In a transverse wave wavelength can be measured as the distance between a. Two adjacent crests. b. A crest and the adjacent trough. c. A crest and the resting position. d.None of the above

Answer:

I think it's A and D.

Answer:

The arrow of the second archer standing on a high building will have more potential energy.

The arrow of the first archer standing on ground will have more kinetic energy.

Explanation:

POTENTIAL ENERGY:

The potential energy of an object depends upon its height, as given in the formula:

P.E = mgh

Hence, the arrow with the greater height will have more potential energy.

Therefore, the arrow of the second archer standing on a high building will have more potential energy.

KINETIC ENERGY:

The kinetic energy of an object depends upon its speed, as given in the formula:

K.E = (1/2)mv²

Hence, the arrow with the greater speed will have more kinetic energy energy. Since, the second arrow is stationary, it will have zero kinetic energy. But, the first arrow will have some K.E due to its speed.

Therefore, the arrow of the first archer standing on ground will have more kinetic energy.

(a)  The second archer standing at the top of the building will have more potential energy.

(b) The first archer on the ground level will have the more kinetic energy.

The potential energy of an object is the energy possessed by the object by virtue its position above above the ground.

P.E = mgh

where;

Thus, the second archer standing at the top of the building will have more potential energy.

(b) The kinetic energy of an object is the energy possessed by the object due to its motion.

K.E = ¹/₂mv²

where;

An object in motion, has zero velocity at maximum height but maximum velocity on the ground level.

Thus, the first archer on the ground level will have the more kinetic energy.

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Answer:B

Explanation:

Mass

More mass is added to a wheelbarrow, and a larger force is required to move it - illustrates Newton's second law.

An object's push or pull is seen as exerting a force. The interaction of the objects produces push and pull. You can also use words like stretch and squeeze to describe force.

The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.

Newton’s second law of motion states that “Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration."

Hence more mass is added to a wheelbarrow, and a larger force is required to accelerate, i.e., move it.

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Answer:

Mass, m = 71.42 kg

Explanation:

Given that,

Normal force acting on a box, F = 700 N

We need to find the mass of a box. Let it is m. Normal force acting on an object is balanced by its weight such that,

F = mg

Where m is the mass of the box

[tex]m=\dfrac{F}{g}\\\\m=\dfrac{700\ N}{9.8\ m/s^2}\\\\m=71.42\ kg[/tex]

So, the mass of the box is 71.42 kg.

Explanation:

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.

Answer:

A)

Explanation:

I took the test and got it right. Flvs

Answer:

Explanation:

The area of the base of the statue can be found by using the formula

[tex]a = \frac{f}{p} \\ [/tex]

f is the force

p is the pressure

From the question we have

[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

1144.8

Explanation:

1 meter per second is equal to 3.6 kilometers per hour

Answer:

45 degrees

Explanation:

a projectile travels the farthest when it is launches at the angle of 45 degree.

Answer:

Explanation:

a. WEIGHT: As we have seen, weight is the gravitational force exerted on an object by the Earth (or any other celestial body). If an object is near the Earth's surface and has mass, then the object has a weight. The magnitude of its weight is w = mg and its direction is toward the center of the Earth.

Answer:

C

Explanation:

Answer:

2.a centripetal force.

Answer:

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Explanation:

We can estimate the absolute uncertainty by the definition of total differential. That is:

[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)

Where:

[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].

[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.

[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.

[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.

[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.

[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.

[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.

Partial derivatives are, respectively:

[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)

[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)

[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)

And we expand (1) as follows:

[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]

[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)

If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:

[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]

[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]

And the expected density is:

[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)

[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]

[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]

The percentage uncertainty in his calculated value of density is:

[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)

If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:

[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]

[tex]\%e = \pm 0.713\,\%[/tex]

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Answer:

plane LA534 will have to decrease it speed by twice that of plane  LA639.

Explanation:

Since both planes want to land there must be a decrease in speed.

We are told that plane  LA534 velocity is twice that of plane LA639.

Now for the two planes to land without a crash. the opposite or reverse will be the case in terms of deceleration, that is

Plane LA639 will have to decrease its speed by twice the decrease in speed of plane  LA534

Answer:

Workdone = 465766038 Joules.

Explanation:

Given the following data;

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

Answer:The answer is A

Explanation: You can instantly eliminate b and d because an ion isn't mentioned, and it is impossible to have a carbon dioxide atom. C is incorrect because there has to be two oxygen's to create a carbon dioxide molecule.

Answer:

D. A carbon dioxide molecule

Explanation:

Answer:

W = 1493.9 J = 1.49 KJ

Explanation:

The work done by the elevator on the object will be equal to the gain in is potential energy:

W = ΔP.E

W = mgΔh

where,

W = Work = ?

m = mass of object = 7.4 kg

g = 9.8 m/s²

Δh = gain in height = 20.6 m

Therefore,

W = (7.4 kg)(9.8 m/s²)(20.6 m)

W = 1493.9 J = 1.49 KJ

Answer:

See below

Explanation:

ICT " Information and the communication technology" can be used in sectors like Telecommunications. ADB support for telecommunications and ICT is changing lives across the region. ...

ICT in Education. ICT can improve the efficiency and quality of education at all levels. ...

