Suppose the width of your fist is 4.1 inches and the length of your arm is 35.4 inches. Based on these measurements, what will be the angular width (in degrees) of your fist held at arm’s length?

[tex]\sqrt{2}[/tex]Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = [tex]\sqrt{2}[/tex] V1

Since momentum = M V  the momentum increases by [tex]\sqrt{2}[/tex]

Answer:

[tex]t=15.3s[/tex]

Explanation:

Hello,

In this case, since the speed is defined in terms of the distance over time:

[tex]V=\frac{x}{t}[/tex]

We can easily solve for the time with the given speed and distance:

[tex]t=\frac{x}{V}=\frac{1500m}{98m/s}\\ \\t=15.3s[/tex]

Regards.

Explanation:

Sum of moments about point A:

∑τ = Iα

-mg (L/2) + Rb x = 0

-(5 kg) (10 m/s²) (0.75 m) + Rb (0.70 m) = 0

Rb = 53.6 N

Sum of forces in the y direction:

∑F = ma

Ra + Rb − mg = 0

Ra = mg − Rb

Ra = (5 kg) (10 m/s²) − 53.6 N

Ra = -3.6 N

Answer:

u = 88.54 m/s

Explanation:

Given that,

A discus thrower achieves a high throw of 100 m.

Angle of projection is 30°

We need to find the initial speed of the discuss.​ It is a cse of projectile motion. The maximum height reached by the discus is given by :

[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]

u is the initial speed of the discus

So,

[tex]u^2=\dfrac{2\times 9.8\times 100}{\sin^2(30)}\\\\u=\sqrt{7840}\\\\u=88.54\ m/s[/tex]

So, the initial speed of the discus is 88.54 m/s.

Explanation:

Average speed = distance / time

|v| = (7 km + 2 km) / (2 hr + 1 hr)

|v| = 3 km/hr

Average velocity = displacement / time

v = (7 km east + 2 km east) / (2 hr + 1 hr)

v = 3 km/hr east

Answer:

0.8216c

Explanation:

Using the relationship

L' = L√(1 - v²/c²)

where

L = original length,

L' = observed length,

v = velocity,

c =speed.

L'/L = 0.57

Then

0.57 = √(1 - v²/c²)

1 - v²/c² = 0.57² = 0.3249

v²/c² = 1 - 0.3249 = 0.6751

v² = 0.6751c²

v = c√0.6751 = 0.8216c

Explanation:

Answer:

The correct answer is B)

Cutting the frequency of the bottom waves in half will cause it to match top waves.

Explanation:

USATESTPREP

Cutting the frequency of the bottom waves will cause it to match the top waves. So the correct option is B.

A propagation of disturbance, from one point to another point is called a wave. Waves are either mechanical or non-mechanical. Electromagnetic waves are non-mechanical waves.

The mechanical waves cannot travel without a medium e.g sound waves. Non-mechanical waves do not require a medium to travel, this means that they can even travel through a vacuum.

Waves are of two types. Transverse waves and longitudinal waves.

If the direction of propagation of wave is perpendicular to the direction of movement of particles of the medium, it results in a transverse wave.

If the direction of propagation of wave is parallel to the direction of movement of particles in a medium, it results in a longitudinal wave.

Five properties of waves are amplitude, frequency, wavelength, time period and speed.

Maximum displacement from the mean position is the amplitude. The number of vibrations in a fixed point in unit time is the frequency. The distance between two identical points is the wavelength. Time taken by a wave to pass through a point is the time period and the distance travelled between particular points in a unit of time is the speed.

Therefore the correct option is B.

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Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

        ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

Answer:

a

  x-component             [tex]20 \ m[/tex]

y-component     [tex]500 - 452 = 48 \ m[/tex]

b

 Magnitude [tex]d = 52 \ m[/tex]

direction is  [tex]\theta = 67.4^o[/tex]

Explanation:

From the question we are told that

   The first  vertical distance is  [tex]y_1 = 500 \ m[/tex]

    The  first horizontal distance  is  [tex]x = 20 \ m[/tex]

    The  second vertical distance is  [tex]y_2 = 452 \ m[/tex]

Generally the displacement is  

x-component             [tex]20 \ m[/tex]

y-component     [tex]500 - 452 = 48 \ m[/tex]

Generally the helicopters displacement is mathematically evaluated as  

       [tex]d = \sqrt{ x- component ^2 + y- component ^2 }[/tex]

      [tex]d = \sqrt{ 20t ^2 + 48 ^2 }[/tex]

      [tex]d = 52 \ m[/tex]

The  direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as

         [tex]\theta = tan ^{-1}[ \frac{48}{20}][/tex]

=>       [tex]\theta = tan ^{-1}[ 2.4 ][/tex]

=>      [tex]\theta = 67.4^o[/tex]

Answer:

bonus questions:becz displacement work only with the help of any force or object and distance is the totl length of two points

Answer:

"However, there is no actual force applied."

