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Apply scientific knowledge and understanding to solve problems
The diagram shows waves being produced on a rope.
The waves are not reflected by the wall.
I. Outline an arrow on the diagram to show the direction in which the waves transfer
energy
Related to the previous diagram, Which one of the arrows, labeled, X, Y, or Z, shows the amplitude of a wave?

Answer:

there it is fella tried on ma own observation

Answer:

Depends: Is it tennis or volleyball?

Explanation:

If it's volleyball, it's C, the other teams point.

If it's tennis, it's A, you get a re-do.

Answer:

a

Explanation:

Fan speed is the answer

metres to kilometres = 1/1000

8 m ⇒ 0.008 km

metres to centimetres = × 100

8 m ⇒ 800 cm

Answer:

1000m=1km

so, 8m=8/1000 = 0.008km

1m=100cm

so, 8m=8×100=800

Answer:

They show the elements that make up a compound.

They show the three-dimensional shape of a molecule.

They show the types of atoms that make up a molecule.

They show the number of each type of atom in a molecule.

Explanation:

#keep learning!!

Answer:

a

Explanation:

Answer:

The human body runs at a constant 37 ºC but the air around you at room temperature is about 20-25 ºC which means heat is constantly leaving your body to warm your surroundings and maintain thermal equilibrium. You don't lose much energy doing this however as air heats reasonably quickly

Explanation:

Answer:

The density of brass is approximately 8,261.73 kg/m³

Explanation:

The percentage composition by mass of brass is given as follows;

The percentage by mass of copper = 67%

The percentage by mass of zinc = 33%

The density of copper, ρ₁ = 8.96 g/cm³

The density of zinc, ρ₂ = 7.133 g/cm³

Therefore, where we have, m = 100 g of brass, we have;

The mass of copper, m₁ = 67 g

The volume of copper, V₁ = m₁/ρ₁

∴ V₁ = 67 g/(8.96 g/cm³) ≈ 7.47767857 cm³

The volume of zinc, V₂ = m₂/ρ₂

∴ V₂ = 33 g/(7.133 g/cm³) ≈ 4.62638411 cm³

The volume of the brass, V = V₁ + V₂

V = 7.47767857 cm³ + 4.62638411 cm³ ≈ 12.104 cm³

The density of brass, ρ = m/V

∴ ρ = 100 g/(12.104 cm³) ≈ 8.26 g/cm³

The density of brass, ρ ≈ 8.26 g/cm³ = 8,261.73 kg/m³

Answer:

a.

[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]

b.

[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]

c.

[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]

d. 1.144 × 10¹³ N

Explanation:

The universal law of gravitation is presented as follows;

[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]

The given mass of the scientific satellite, m = 1,300 kg

The height of the orbit of the satellite, r = 200 km above the Earth's surface

The length of the radius of the Earth, R = 6378 km

The mass of the Earth = 5.97 × 10²⁴ kg

a. The formula for the universal law of gravitation is presented as follows;

[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]

Where;

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the satellite = 1,300 kg

r = The distance between the Earth and the satellite = R + r = 6,378 km + 200 km = 6,578 km

G = The Gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

b. Plugging in the values from the problem into the formula gives;

[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]

c. Solving gives;

[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]

The force acting between the Earth and the satellite, F ≈ 1.144 × 10¹³ N

d. 1.144 × 10¹³ N

Answer:

C) 2.00 m/s^2

Explanation:

F = m*a

150N = 75kg(a)

a = 150N/75kg

a = 2.0m/s²

Explanation:

area=length(m) ×breadth(m) . The unit of area is expressed in terms of fundamental units m^2.thus it is derived unit

Answer:

The energy will be increased by a factor 16

Explanation:

Mathematically, we have it that the energy carried by the sound wave is directly proportional to the square of the amplitude

So we have it that;

E = k * A^2

where E is the wave energy

K is the constant value

A is the amplitude value

So;

If E1 = k * (A1)^2

E2 = k * (A2)^2

But A2 = 4A1

E2 = k * (4A1)^2

E2 = k * 16(A1)^2

Divide E2 by E1

E2/E1 = k/k * 16(A1)^2/(A1)^2

E2/E1 = 16

E2 = 16 * E1

so simply, the energy will be increased by a factor of 16

no net force axting on it

Answer:

I think none of above

Explanation:

hioe it's right answer

Answer:

the energy produced in a body due to its motion is kinetic energy.

energy produced in a body due to its position is potential energy.

Explanation:

kinetic energy is the energy possessed by a body in motion

Potential energy is the energy possessed by a body by virtue of its position with respect to a reference level

Answer:

1= Kidneys

2= Ureters

3=Bladders

4=Urethra

Explanation:

Functions of kidney: Regulations of water balances, excretions of nitrogenous wastes(urea)

Functions of Liver: Detoxifications, protein synthesis

Functions of Skin: Temperature regulations, Metabolic functions, Excretion

Functions of lungs: Respiration, Water balance regulations

Tip: If you want faster and correct answers you should label your questions under the correct subject.

Answer:

wats the following items

Answer:

The untrue option is C: "More arrowheads on lines mean a brighter light"

Explanation:

Ray diagrams are used to show how the light behaves with things like mirrors or lenses. Where we only study how the direction of the light changes when it interacts with these objects.

The "light" is represented with arrows, where again, the only thing we care is the direction of the light, so the first statement is true, the arrowheads show the direction of the light, and only that.

