Which statement best explains how wave one compares with wave 2
A. wave one has louder sound
B. wave one has softer sound
C. wave one had higher pitch
D. wave one has a lower pitch​

Answer: thermal conductivity

The difference in the rate at which these two materials transfer thermal energy throughout the cups is known as thermal conductivity.

It is the rate where the heat is to be transferred from conduction via the cross section area of the unit with respect to the material at the time when the temperature should be gradient and existed with the perpendicular to the area.

Also, in this, two material is to be transferred via the thermal energy with the help of cups.

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Answer:

Left to right and top to bottom

Explanation:

On the periodic table, the properties repeat from left to right and from top to bottom.

Periodic properties have a pattern from the top to the bottom or down a group or family.

Also, across the period from left to right, they also show a repeating pattern.

Answer:

D) Thermal energy will move from the water bottle to the ice.

Explanation:

Overtime, what happens is that thermal energy will move from the water bottle to the ice.

The water bottle is at a higher temperature compared to the ice. So, thermal energy will move from a place at higher temperature to one with lower temperature.

Answer:

6.8*10^-8C

Explanation:

The formula for calculating the Electric field on the sphere is expressed as;

E = Sum of force on the sphere/charge q

q is the charge on the sphere

E is the electric field

Given

E = 5000N/C

T = 2.3 x 10^-3 N.

q = ?

W  = 0.0002 * 9.8

W = 0.00196N

W = 1.96*10^-3N

From the formula;

q = T-W/E

W is the weight of the sphere

T is the tension in the string

q  =  (2.3*10^-3 - 1.96 x 10^-3)/5000

q = 0.00034/5000

q = 6.8*10^-8C

Hence the charge on the sphere is 6.8*10^-8C

Answer:

C) The natural frequency of the building should be very different from typical earthquake frequencies

Explanation:

We shall apply the concept of resonance in this problem .

When a body is applied an external harmonic force ( forced vibration) such that natural frequency of body is equal to frequency of external force or periodicity of external force , the body vibrates under resonance ie its amplitude of vibration becomes very high .

In the present case if natural frequency of building becomes equal to the earthquake's frequency ( external force ) , the building will start vibrating with maximum amplitude , resulting into quick collapse of the whole building . So to avoid this situation , natural frequency of building should be very different from typical earthquake frequencies .

Answer:

Both balloons experiences same buoyant force.

Explanation:

Buoyant force is defined as the upward force that is exerted on an object which is wholly or partly immersed in a fluid. We can also define it as the upward force exerted by any fluid upon a body immersed in it. We may also refer to this buoyant force as the upthrust.

This buoyant force is the push of air on the balloon and it is independent of the contents of the balloons. Hence, both balloons experiences same buoyant force.

Answer:

a. Q = 1881.73 x [tex]10^{-13}[/tex] C

b. As battery is not removed so, potential difference will remain same.

c. E = 21.42 x [tex]10^{3}[/tex] V/m

d. Q = 895.5 x [tex]10^{-13}[/tex] C

e. Again the potential difference will not change it will remain same as 9 V

f. E = 45 x [tex]10^{3}[/tex] V/m

Explanation:

Solution:

Here, Teflon is used so, the dielectric constant of the Teflon K = 2.1

Diameter = 2.1 cm

Radius = 2.1/2 cm

Radius = 1.05 cm

Radius = 0.015 m

Now, we need to find the area of each plate:

A = [tex]\pi r^{2}[/tex]

A = (3.14) ([tex]0.015^{2}[/tex])

A = 0.000225 [tex]m^{2}[/tex]

A = 2.25 x [tex]10^{-4}[/tex] [tex]m^{2}[/tex]

We are given the thickness of the plate which equal to the distance between the two plates.

d = 0.20 mm = 0.2 x [tex]10^{-3}[/tex] m

d = 0.2 x [tex]10^{-3}[/tex]  m = distance between two plates.

Hence, the capacitance of the dielectric without the dielectric

C = [tex]\frac{E.A}{d}[/tex]

Putting up the values we get,

E = 8.85 x [tex]10^{-12}[/tex]

C = [tex]\frac{8.85 . 10^{-12} x 2.25 . 10^{-4} }{0.002}[/tex]    

C = 99.5 [tex]10^{-13}[/tex]

If dielectric is included then,

[tex]C^{'}[/tex] = K C

[tex]C^{'}[/tex] = (2.1) ( 99.5 x [tex]10^{-13}[/tex])

[tex]C^{'}[/tex] = 209.08 x [tex]10^{-13}[/tex] F

As we know the voltage of the battery V = 9V So,

a) Charge before the Teflon is removed:

Q = CV

Q = [tex]C^{'}[/tex]V

Q = (209.08 x [tex]10^{-13}[/tex] F) (9V)

Q = 1881.73 x [tex]10^{-13}[/tex] C

b) Potential Difference before the Teflon is removed = ?

As battery is not removed so, potential difference will remain same.

c) Electric Field =?

