Physics High School

what is latent heat of vaporization?

Answers

Answer 1

The amount of energy required for a unit mass of a substance to undergo a phase change from liquid to gas.

Related Questions

In Boolean Algebra zero represent
1) Zero potential
2) Ground potential
3) low potential
4) Both 1 &2

Answers

Answer:

Option 1 & 2

Explanation:

The area of mathematics known as Boolean algebra concerns with operations on logical quantities with binary variables. To express truths, Boolean features are transformed as binary numbers: 1 = true and 0 = false. Boolean algebra concerns with logistical processes, whereas fundamental algebra deals with machine based.

4. An object is thrown from from the ground upward with an initial speed of 3.75 m/s. How long will the object be in the air before it lands on the ground?

Answers

Answer:

Explanation:

There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.

Going with the physics version first, here's what we know:

a = -9.8 m/s/s

v₀ = 3.75 m/s

t = ??

That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:

v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:

v = v₀ +at and filling in:

0 = 3.75 + (-9.8)t and

-3.75 = -9.8t so

t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:

2(.38) = .76 seconds. Now onto the calculus way:

The position function is

[tex]s(t)=-4.9t^2+3.75t[/tex] The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:

v(t) = -9.8t + 3.75 and

0 = -9.8t + 3.75 and

-3.75 = -9.8t so

t = ,38 and multiply that by 2 to find the time the whole trip took:

2(.38) = .76 seconds.

Your parallel capacitors are 15 μf and 20 μf. The series capacitors are 10 μf and 12 μf. This circuit is connected to a 14 v battery, also determine the potential energy and the voltage across each capacitor

Answers

Answer:

a. i. 6.608 V ii. 5.507 V iii. 1.89 V iv. 1.89 V

b. i. 0.22 mJ ii. 0.182 mJ iii. 0.027 mJ iv. 0.036 mJ

Explanation:

a. The voltage across each capacitor

Since the 15 μf and 20 μf capacitors are in parallel, their total capacitance is C = 15 μf + 20 μf = 35 μf.

Also, since C is in series with the 10 μf and 12 μf which are in series, their total capacitance, C' is gotten from 1/C' = 1/10 μf + 1/12 μf + 1/35 μf

1/C' = (12 + 42 + 35)/420 /μf

1/C' = 89/420 /μf

C' = 420/89 μf

C' = 4.72 μf

The total charge in the circuit' is thus Q = C'V where V = voltage = 14 V

So, Q = C'V = 4.72 μf × 14 V = 66.08 μC

Since the 10 μf and 12 μf are in series, Q is the charge flowing through them.

Since Q = CV and V = Q/C

i. The voltage across the 10 capacitor is

V = 66.08 μC/10 μF = 6.608 V

ii. The voltage across the 12 capacitor is

V = 66.08 μC/12 μF = 5.507 V

The voltage across the 15 μF and 20 μF capacitors.

Since the capacitors are in parallel, the voltage across them is the voltage across their combined capacitance, C

So, V = Q/C = 66.08 μC/35 μF = 1.89 V

iii. The voltage across the 15 μF capacitor is 1.89 V

iv. The voltage across the 20 μF capacitor is 1.89 V

b. The potential energy of each capacitor

i. The potential energy of the 10 μF capacitor

E = 1/2CV² where C = Capacitance = 10 μF = 10 × 10⁻⁶ F and V = voltage across capacitor = 6.608 V

E = 1/2CV²

E = 1/2 × 10 × 10⁻⁶ F(6.608 V)²

E = 5 × 10⁻⁶ F(43.666) V²

E = 218.33 × 10⁻⁶ J

E = 0.21833 × 10⁻³ J

E = 0.21833 mJ

E ≅ 0.22 mJ

ii. The potential energy of the 12 μF capacitor

E = 1/2CV² where C = Capacitance = 12 μF = 12 × 10⁻⁶ F and V = voltage across capacitor = 5.507 V

