how to answer thiss helppp​

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:

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.

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.

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]

Answer:

The second throw shows both accuracy and precision.

Explanation:

The second throw shows both accuracy and precision because he knew in which degree and the amount of force that can be applied to the dart so that it hits the center of the bullseye or close to it. Due to knowing of these two things the Carlos is able to hit the dart close to the target at the second time so in this way he got both accuracy and precision at the second turn of throwing darts.

Answer:

B-The second throw shows accuracy but poor precision.

Explanation:

Answer:

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

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

[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

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

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

Answer:

First Condition of Equilibrium

For an object to be in equilibrium, it must be experiencing no acceleration. This means that both the net force and the net torque on the object must be zero. Here we will discuss the first condition, that of zero net force. ... Fnet=0 F net = 0.

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.

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)

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

Explanation:

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.

Answer:

The force of gravitational attraction between them also decreas

Answer:

d=1/2 (a)(t^2)

D = distance

A = acceleration

T = time

acceleration due to gravity is 32 ft/second

so, d=1/2 (32)(5^2)

d=16(25)

d=400

Explanation:

Answer:

400 ft.

Explanation:

D= 1/2 gt^2

=1/2(-32 ft/sec^2)(5 sec^2)

= -(1/2)(32)(25) ft

D= -400 ft, down

Answer:

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

Answer is A it will increase

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^)／

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

[tex]F_{g0}=G \frac{mM_{E0}}{r^{2}}[/tex]

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=2M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{2mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=2[/tex]

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=3M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{3mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=3[/tex]

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

[tex]M_{Ef}=\frac{M_{E0}}{4}[/tex]

so:

[tex]F_{gf}=G \frac{mM_{E0}}{4r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{1}{4}[/tex]

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

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

Answer:

C. Basic research provides fundamental knowledge that can be used to conduct applied research.

Explanation:

Applied research is used to solve practical and specific problems. Basic research is used to advance theories and scientific knowledge; it might not result in an invention or a solution, but can be used to gain knowledge about a specific subject.

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.

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.

Answer:

second column

Explanation:

Answer:

Trial 3 is the answer.

Explanation:

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

Answer:

The ray box used as the source of light

Answer: its A

Explanation: i just did the test and got it right also known. The ray used as the source of light

Answer:

A chemical change has taken place to form nitric oxide.

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.

Answer:

5.33ms-¹

Explanation:

that is the procedure above

Answer:

b. they move in convection currents.

Explanation:

i learned this in 4th grade

Answer:

4V

Explanation:

First, we calculate the total resistance to the given battery cell of emf 5V. The total resistance is the sum of all the resistance in the cell i.e.

Total resistance = 2Ω + 8Ω = 10Ω

Using ohms law equation to calculate the current passing through the battery cell:

V = IR

Where; V = voltage, I = current, R = resistance

5V = I × 10Ω

I = 5/10

I = 0.5A

Terminal voltage is calculated by the us of the following equation:

V=emf−IR

Where; R is internal resistance

V = 5 - (0.5 × 2)

V = 5 - 1

V = 4V

Therefore, the potential difference across the terminals of the battery cell is 4V

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 .