Which is not true about waves?
O Compression and rarefaction are found in longitudinal waves.
O Sound waves are transverse waves.
O Amplitude, wavelength, and troughs all describe parts of a wave.
O Electromagnetic waves are transverse waves.

Answer:

a = 50 [m/s²]

Explanation:

This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force =25 [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

[tex]a=F/m\\a=25/0.5\\a= 50 [m/s^{2}][/tex]

Answer:

(1) 180 J

(2) 60 m/s

Explanation:

(1) From the question,

Amount of work performed by the pitcher on the baseball = Force × distance.

W = F×d............... Equation 1

Given: F = 90 N,  and d = 2.0 m.

Substitute into equation 1

W = 90×2

W = 180 Joules.

(2)

F = ma........... Equation 2

Where a = acceleration of the ball.

a = F/m

Given: F = 90 N, m = 0.1 kg.

therefore,

a = 90/0.1

a = 900 m/s².

Using,

v² = u² + 2as ............ Equation 3

Where u = 0 m/s, a = 900 m/s², s  = 2 m.

substitute into equation 3

v² = 0² + 2×900×2

v² = 3600

v = √3600

v = 60 m/s

Answer:

c. 0.501 kg

Explanation:

Given;

volume of the denoted blood, V = 473 ml = 0.000473 m³

density of the denoted blood, ρ = 1060 kg/m³

The mass of donated blood is given as;

mass of the donated blood = density of the denoted blood x volume of the denoted blood

m = 1060 x 0.000473

m = 0.501 kg

Therefore, the mass of donated blood is 0.501 kg

Answer:

velocity and wavelength likely to change in the process.

Explanation:

When an object passes from one medium to another, refraction occurs. Due to refraction, it changes iits direction. Some of the properties is changed when it undergoes refraction. The velocity of a wave and wavelength change in this process.

Hence, the correct option is (a) "velocity and wavelength".

Answer:

angle of reflection and angle of incident is always equal

Answer:

let M be the mass of brick

let m be the mass of the pebble

.5M(3)^2 =KE

.5m(5)^2= KE

.5M9 = .5m25

9M = 25m

(9/25)M = m

Answer:

A

Explanation:

e=1/2M[tex]V^{2}[/tex]

1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25

[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]

Answer:

it is a ionic bond because electrons are transferred from chlorine to strontium

Strontium chloride, SrCl₂, is an ionic compound because Ionic compounds are formed through the transfer of electrons from one atom to another, resulting in the formation of positively charged ions (cations) and negatively charged ions (anions).

In the case of strontium chloride, strontium (Sr) is a metal, and chlorine (Cl) is a non-metal. Strontium readily donates two valence electrons to chlorine, which has a strong affinity for gaining one electron to achieve a stable electron configuration. The electron transfer between strontium and chlorine leads to the formation of S[tex]r^2^+[/tex] cations and C[tex]l^-[/tex] anions. The resulting ionic bond holds the cations and anions together in a crystal lattice structure. The attraction between the oppositely charged ions creates the ionic bond, making strontium chloride an ionic compound.

Learn more about the ionic compound here.

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#SPJ6

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

Sprint constant = k = 160 N / m

x = 0.15m

Now the force on the block is

F = kx

= (160) (0.15)

= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

Answer:

The answer is "0.5555 m"

Explanation:

Where the reference leaves the list and the viewer is at rest:

[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]

   [tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]

   [tex]=0.5555 \ m[/tex]

Answer:

1.8 m/s

Explanation:

Here is the complete question

The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.

What is the rate of discharge at point B?