E = 1/2CV²

E = 1/2 × 12 × 10⁻⁶ F(5.507 V)²

E = 6 × 10⁻⁶ F(30.327) V²

E = 181.96 × 10⁻⁶ J

E = 0.18196 × 10⁻³ J

E = 0.18196 mJ

E ≅ 0.182 mJ

iii. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 15 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 15 × 10⁻⁶ F(1.89 V)²

E = 7.5 × 10⁻⁶ F(3.5721) V²

E = 26.79 × 10⁻⁶ J

E = 0.02679 × 10⁻³ J

E = 0.02679 mJ

E ≅ 0.027 mJ

iv. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 20 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 20 × 10⁻⁶ F(1.89 V)²

E = 10 × 10⁻⁶ F(3.5721) V²

E = 35.721 × 10⁻⁶ J

E = 0.035721 × 10⁻³ J

E = 0.035721 mJ

E ≅ 0.036 mJ

what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s

Answers

Answer:1200

Explanation:

F=ma =m(Vf-Vi)/t

F=120(25-5)/2 =1200N

Answer:

[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]

in parallel combination of electrical appliances Total Electric Power a. increase b. decrease c. remain same

Answers

Answer:

In a parallel combination of electrical appliances total electric power will increase

Answer is A it will increase

How many more neutrons are in a I SOTOPE of copper-14 than in standard carbon atom

Answers

Answer:

2 more neutrons

Explanation:

To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:

For carbon–14:

Mass number = 14

Proton number = 6

Neutron number =?

Mass number = Proton + Neutron

14 = 6 + Neutron

Collect like terms

14 – 6 = Neutron

8 = Neutron

Neutron number = 8

For carbon–12:

Mass number = 12

Proton number = 6

Neutron number =?

Mass number = Proton + Neutron

12 = 6 + Neutron

Collect like terms

12 – 6 = Neutron

6 = Neutron

Neutron number = 6

SUMMARY:

Neutron number of carbon–14 = 8

Neutron number of carbon–12 = 6

Finally, we shall determine the difference in the neutron number. This can be obtained as follow:

Neutron number of carbon–14 = 8

Neutron number of carbon–12 = 6

Difference =?

Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)

Difference = 8 – 6

Difference = 2

Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)

Why would an airplane flying at 10,000 meters above the ground have more gravitational potential
energy than the same airplane flying at 1,000 meters above the ground?

Answers

Answer:

The gravitational potential energy (gpe) possessed by an object or body is directly proportional to the height of the object or body.

Explanation:

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Generally, the gravitational potential energy (gpe) of an object or body is directly proportional to the height of the object or body. Thus, the gravitational potential energy of a body increases as the height of the body increases.

In conclusion, an object with a higher height would have a higher gravitational potential energy.

what is gamma rays brust

Answers

Explanation:

A gamma - Ray Brust (GRB) is often the brightest source in the sky, often brighter than everything else combined.

Answer:

A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.

A ball is at the top of the hill. As the ball rolls down the hill, its total mechanical energy will:

Answers

Answer:

To explain what happens with the ball we must remember the Law of Conservation of Energy.

This law states that the energy can be neither created nor destroyed only converted from one form of energy to another.

Then,

At the top of the hill, the potential energy is maximum and the kinetic energy equals to zero.

When the ball starts to roll down the potential energy will be lower and the kinetic energy will have a low value.

At the middle of the hill, both energies have the same values.

At the end of the hill, the potential energy will be equal to zero and the kinetic energy will be maximum.

Correct me if I'm wrong...

hope it helps...

During the rainy season, we can observe lighting in the sky. Due to lighting, the atmospheric
nitrogen combines with atmospheric oxygen to form nitric oxide. Which among the following is a
correct statement concerning this process?

O A physical change has taken place in the atmosphere during lighting.
Ques
O No change occurred, as lighting is natural event.
A chemical change has taken place to form nitric oxide.
O All of the choices

Answers

Answer:

A chemical change has taken place to form nitric oxide.

A car moving along a racetrack has a centripetal acceleration of 11.3 m/ s2. If the speed of the car is 30.0 m/s, what is the distance between the car and the center of the track?

Answers

Answer:

Explanation:

The formula for centripetal acceleration is

[tex]a_c=\frac{v^2}{r}[/tex] and filling in:

[tex]11.3=\frac{(30.0)^2}{r}[/tex] and solving for r:

[tex]r=\frac{(30.0)^2}{11.3}[/tex] gives us that

r = 79.6 m

The area of a position-time graph is the

Answers

Answer:

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Explanation:

..........

how to solve for resistors

Answers

9514 1404 393

Answer:

A1 = 3A, A2 = 1.5A

Effective resistance = 2Ω

Explanation:

When the switch is closed, the voltage across each resistor is 6V, so the current through it (A2) is ...

A2 = 6V/(4Ω) = 1.5A

There are two parallel paths, each with that current, so the current from the battery is ...

A1 = A2 +A2 = 1.5A +1.5A = 3.0A

Then the effective resistance is ...

Reff = 6V/(3.0A) = 2.0Ω

The solution to the circuit is ...

A1 = 3A, A2 = 1.5A

Effective resistance = 2Ω

Example Problem
The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester

Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark

Answers

Answer:

Part A

a) F = -16x + 4, b) x = 0.25 m, c) STABLE

Explanation:

Part A

a) Potential energy and force are related

F = [tex]- \frac{dU}{dx}[/tex]- dU / dx

F = - (8 2x -4)

F = -16x + 4

b) The object is in equilibrium when the forces are zero

0 = -16x + 4

x = 4/16

x = 0.25 m

c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.

In this case there is only one equilibrium point

by changing the position a bit

x ’= x + Δx

we substitute

F ’= - 16 x’ + 4

F ’= - 16 (x + Δx) + 4

F ’= (-16x +4) - 16 Δx

at equilibrium position F = 0

F ’= 0 - 16 Δx

we can see that the body returns to the equilibrium position, therefore it is STABLE

PART B

This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved

initial instant. Before the shock

p₀ = m v

final instant. After the crash

p_f = (m + M) v_f

We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic

p₀ = pf

mv = mv ’+ M v_f

in the case of the elastic collision, the kinetic energy is conserved

K₀ = K_f

½ m v² = ½ m v’² + ½ M v_f²

we write the system of equations

mv = mv ’+ M v_f (1)

m (v² -v'²) = M v_f ²

m (v - v ’) = M v_f

m (v-v ’) (v + v’) = M v_f

v + v ’= v_f

we substitute in equation 1 and solve

v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]

v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]

the mechanical energy of the neutron is

initial

Em₀ = K = ½ m v²

final moment

Em_f = K + U = ½ m v_f ² + U

U is the energy lost in the collision

total energy is conserved

Em₀ = Em_f

½ m v² = ½ m v_f ² + U

U = ½ m (v² -v_f ²)

U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]

U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]

U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]

U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]

Let's do the same calculations for the nucleus

initial Em₀ = 0

final Em_f = K + U = ½ M v_f ² + U

Em₀ = Em_f

0 = K + U

U = -K

U = - ½ M v_f ²

U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²

U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]

We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.

b) the fraction of energy lost

f = U / Em₀

f = 4 m M / m + M

c) let's calculate the fraction of energy lost in a collision

m = 1.67 10⁻²⁷ kg

M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg

f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷

f = 5.92 10⁻²⁷ J

the energy of a fast neutron is greater than 1 eV

Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J

Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo

#_collisions = 0.95 Eo / f

#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷

#_collisions = 2.7 10⁷ collisions

How long will it take a car to acceleration from 15.2ms to 23.Ms if the car has an average acceleration of 3.2m\s

Answers

Answer: 2.43 s

Explanation:

Initial velocity is [tex]u=15.2\ m/s[/tex]

Final velocity [tex]v=23\ m/s[/tex]

Average acceleration is [tex]a_{avg}=3.2\ m/s[/tex]

Average acceleration is change in velocity in the given amount of time

[tex]\therefore a_{avg}=\dfrac{v-u}{t}\\\\\Rightarrow 3.2=\dfrac{23-15.2}{t}\\\\\Rightarrow t=\dfrac{7.8}{3.2}\\\\\Rightarrow t=2.43\ s[/tex]

Thus, 2.43 s is required to acquire that average acceleration with 23 m/s velocity .

what is physical change ?