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( hope it helped <3 )

Answer:

The velocity the stone must be thrown with is 20 m/s.

Explanation:

Given;

time taken for the stone to reach it highest point, t = 2 s

acceleration due to gravity, g = 10 m/s²

The maximum height reached by the stone at the given time is calculated as;

[tex]h = \frac{1}{2}gt^2\\\\h = \frac{1}{2}*10*2^2\\\\h = 20 \ m[/tex]

Determine the initial of the stone;

v² = u² + 2gh

where;

v is the final velocity of the stone at maximum height = 0

u is the initial velocity of the stone = ?

0 = u² - 2(10 x 20)

0 = u² - 400

u² = 400

u = √400

u = 20 m/s

Therefore, the velocity the stone must be thrown with is 20 m/s.

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, [tex]\theta=35^{\circ}[/tex]

Time of flight :

[tex]t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s[/tex]

Horizontal distance traveled  is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

[tex]H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m[/tex]

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

Answer:

Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.

Answer:

Explanation:

Given

the range = 5.00m (distance moved in horizontal direction)

Height = 0.150m

Required

Initial velocity of the dart

Using the formula for calculating range of a projectile;

R = U√2H/g

5 = U√2(0.15)/9.8

5 = U√0.0306

5 = 0.1749U

U = 5/0.1749

U = 28.59m/s

Hence the initial horizontal velocity of the dart is 28.59m/s

Answer:

28.59

Explanation:

khan academy

Answer:

B. An opposing force caused by friction produced a lower acceleration than calculated

Explanation:

Answer:

An opposing force caused by friction produced a lower acceleration than calculated.

Explanation:

Answer:

12.18

Explanation:

The maximum height that the orange will reach is 12.18 m.

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion.

You throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s.

If u is the initial speed, the maximum height of the ball from the point of throwing is

H = u² sin²θ / 2g

Put the values, we get

H = 3.5² sin²32° / (2x9.81)

H = 0.1753 m

So, the maximum height calculated from the ground is

12 + 0.18 = 12.18 m

Thus ,the maximum height that the orange will reach is 12.18 m.

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Answer:

0.364

I believe... Good luck!

Answer:

The car that is accelerating is B a car that rounds a curve at a constant speed

Explanation:

Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.

Answer:

its B (:

Explanation:

i  took the test :)

Answer:

Explanation:

Gneiss is a high grade metamorphic rock, meaning that it has been subjected to higher temperatures and pressures than schist. However, unlike slate and schist, gneiss does not preferentially break along planes of foliation because less than 50% of the minerals formed during the metamorphism are aligned in thin layers.

Answer:

A = (27.95 N, 21 N)

Explanation:

The polar co-ordinates are given as:

(r,θ) = (35 N, 37°)

Now, to convert this into polar co-ordinates (x, y), we will use following relations:

r² = x² + y²

(35)² = x² + y²

1225 = x² + y²  ----------- equation (1)

and

tan θ = y/x

tan 37° = y/x

y =  0.753 x   ------------------- equation (2)

Substituting this value in equation (1):

1225 = x² + (0.753 x)²

1225 = 1.567 x²

x² = 1225/1.567

x = √781.32

x = 27.95 N

using this value in equation (2)

y = (0.753)(27.95 N)

y = 21 N

Therefore, the vector can be represented in polar co-ordinates as:

A = (27.95 N, 21 N)

Answer:

a) El número de neutrones de Cl-35 es 18 y del Cl-37 es 20.

b) La masa atómica media del cloro es 35.5 uma.

c) En 15 g de cloro hay 2.55x10²³ átomos.    

Explanation:

a) Para calcular los protones y neutrones de los isótopos Cl-35 y Cl-37 debemos usar la siguiente fórmula:

[tex] A = Z + N [/tex]

En donde:

Z: es el número de protones = 17

N: es el número de neutrones

A: es la masa atómica

Los isótopos de un elemento dado tienen el mismo número de protones (número atómico) y en ellos sólo varia la la masa atómica (y por ende el número de neutrones); por lo que el Cl-35 y Cl-37 tienen igual Z (17).

Para el Cl-35 tenemos:

[tex] N = A - Z = 35 - 17 = 18 [/tex]

Y para el Cl-37 tenemos:

[tex]N = 37 - 17 = 20[/tex]

Por lo tanto, el número de neutrones de Cl-35 es 18 y del Cl-37 es 20.

b) La masa atómica media del cloro está dada por:

[tex]M_{Cl} = A_{Cl-35}*\%_{Cl-35} + A_{Cl-37}*\%_{Cl-37} = 35 u*\frac{75}{100} + 37 u*\frac{25}{100} = 35.5 u[/tex]  

Entonces, la masa atómica media del cloro es 35.5 uma.

c) Para encontrar el número de átomos de Cl que hay en 15 g debemos usar el número de Avogadro (6.022x10²³ átomos/mol):

[tex] N = \frac{1 mol}{35.5 g}*15 g*\frac{6.022 \cdot 10^{23} atomos}{1 mol} = 2.55 \cdot 10^{23} atomos [/tex]

Finalmente, en 15 g de cloro hay 2.55x10²³ átomos.

Espero que te sea de utilidad!