As we know,

E = V/(K.d)

E = 9V/(2.1 x 0.2 x [tex]10^{-3}[/tex])

E = 21.42 x [tex]10^{3}[/tex] V/m

d) After the Teflon is removed

Q = CV

Q = (99.5 [tex]10^{-13}[/tex] ) ( 9)

Q = 895.5 x [tex]10^{-13}[/tex] C

e) Again the potential difference will not change it will remain same as 9 V

f) Electric Field = ?

E = [tex]\frac{V}{d}[/tex]  (Teflon is removed)

E = 9/0.2 x [tex]10^{-3}[/tex]

E = 45 x [tex]10^{3}[/tex] V/m

Answer:

10. 65 N

11. 35 N

Explanation:

10. Determination the force that sends the soccer ball towards the goal.

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Initial velocity (u) of the ball = 0 m/s

Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s

Acceleration (a) of the ball =?

a = (v – u) /t

a = (32.5 – 0) / 0.21

a = 32.5 / 0.21

a = 154.76 m/s²

Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:

Mass (m) of the ball = 0.42 kg

Acceleration (a) of the ball = 154.76 m/s²

Force (F) =?

F = ma

F = 0.42 × 154.76

F = 65 N

Thus, the force that sends soccer ball towards the goal is 65 N

11. Determination of the force exerted by the rocket on the satellite.

We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:

Initial velocity (u) of satellite = 0 m/s

Final velocity (v) of satellite = 0.63 m/s

Time (t) = 1296 s

Acceleration (a) of the satellite =?

a = (v – u) /t

a = (0.63 – 0) / 1296

a = 0.63 / 1296

a = 4.861×10¯⁴ m/s²

Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:

Mass (m) of the satellite = 72000 Kg

Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²

Force (F) =?

F = ma

F = 72000 × 4.861×10¯⁴

F = 35 N

Thus, the force exerted by the rocket on the satellite is 35 N

Answer:

P = 490500 [Pa]

Explanation:

The pressure at the bottom of a vessel and even of a lake or sea can be calculated by means of the following hydrostatic equation.

[tex]P=Ro*g*h[/tex]

where:

P = pressure [Pa] (units of pascal)

Ro = water density = 1000 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 50 [m]

Now replacing:

[tex]P=1000*9.81*50\\P=490500[Pa][/tex]

Answer:

The answer is below

Explanation:

Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.

Momentum = mass * velocity

The principle of conservation of momentum states that momentum cannot be created or destroyed but  can be transferred. Therefore the momentum before and after an action is equal.

Initial momentum = Final momentum

Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.

m = 71 kg, M = 500 kg, v = 6 m/s

Initial both the raft and diver are at rest, hence u and U is zero, hence:

mu + MU = mv + MV

71(0) + 500(0) = 71(6) + 500(V)

0 = 426 + 500(V)

500(V) = -426

V = -426/500

V = -0.852 m/s

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the  Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2              Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

Answer: B

Explanation: The forces are equal and opposite each other.

When two objects are interacting but stay stationary, then the forces are equal and opposite each other.

Force may be defined as the process of pushing and pulling an object with an actual mass that stimulates its velocity to be changed.  It is a type of vector quantity because it has both magnitude and direction.

It is the simple and fundamental concept of physics that when two or more objects are interacting with one another but do stimulate any change in their position, the forces among them are definitely equal and opposite to one another. It is the most plausible explanation of Newton's third law of motion.

Therefore, when two objects are interacting but stay stationary, then the forces are equal and opposite each other.

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Your question seems incomplete. The most probable complete question is as follows:

Answer:

[tex]|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C[/tex]

[tex]2v(t)\cdot \frac{dv(t)}{dt}=0[/tex]

[tex]v(t)\cdot a(t)=0[/tex]

Explanation:

Let's start with the definition of a constant velocity.

If the velocity magnitude, in three dimensions,  is a constant value (C) we have a constant velocity, which means.

[tex]|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C[/tex]

Now, we know that the dot product between v(t) and v(t) is the |v|².

[tex]v(t)\cdot v(t)=|v|^{2}(t)[/tex]

If we take the derivative whit respect to time in both sides of this equation we will have:

[tex]\frac{d}{dt}(v(t)\cdot v(t))=\frac{d}{dt}|v|^{2}(t)[/tex]

We apply the product rule on the left side and the right side will zero because the derivative of a constant is 0.

[tex]\frac{dv(t)}{dt}\cdot v(t)+v(t)\cdot \frac{dv(t)}{dt}=0[/tex]

[tex]2v(t)\cdot \frac{dv(t)}{dt}=0[/tex]

We know that dv(t)/dt = a(t) (using the acceleration definiton)

Therefore, we conclude:

[tex]v(t)\cdot \frac{dv(t)}{dt}=0[/tex]

[tex]v(t)\cdot a(t)=0[/tex]

If the dot product is 0, it means that v(t) and a(t) are orthogonal.

I hope it helps you!