"The passenger is merely continuing in a straight direction."

Explanation:

Am AP Phys student

Answer:

The  value is  [tex]v = -0.04 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass  of the block is  [tex]m = 2.0 \ kg[/tex]

   The  force constant  of the spring is  [tex]k = 590 \ N/m[/tex]

   The amplitude  is  [tex]A = + 0.080[/tex]

   The  time consider is  [tex]t = 0.10 \ s[/tex]

Generally the angular velocity of this  block is mathematically represented as

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

=>   [tex]w = \sqrt{\frac{590}{2} }[/tex]

=>   [tex]w = 17.18 \ rad/s[/tex]

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         [tex]v = -A w sin (w* t )[/tex]

=>       [tex]v = -0.080 * 17.18 sin (17.18* 0.10 )[/tex]

=>       [tex]v = -0.04 \ m/s[/tex]

The velocity of the block at the given time is -0.04 m/s.

The angular speed of the block is determined by using the following wave equation;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{590}{2} } \\\\\omega = 17.176 \ rad/s[/tex]

The velocity of the block at the given time is calculated as follows;'

[tex]v = - A sin(\omega t)\\\\v =- 0.08 \times sin(17.176 \times 0.1)\\\\v = -0.04 \ m/s[/tex]

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Answer:

I = 6 mA

Explanation:

Given that,

Number of strands are 125

Resistance of each strand is 2.65 mΩ

The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A.

We need to find the current in each strand.

Total current is 0.75 A

Number of strands are 125

So, current in each strand :

[tex]I=\dfrac{0.75}{125}\\\\I=0.006\ A\\\\I=6\ mA[/tex]

So, 6 mA of current flows in each strand.

Answer:

For solar energy, I would show them how a magnifying glass works when exposed to the sun.

For wind energy, I would teach them how to make a paper windmill and explain how it works.

For the hydroelectric energy, I would have them make a plastic turbine and explain to them how to use it in rivers or streams.

For electromagnetic energy, I would tell them to rub a balloon until their hairs stand on end.

And for electricity, I would teach them how the other energy sources create electricity and what electricity works for in these times.

Explanation:

To explain something so complicated to a child is not as easy as it would be with a teenager or an adult.

To make the children learn about the forms of energy, I would use the nemotechnique rule, using short and easy-to-remember sentences and explaining with many examples about how to get each type of energy and its use, in addition to adding didactic, visual and auditory content, which are the most common types of learning in children.

Answer:

c. both have same energy

Explanation:

The complete question is

suppose you have two cans, one with milk, and the other with refried beans. The cans have essentially the same size, shape, and mass. If you release both cans at the same time, on a downhill ramp, which can has more energy at the bottom of the ramp? ignore friction and air resistance..

a. can with beans

b. can with milk

c. both have same energy

please explain your answer

Since both cans have the same size, shape, and mass, and they are released at the same height above the ramp, they'll possess the same amount of mechanical energy. This is because their mechanical energy, which is the combination of their potential and kinetic energy are both dependent on their mass. Also, having the same physical quantities like their size and shape means that they will experience the same environmental or physical factors, which will be balanced for both.

Answer:

La energía potencial elástica es la energía asociada con los materiales elásticos. Por ejemplo, un resorte al ser comprimido o elongado almacena energía potencial elástica y, al ser soltado, puede realizar trabajo sobre un objeto.

Para mantener el resorte comprimido o alargado una cierta longitud x, a partir de su largo natural, es necesario que, en este caso, la mano aplique una fuerza F_{M} sobre el resorte; esta fuerza es directamente proporcional a x.

Explanation:

ón conocida como ley de Hooke.

Para encontrar una expresión que describa la energía potencial asociada con la fuerza del resorte, se determina el trabajo que se requiere para comprimir el resorte desde su posición de equilibrio hasta cierta posición final arbitraria x. Debido a que la fuerza varía desde O hasta kx, se utiliza la fuerza promedio \frac{(F_{0}+F_{X})}{2}.

 \[ \bar{F}=\frac{0+K X}{2}=\frac{1}{2}kx \]

fuerza-sobre-un-resorte

Fuerza sobre un resorte. La fuerza para estirar un resorte aumenta linealmente con su elongación .