The intensity of the light, in this context, has no effect on how light behaves, so there is not a necessity of representing the intensity of the light, thus, more arrowheads on lines do not mean a brighter light. It may only be used to represent changes in direction of the light.

Finally, we know that light travels in straight pats (the pats can be curved in some cases, like with large gravitational fields, but this is not the case of a ray diagram) so the lines that represent the light should always be straight, thus option B is also true.

The untrue option is C: "More arrowheads on lines mean a brighter light"

Answer:

14000 J / 0.40 J/s = 35 000 s or about 9.75 hrs

Answer:

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

Explanation:

The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x - axis is

E = Qx/4πε₀[√(x² + R²)]³

Since the rings are separated by a distance L, the electric field at point x due to the second ring is E' = -Q(L - x)/4πε₀[√((L - x)² + R²)]³. It is negative since it points in the negative x - direction.

So, the resultant electric field at x is E" = E + E' = Qx/4πε₀[√(x² + R²)]³ + {-Q(L - x)/4πε₀[√((L - x)² + R²)]³}

E" =  Qx/4πε₀√[(x² + R²)]³ - Q(L - x)/4πε₀√[((L - x)² + R²)]³

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[((L - x)² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L² - 2Lx + x² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L - 2x)L + (x² + R²)]³})

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

So, the electric field at points along the x axis is

E" =  Q/4πε₀√[(x² + R²)]³(x - {(L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

Answer:

O C. A person absorbs most of the light that hits him or her.

Explanation:

Answer:

Answer:

The magnitude of the force acting on the sled is 60.5 newtons.

Explanation:

The Work-Energy Theorem states that the work done by the external force applied on the sled ([tex]W[/tex]), in joules, is equal to the change of its translational kinetic energy ([tex]\Delta K[/tex]), in joules:

[tex]W = \Delta K[/tex] (1)

By definitions of work and translational kinetic energy we expand the equation above:

[tex]F\cdot s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex] (1b)

Where:

[tex]F[/tex] - External force applied on the sled, in newtons.

[tex]s[/tex] - Travelled distance, in meters.

[tex]v_{1}, v_{2}[/tex] - Initial and final velocities, in meters per second.

If we know that [tex]m = 11\,kg[/tex], [tex]v_{1} = 4\,\frac{m}{s}[/tex], [tex]v_{2} = 7\,\frac{m}{s}[/tex] and [tex]s = 3\,m[/tex], then the external force applied on the sled is:

[tex]F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot s}[/tex]

[tex]F = \frac{(11\,kg)\cdot \left[\left(7\,\frac{m}{s} \right)^{2}-\left(4\,\frac{m}{s} \right)^{2}\right]}{2\cdot (3\,m)}[/tex]

[tex]F = 60.5\,N[/tex]

The magnitude of the force acting on the sled is 60.5 newtons.

Answer:

Explanation

The most famous and common example is Christmas tree lights. You can't tell easily by looking at them whether they are in series or parallel. But you sure know the difference when one of them burns out. When that happens, the whole string goes dead. No matter what you do (other than find out which bulb burned out) will not fix the problem.

Another example is anything that is temperature controlled. For example a furnace is controlled by a thermostat. When the room temperature reaches a certain point, the thermostat is constructed in a certain way so that it forms an open circuit and no current can flow through it. The furnace motor turns off and the furnace stops pumping hot air into a room.

Answer:

True

Explanation:

The reproductive success of any species is the capability to produce their offspring per breeding lifetime or event.

Most of the species have to attract their partners by their physical capability and build up so that the mother can choose her partner in order to breed the best kind of off spring.

In the context, in case of deer, the size of the their antlers as well as behavior in herding is considered as the best chances for a successful reproduction to compete among the males and find their breeding mate.

Thus the answer is TRUE.

Answer:

DONT KNOW SORRY XD

Explanation:

BEC I AM BRAINLESS USING BRAINLY

Answer:

Ok, so at point 20 on the x axis you put 12 on the y-axis, then draw a line from 0, 0 to that point, this demonstrates the acceleration to 12 m/s.

Next, calculate the time he cycled for if he suckled the distance of 4800m :

4800/12 = 400 s

Put a dot at 400, 12. The time changed but the speed remained constant, line is horizontal.

For the last one, he had to decelerate at an equal speed. Since there is 100 s left of the journey at that stage, You need to draw a straight line from the previous point to 500, 0. So at time 500 he was not accelerating at all, he came to rest.

a) ?

b) -0.16 meters a second acceleration

Answer:

34902

Explanation:

Pv=nRt

v=nrt/p

subsitute

v=15*831*280/100

v=34902

When the brake is applied to the car and comes to rest after some time is an example of an object that is slowing down.

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

1. When the brake is applied to the body and comes to rest after some time is an example of an object that is slowing down.

2. When a person accelerates the car and the velocity is increasing is an example of speeding up.

3. An object moves in a circular motion moving at a constant speed but changing direction,

4. Acceleration is the rate of change of velocity. If the change in the velocity is constant the acceleration is zero is an example of moving at a constant velocity with zero acceleration.

To learn more about acceleration, refer to the link;

https://brainly.com/question/2437624

#SPJ1

Answer:

B) 2.7 g of aluminium has a volume of 1 cm^3

Explanation:

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

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

If the density of aluminum is 2.7 g/cm³, it simply means that 2.7 g of aluminium has a volume of 1 cm³

Check:

Given the following data;

Mass = 2.7 grams

Volume = 1 cm³

Substituting into the formula, we have;

[tex]Density = \frac{2.7}{1}[/tex]

Density = 2.7 g/cm³