Answer:

[tex]m_2=6.3\:\mathrm{kg}[/tex]

Explanation:

In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:

[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]

Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].

Plugging in all given values, we have:

[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot9.2^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot m_2\cdot 6.7^2,\\m_2=\fbox{$6.3\:\mathrm{kg}$}[/tex].

Answer:

c

Explanation:

im smart....................... i think

This law is about inertia, and the law displayed in A is Newton's third law of equal and opposite reactions, so option A is correct.

The basis of classical mechanics is laid out in three assertions known as Newton's laws of motion, which were first articulated by English physicist and mathematician Isaac Newton. These laws describe the relationships between forces acting on a body and its motion.

Unless a force acts on a body that is at rest or moving in a straight line at a constant speed, Newton's first law asserts that it will continue to be at rest or move in that direction.

This law is about inertia (an object wanting to stay in its state of motion) and the law displayed in A is Newton's third law of equal and opposite reactions, therefore, it is not an example of  Newton’s first law of motion.

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Answer:

-2 South

Always subtract where they ended first is what my teacher said

Answer:

14.6m/s

Explanation:

Given parameters:

Mass of the student  = 62kg

Initial velocity  = 0m/s

Height of slide = 10.6m

g  = 10m/s²

Unknown:

Speed at the bottom of the slide = ?

Solution:

The speed at the bottom of the slide is the final velocity;

             v ² = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

             v² = 0²  +  2x 10 x 10.6

             v²  = 212

              v  =  14.6m/s

Answer:

Data

Explanation:

Data is factual information not subject to bias.

This ultimately implies that, data connotes fact, thus, it is an information that is credible, accurate, a statement of truth, evidential and proven.

In Computer programming, a data dictionary can be defined as a centralized collection of information on a specific data such as attributes, names, fields and definitions that are being used in a computer database system.

In a data dictionary, data elements are combined into records, which are meaningful combinations of data elements that are included in data flows or retained in data stores.

This ultimately implies that, a data dictionary found in a computer database system typically contains the records about all the data elements (objects) such as data relationships with other elements, ownership, type, size, primary keys etc. This records are stored and communicated to other data when required or needed.

Answer:

C

Explanation:

Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.

Answer:

Given :  A bus of mass 760 kg requires 120 m to reach certain velocity value Vf.

the bus engine exerts a constant forward force F.

To Find : When the bus is towing a 330-kg small car, how long distance needed to reach same Vf?

Solution:

V²  - U² = 2aS

V = Vf

U = 0

S = 120 m

=> Vf² - 0 = 2a(120)

=> Vf² = 240a

m = 760  kg

Force = F

F = ma

=> F =760 a

=> a = F/760

Vf² = 240F/760

Case 2  :When the bus is towing a 330-kg small car,

m = 760 + 330 = 1090 kg

a = F/1090

Vf²  = 2aS

=> 240F/760  = 2 (F/1090) S

=> S = 120 x 1090 /760

=> S = 172.1   m  

172.1   m   distance needed to reach same Vf

Explanation:

Answer:

[tex]12\:\mathrm{N}[/tex]

Explanation:

Force is given by the equation [tex]F=ma[/tex].

Plugging in given values, we have:

[tex]F=ma=2.6\cdot 4.5=11.7=\fbox{$12\:\mathrm{N}$}[/tex] (two significant figures).

Answer:

[tex] \underline{ \boxed{ yes}}\\[/tex]

Explanation:

[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]

Answer:

[tex]0.15\: \mathrm{Hz}[/tex]

Explanation:

The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.

Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.

Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).

The radius of the track is irrelevant in this problem.

Answer:

(a) The two balls collide [tex]2\; \rm s[/tex] after launch.

(b) The height of the collision is [tex]4\; \rm m[/tex].

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].

Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].

On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].

Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:

[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].

For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].

For the ball thrown downwards:

[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)

Similarly, for the ball thrown upwards:

[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)

Equate the two expressions and solve for [tex]t[/tex]:

[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].

[tex]t = 2[/tex].

Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.

Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:

[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].

In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].

The time the two balls collide is 0.4 seconds while the height at which they collide is 4m

The given parameters are :

Initial Velocity U = 8m/s

Height H = 40m

For the second ball, the initial velocity = 12m/s

a.) For the first ball, the height attained at the point of collision will be

h = ut + 1/2gt^2

h = 8t + 1/2 x 10t^2 ........ (1)

For the second ball, the height attained at the point of collision will be

h = 12t - 1/2 x 10t^2 .........(2)

Since the height will be the same for the two balls, equate the two equations

8t + 10t^2 = 12t - 10t^2

Collect the like term

8t - 12t = -5t^2 - 5t^2

-4t = -10^2

10t = 4

t = 4/10

t = 0.4s

b.) Substitute time t in any of the equation to find the height

h = 12(0.4) - 0.5 x 10(0.4)^2

h = 4.8 - 0.8

h = 4m

Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m

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Answer:

Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

Explanation: