A school bus's velocity decreases from 15 m/s to 5 m/s forwards over a period of 10 s. What is the bus's average acceleration? [Vf=v,+at]

Answer:

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

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

Answer:

15.3 sorry if i got it wrong

Explanation:

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:

100 m

Explanation:

Answer:

Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]

The energy required comes in three parts:

The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:

[tex]Q = c \cdot m \cdot \Delta T[/tex],

where

For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].

Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].

The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:

[tex]Q = m \cdot L_\text{f}[/tex].

Apply this equation to find the size of the second part of energy input:

[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].

Find the sum of these three parts of energy:

[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].

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: 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.

Solar Energy is the answer to the question tell me if i`m wrong

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:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

The ball momentum is

[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]

As there is a similar momentum

So,

[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 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 ❄️

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]

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:

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:

speed = distance/time

Explanation:

distance -> s

speed -> v

time -> t

The calculation of speed using time and distance shows that speed is equal to distance/time.

Speed refers to the rate of movement of a vehicle or person.

Thus, the speed can be computed by dividing the total distance by the time consumed.

Learn more about speed, distance, and time at https://brainly.com/question/4931057

Answer:

m = 4.4 × 10³ kg

Explanation:

Given that:

The total yearly energy is 4.0 × 10²⁰ J

The amount of mass that provides this energy can be determined by using the formula:

E = mc²

where;

c = speed of light in free space = (3 × 10⁸)

4.0 × 10²⁰ = m × (3 × 10⁸)²

[tex]m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}[/tex]

m = 4.4 × 10³ kg

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:

sound waves are transverse waves

Answer:

Gold = 19.3g/cm³

Explanation:

Given the following data;

Mass = 38.6g

Volume = 2cm³

To find the density;

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]

Substituting into the equation, we have;

[tex]Density = \frac{38.6}{2}[/tex]

Density = 19.3g/cm³

Therefore, from the table the material is Gold.

Gold is a chemical element and it is the element 79 on the periodic table. Thus, it has an atomic number of 79. The chemical symbol for Gold is Au, it is chemically classified as a transition metal and as a solid at room temperature.

Generally, the chemical element Gold is known to have the following physical properties;

I. Malleable.

II. Ductile.

III. A good conductor of electricity and heat.

Also, it is a non-toxic chemical element with a beautiful lustrous sheen.

Answer:

Density = 5 g/m³

Explanation:

Given:

Mass of block = 20 gm

Volume of block = 4 m³

Find:

Density

Computation:

Density = Mass / Volume

Density = 20 / 4

Density = 5 g/m³

5 kg/m3

Explanation:

The formula to find density is (d) = mass (m) ÷ volume (v)

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: Biosensors are modern electronic devices which are made up of biological recognition system and a transducer, for processing of signals and to quantify a particular analyte.

Explanation:

In modern medicine, a Biosensors are used to replace some clinical laboratory investigations of biological fluids with the advantages of being easy to use, rapid and robust as well as offering multianalyte testing capability. They are classified based on their biological recognition elements which includes:

--> enzymatic biosensors

--> immuno biosensors

--> DNA and whole cell biosensors.

They can also be classified according to their signal transduction methods. These includes:

--> electrochemical Biosensors,

--> optical biosensors,

--> thermal and mass-based biosensors;

Some of the HEALTH APPLICATIONS of a modern biosensor includes:

--> Monitoring of glucose levels in diabetic patients which makes use of enzyme based sensors to monitor concentration of glucose in patients.

--> Detection of general metabolic status of bacteria, fungi, yeast, animal or plant cells.

--> Monitoring of cholesterol levels to prevent cardiovascular diseases. Cardiac troponin and. C - Reactive Proteins are biomarkers that are easily detected using biosensors.

--> They are used for Cancer clinical testings

Answer:

16m/s

Explanation:

[tex]v_{f}=v_{i}+at[/tex]

[tex]v_{f}=0+2\cdot8[/tex]

[tex]v_{f}=16\ \frac{m}{s}[/tex]

Therefore,  the speed after 8 seconds is 16m/s

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:

+ 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:

(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:

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.

Learn more about the torque here:

https://brainly.com/question/19247046