Answers

Answer:

Physical Changes :- The substance in which no new substance is formed are called physical changes.

The molecular composition of the substance are totally same.

For example :- Crushing a mineral into powder.

what happens to gravitational force when distance is quarter from the original
please answer it faster help me

Answers

Answer:

The force of gravitational attraction between them also decreas

A car moving east at a velocity of 16.0 m/s collides with a stationary truck with exactly twice the mass. If the two vehicles lock together, calculate the velocity of their combined mass immediately after collision

Answers

Answer:

5.33ms-¹

Explanation:

that is the procedure above

A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s2. How much time is required for the car to reach a speed of 90 km/h

Answers

Answer:

Time, t = 4.165 seconds.

Explanation:

Given the following data;

Initial velocity = 60 km/h

Final velocity = 90 km/h

Acceleration = 2 m/s²

Conversion:

60 km/hr to meters per seconds = 60*1000/3600 = 60000/3600 = 16.67 m/s

90 km/hr to meters per seconds = 90*1000/3600 = 90000/3600 = 25 m/s

To find the time, we would use the first equation of motion;

[tex] V = U + at[/tex]

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Making time, t the subject of formula, we have;

[tex] t = \frac{V - U}{a}[/tex]

Substituting into the equation, we have;

[tex] t = \frac{25 - 16.67}{2}[/tex]

[tex] t = \frac{8.33}{2}[/tex]

Time, t = 4.165 seconds.

Question 23 of 23
Suppose a current flows through a copper wire. Which two things occur?
O A. The field is parallel to the direction of flow of the current.
B. An electric field forms around the wire.
OC. A magnetic field forms around the wire.
U
D. The field is perpendicular to the direction of flow of the current.
SUBM

Answers

Answer:

The field is parallel to the direction of flow of the current.

Define Circular motion and its types !

Answers

[tex] \green{\huge{\red{\boxed{\green{\mathfrak{QUESTION}}}}}} [/tex]

Define Circular motion and its types

[tex] \huge\green{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}}}[/tex]

[tex]{ \red{ \bold {\mathbb {\textit{Circular \: motion}}}}}[/tex]

MOTION OF THE BODY CIRCULAR PATH IS CALLED CIRCULAR MOTION

‌A tengential is always responsible for change in speed‌If the Direction of acc and velocity is same then body's speed inc‌If they r in oppsite direction then Body's speed dec .‌If speed becomes constant then tangential acc become zero .‌By the basis of tangial acc the circular motion is divided into two types uniform circular motion and non uniform circular motion

[tex]{ \blue {\bold{UNIFORM \: CIRCULAR}}} \\ \green{ \bold{ MOTION}}[/tex]

‌when object in a circular path in a constant speed or constant circular speed is called uniform circular motion. ‌In case of uniform circular motion acc tendial will be zero ‌In case of uniform circular motion acc centripetal will be equal to net acc.

[tex] { \green{ \bold{ NON UNIFORM}}} \\ {\blue{ \bold{CIRCULAR \: MOTION}}}[/tex]

‌Centripetal acc is not equal net acc ‌Circular acc and tendial acc is not equal to zero .

circular motion is a movement of an object along the circumference of a circle or rotation along a circular path.
They are two types named Uniform circular motion and Non-uniform circular motion.

Tony walks at an average speed of 70 m/min from his home to
school. If the distance between his home and the school is
2100 m, how much time does it take for Tony to walk to
school?
min

Answers

Answer:

The answer is 30 min

Explanation:

t = s/v

v = 70 m/min, s = 2100 m

t = 2100/70 = 30 min.

Hope it helps you! ＼(^ᴥ^)／

5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura

Answers

Answer:

1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.

2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.

3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.

4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.

5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.

6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.

7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.

8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.

9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.

10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.

11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.

12) Quemar basura - Cambio químico - Reacción de combustión.

Explanation:

A continuación, veremos que representa cada caso: