 10⁻²⁷ kg.

From the above observation ,

The average mass of an atom of the bacterium is ten times the mass of a hydrogen atom.

Atomic mass 1 bacterium atom = 10 x mass of hydrogen atom

                                                  = 10 x 10⁻²⁷ kg.

                                                  = 10⁻²⁶ kg.

Thus,

The number of atoms in a bacterium

=  [tex]\frac{Total mass}{Atomicmass of 1 bacterium}[/tex]

= [tex]\frac{10^{-15} }{10^{-26} }[/tex]

=  [tex]10^{11}[/tex]

The approximate number of atoms in a bacterium  is [tex]10^{11}[/tex].

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