Solution

Using Bernoulli's equation,

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)

(h₂ - h₁) = 5 m

Substituting the values of the variables, we have

700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)

36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)

- 13000 Pa = 500 kg/m³(v₂² - v₁²)

(v₂² - v₁²) = - 13000 Pa/500 kg/m³

(v₂² - v₁²) = -26 m²/s²

By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B

πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m

d₁²v₁ = d₂²v₂

v₁  = v₂(d₂/d₁)²

= v₂(0.15/0.05)²

= v₂(3)²

= 9v₂

So, substituting v₁ = 9v₂ into the above equation, we have

(v₂² - v₁²) = -26 m²/s²

v₂² - 9v₂² = -26 m²/s²

- 8v₂² = -26 m²/s²

v₂² = -26 m²/s² ÷ (-8)

v₂² = 3.25 m²/s²

taking square root of both sides,

v₂ = √(3.25 m²/s²)

= 1.8 m/s

So, the rate of discharge at point B is 1.8 m/s

Answer:

The cork moves up and down.

Explanation:

The cork moves up and down because of the waves produce due to tapping the surface of water. The cork experience movement due to placed on the water surface so when the wave passes through the medium by tapping of water, movement of cork also occur. Cork has high volume and lower mass so it floats on the water surface and experience movement due to waves.

Answer: 3) work

Explanation:

Work is doing something that requires you to expend energy. In walking across the room while your hands were carrying the stack of books, you expended energy through them which means that they did work.

We can neither tell the efficiency nor the power taken to do the work above from the given information. And by moving and holding books, there was definitely work done which means that all options apart from 3 are false.

Answer:

boiling water because salt dissolves quicker in hot Water and the hottest is boiling

Answer:

Explanation:

Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees

Answer:

well i tough it was infinte but its not an the layer is a d         i       c               k  

Explanation:

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 74 \times {11}^{2} \\ = 37 \times 121[/tex]

We have the final answer as

Hope this helps you

Answer:-1 m/s/s

Explanation:

The acceleration of the bus when its velocity changes from 15 m/s to 5 m/s will be a=1 m/s^2

Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.

Now from the first equation of motion we have,

[tex]a= \dfrac{v-u}{t}[/tex]

we have V=15 m/s, u=5 m/s , t=10 seconds

[tex]a=\dfrac{15-5}{10}=1\ \frac{m}{s^2}[/tex]

Hence the acceleration of the bus will be a=1 m\s^2

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Answer:

Displacement

Explanation:

I took the test.

Answer:

For the aceleration we have:

Vf = Vo + a * t

Clearing "a":

a = (Vf - Vo) / t

Replacing and resolving:

a = (34 m/s - 0 m/s) / 21 s

a = 34 m/s / 21 s

a = 1,61 m/s^2

The aceleration of the vehicle is 1,61 meters per second squared

Answer:12.1 j

Explanation:

Answer:

0.8 m/s

Explanation:

F(avg) * Δt = I, where

F(avg) = Average force

Δt = change in time

I = impulse

From the question, we know that the

average force is given as -17600 N

time change is given as 55 milliseconds

speed of the player, v = 8 m/s

Mass of the player is given as 110 kg

Impulse, I = F(avg) * Δt

I = -17600 * 0.055

I = -968

Now, using the Impulse-Momentum theory, we have that

Δp = I and thus,

Δp = m [v(f) - v(i)], where

Δp = change in the momentum

v(f) = final speed of the player

v(i) = initial speed of the player

Substituting the values, we have

I = m [v(f) - v(i)]

-968 = m.v(f) - m.v(i)

m.v(f) = m.v(i) - 968

110v(f) = 110 * 8 - 968

110v(f) = 880 - 968

110v(f) = -88

v(f) = -88 / 110

v(f) = -0.8 m/s

Answer: Induced emf is given by:

ε= Bvl

On putting the values we get

=5×10

−5

×1.50×2

=0.15mV

Explanation:

Hope these helped have a wonderful Christmas break ❄️

The combination of decay processes that would convert U-239 into PU- 239 is beta decay followed by another beta decay. There are two different types of beta decay (positive and negative beta decays) they both occur simultaneously in order to move the atoms towards stability.

Work done against gravity becomes potential energy of the object-Earth system. Its given by:

W = Ep

W = mgh  but weigh(w) = mg

W = wh

W = 350 N × 5 m

Power is the amount of energy transferred or converted per unit time. This is:

P = W/t

P = 1750 J / 20 s

Answer:

330

Explanation:

(a) The torque the net thrust produces about the center of the circle is of 20 N-m.

(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².

Given data:

The mass of model airplane is, m = 1.0 kg.

The radius of horizontal circle is, r = 20.0 m.

The magnitude of net thrust by engine is, F = 1.0 N.

(a)

The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,

T = F × r

Solving as,

T = 1.0 × 20.0

T = 20 N-m

Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.

(b)

The expression for the angular acceleration of airplane during the horizontal flight is given as,

[tex]T = I \times \alpha[/tex]

Here, I is the moment of inertia of airplane and its value is,

[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]

So, the angular acceleration is,

20 = 200 × α

α = 20/200

α = 0.1 rad/s²

Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².

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Answer:

this point we have a destructive inference  m=1

Explanation:

This is an interference exercise, where the light has two possible states one of constrictive interference

            d sin θ = m λ

and another of destructive interference

           d sin θ = (m + ½) λ

the term on the left is the difference in trajectory that tells us that it is worth

            d sin θ = 300 nm = 300 10⁻⁹ m

We can find what type of interference we have by finding m, which must be an integer

constructive interference

           m = d sin θ /λ

let's calculate

          m = 300 10⁻⁹ / 500 10⁻⁹

          m = 0.6

this value is not possible because m must be an integer

destructive interference

          m + ½ = d sin θ /λ  

          m = d sin θ /λ  + ½

          m = 0.6 + 0.5

         m = 1.1

that can be approximated to m = 1

this value if possible, so at this point we have a destructive inference

Complete Question

(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .

(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.

Answer:

A

  [tex]\Delta E = 337 \ eV[/tex]

B

  [tex]\lambda = 3.439 *10^{-9} \ m[/tex]

Explanation:

Considering question a

From the question we are told that

 The width of the box is  [tex]w = 0.125 \ nm = 0.125 *10^{-9} \ m[/tex]

Generally the energy level of a particle confined to a box is mathematically represented as

         [tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]

Generally the excitation energy is mathematically represented as

         [tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]

From the question  [tex]n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1[/tex]  

   Here h  is the Planck's constant with a value  of  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

         m is the mass of  electron with value  [tex]m = 9.11 *10^{-31} \ kg[/tex]

So

        [tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ][/tex]

=>     [tex]\Delta E = 539 *10^{-19} \ J[/tex]

=>     [tex]\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J[/tex]

=>     [tex]\Delta E = 337 \ eV[/tex]

Considering question b

Generally the energy level of a particle confined to a box is mathematically represented as

         [tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]

Generally the excitation energy is mathematically represented as

         [tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]

From the question  [tex]n_2 = 4 \ \ and \ \ n_1 = 1[/tex]

 Here h  is the Planck's constant with a value  of  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

         m is the mass of  electron with value  [tex]m = 9.11 *10^{-31} \ kg[/tex]

So

        [tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ][/tex]

=>      [tex]\Delta E = 578 *10^{-19} \ J[/tex]

=>      [tex]\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}[/tex]  

=>      [tex]\Delta E = 361.45 \ eV[/tex]  

Gnerally the wavelength is mathematically represented as

          [tex]\lambda = \frac{hc}{\Delta E }[/tex]

=>       [tex]\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }[/tex]

=>       [tex]\lambda = 3.439 *10^{-9} \ m[/tex]

Answer:

15.3 sorry if i got it wrong

Explanation:

There are

[tex]6.023 \times {10}^{23} [/tex]

electrons in a mole. So charge of 1 mole electron is

[tex]6 \times {10}^{23} \times ( - 1.6022) \times {10 }^{ - 19} [/tex]

Im not sure what are you asking about. But in physic/ mechanic usually when the object against gravity, its deceleration but when the object follow the gravity its acceleration

Also usually when the object move to the left side the speed will be negative (-) this is probably because of deceleration? since deceleration = -acceleration.