El trabajo realizado por la fuerza aplicada será: W=\bar{Fx}=\frac{1}{2}kx^{2}

El trabajo realizado se almacena en el resorte comprimido en forma de energía potencial elástica como:

 \[ \boxed{ Ep_{elas}=\frac{1}{2}kx^{2}} \]

Una vez que se ha comprimido o estirado el resorte respecto a su posición de equilibrio, la energía potencial elástica se puede considerar como la energía almacenada en el resorte deformado. Esta energía siempre es positiva en un objeto deformado al depender de x^{2}.

Por ejemplo, en la figura se observa que un resorte realiza trabajo sobre un bloque. El resorte que se encuentra sin deformar (a) cuando es empujado por un bloque de masa m, se comprime una distancia x (b). Cuando el bloque se suelta (c), partiendo del reposo, la energía potencial plástica almacenada en el sistema se transforma en energía cinética del bloque.

energia-potencial

Answer:

The  value is  [tex]a = - 4.45 m/s^2[/tex]

Explanation:

From the question we are told that  

       The  initial speed is  [tex]u = 28 \ m/s[/tex] at a distance of  [tex]s_1 = 0 \ m[/tex]

        The  final speed is  [tex]v = 0 \ m/s[/tex]    at a distance of  [tex]s_2 = 88 \ m[/tex]

Generally  from the  kinematic equation we have that

       [tex]v^2 = u^2 +2as[/tex]

=>   [tex]a = \frac{v^2 - u^2 }{ 2(s_2 - s_1 )}[/tex]

=>  [tex]a = \frac{0 - 28^2 }{ 2(88 - 0 )}[/tex]

=>   [tex]a = - 4.45 m/s^2[/tex]

The negative sign shows that it is decelerating

Answer:

ΔE = 59.75 A,

Explanation:

Titanium has 3 electrons in its last shell, as it is doubly ionized, it is left with a single electron in this shell, which is why it behaves like a hydrogen-type atom, consequently we can use Bohr's atomic theory

                  rₙ = a₀ /Z     n²

                 Eₙ = 1k e² / 2a₀ (Z² / n²)

Where a₀ is Bohrd's atomic radius so  = 0.529 núm

Let's find out what quantum number n has each orbit

rn = 13.25 A = 1.325 nm

for Titanium with atomic number 22

            n² = Z rₙ / a₀

           n = √ (22 (1.325 / 0.529))

           n = 7.4

since N is an entry we take

           n = 7

rn = 2.12 A = 0.212 nm

           n = √ (22 / 0.529) 0.212

           n = 3

With these values ​​we can calculate the energy of the transition from level ne = 7 to level no = 3

         ΔE = ka e2 Z2 / 2ao (1n02 - 1 / nf2)

          ΔE = 9 10⁹ 1.6 10⁻¹⁹ 22² (2 0.529 10⁻⁹) (1/3² - 1/7²)

          ΔED = 6.5875 10² (0.111 - 0.0204)

          ΔE = 59.75 A

let us be the Planck relation between energy and frequency

          E = h f

the frequency is related to the speed of light

           c = λ f

            f = c / λ

we substitute

           E = h c /y

           E = ΔE

           h c /λ = E

           λ  = 6.63 10-34 3 108 / 59.75

           λ= 3.01939 10⁻²⁴ m

          λ = 3.01939 10⁻²² cm

Answer:

Explanation:

Coefficient of static friction μs = .830216

Coefficient of kinetic friction μk = .326245

a ) The angle at which the box starts sliding depends upon coefficient of static friction . If θ be the required angle

tanθ = μs

tanθ = .830216

θ = 39.7°

b )

When the box starts sliding , kinetic friction will be acting on it .

frictional force on the box = μk mg cos 39.7

net force on the box

= mg sin39.7 -  μk mg cos 39.7

Applying Newton's law of motion

mg sin39.7 -  μk mg cos 39.7  = m a

a = g sin39.7 -  μk g cos 39.7

= 9.8 x sin 39.7 - .326245 x 9.8 x cos 39.7

= 6.26 - 2.46

= 3.8 m /s² .

Answer:

a) Pb= 200 PA

b).work done= -3600 joules

c).3600joules

D).the system works under isothermal condition so no heat was transferred

Explanation:

2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.

a). PbVb= PaVa

Pb= (PaVa)/VB

Pb= (600*3)/9

Pb= 1800/9

Pb= 200 PA

b). work done= n(Pb-Pa)(Vb-Va)

Work done= 2*(200-600)(9-3)

Work done= -600(6)

Work done=- 3600 Pam³

work done= -3600 joules

C). Change in internal energy I the work done on the system

= 3600joules

D).the system works under isothermal condition so no heat was transferred

Answer:

17,300 m

Explanation:

Using kinematic equations, first find the time it takes to land.

Δy = v₀ t + ½ at²

0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s or 68.5 s

The horizontal distance it moves in that time is:

Δx = v₀ t + ½ at²

Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²

Δx = 17,300 m

Alternatively, you can use the range equation:

R = v₀² sin(2θ) / g

R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)

R = 17,300 m

The distance a cannonball will land if it is fired on flat ground at 420 m/s at a 53.0° angle is 17,300 meters.

The complete movement of an object, regardless of direction, is referred to as distance. The amount of ground a thing travels from its starting point to its destination is also referred to as distance.

Given:

A cannonball is fired on flat ground at 420 m/s at a 53.0° angle,

Calculate the time to land on the ground as shown below,

[tex]\Delta y = v_o t +1/2 at^2[/tex]

0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²

t = 0 s or 68.5 s

Calculate the distance as shown below,

[tex]\Delta x = v_o t +1/2 at^2[/tex]

Δ x = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²

Δ x = 17,300 m

Thus, the total distance covered by the cannonball fired with a speed of 420 m/s is 17300 meters.

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Answer:

0.21m

Explanation:

Note that the 200 J before breaking legs is Kinetic energy

So if potential energy= kinetic energy

Then 200J = mgh

h= 200/75x 9.8

= 0.21m so the maximum height will be 0.21m

Answer:

Choice a. [tex]70\; \rm N[/tex], assuming that the skating rink is level.

Explanation:

There are two horizontal forces acting on the boy:

The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.

The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.

[tex]\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N[/tex].

The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.

However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.

Therefore, the (combined) net force on this boy would be:

[tex]\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N[/tex].

Answer:

d apple's are the best fruit

Answer:

more permeable

Explanation:

no idea, i just remember learning this in school lol.

... noisier when it's in a paper bag ...

Explanation:

- Victor pretends he can speak French > Rising action.

- Victor gets his school schedule > Exposition.

- Victor tries to get Teresa's attention after homeroom and at lunch > Rising action.

- Teresa asks Victor if he will help her in French > Falling action.

- Victor checks out books to learn French and help Teresa > Climax

Answer:

Answer:

E) 80 N/m

Explanation:

Given;

mass of the block, m = 4.8 kg

displacement of the block, x = -0.5 m

velocity of the block, v = -0.8 m/s

acceleration of the block, a = 8.3 m/s²

From Newton's second law of motion;

F = ma

Also, from Hook's law;

F = -Kx

where;

k is the force constant

Thus, ma = -kx

k = -ma/x

k = -(4.8 x 8.3) / (-0.5)

k = 79.7 N/m

k ≅ 80 N/m

Therefore, the force constant of the spring is closest to 80 N/m

Answer:

The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.

Explanation:

Given;

weight of the man, W = 700 N

Weight of the woman, W = 440 N

momentum is given by;

[tex]P = mv\\\\v = \frac{P}{m}[/tex]

Kinetic energy of the man;

[tex]K_m = \frac{1}{2}m_m(\frac{P_m}{m_m})^2 \\\\K_m = \frac{P_m^2}{2m_m}[/tex]

Momentum of the man is calculated as;

[tex]P_m^2 = 2m_mK_m[/tex]

The kinetic energy of the woman is given by;

[tex]K_w = \frac{P_w^2}{2m_w}[/tex]

The momentum of the woman is given;

[tex]P_w^2 = 2m_wK_w[/tex]

Since, momentum of the man = momentum of the woman

[tex]P_m^2 = P_w^2[/tex]

[tex]2m_mK_m = 2m_wK_w\\\\\frac{K_m}{K_w} = \frac{2m_w}{2m_m}\\\\\frac{K_m}{K_w} = \frac{m_w}{m_m}[/tex]

mass of the mas = 700 / 9.8 = 71.429

mass of the woman is = 440 / 9.8 = 44.898

[tex]\frac{K_m}{K_w} = \frac{44.898}{71.429}\\\\\frac{K_m}{K_w} =0.629[/tex]

Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.