How does a circuit breaker protect a refrigerator?

Answer:

The given potential energy of the spring is expressed as follows;

ΔUsp = -(WB + WW)

Where;

WB = Th work done by the spring on the block to which it is attached

WW = The work done by the spring on the wall

We recall that work done, W = Force applied × Distance moved in the direction of the force

The work done by the spring on the block, WB = The spring force × The distance the block moves

The work done by the spring on the wall, WW = The spring force × The distance the wall moves

However, given that the wall does not move, we have;

The distance the wall moves = 0

∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J

Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps

Explanation:

Answer:

B. Current in the circuit is 5.

Ci. Voltage across 7.2 Ω (R₁) is 36 V

Cii. Voltage across 6.2 Ω (R₂) is 31 V

Ciii. Voltage across 8.6 Ω (R₃) is 43 V

Explanation:

We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 7.2 Ω

Resistor 2 (R₂) = 6.2 Ω

Resistor 3 (R₃) = 8.6 Ω

Equivalent Resistance (R) =?

Since the resistors are in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 7.2 + 6.2 + 8.6

R = 22 Ω

B. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 22 Ω

Current (I) =?

V = IR

110 = I × 22

Divide both side by 22

I = 110 / 22

I = 5 A

Therefore, the current in the circuit is 5.

Ci. Determination of the voltage across 7.2 Ω (R₁)

Resistor 1 (R₁) = 7.2 Ω

Current (I) = 5 A

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 5 × 7.2

V₁ = 36 V

Therefore, the voltage across 7.2 Ω (R₁) is 36 V

Bii. Determination of the voltage across 6.2 Ω (R₂)

Resistor 2 (R₂) = 6.2 Ω

Current (I) = 5 A

Voltage 2 (V₂) =?

V₂ = IR₂

V₂ = 5 × 6.2

V₂ = 31 V

Therefore, the voltage across 6.2 Ω (R₂) is 31 V

Ciii. Determination of the voltage across 8.6 Ω (R₃)

Resistor 3 (R₃) = 8.6 Ω

Current (I) = 5 A

Voltage 3 (V₃) =?

V₃ = IR₃

V₃ = 5 × 8.6

V₃ = 31 V

Therefore, the voltage across 8.6 Ω (R₃) is 43 V

Answer:

hard2

Explanation:

Answer:

A and C

Explanation:

Covalent bonding involves sharing by the atoms involved

Statements that can be considered as true statement about covalent bonding between two atoms are:

A. Electrons are shared.

B. The electronegativities of the two atoms are close to each other.

C. The two atoms can be of the same element.

Therefore, option A,B,C are correct.

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Answer:

P = V * Q      potential energy = potential * charge

V = =3.09 J / 6.93 * 10E-4 C = 4460 Joules / Coulomb

The electric potential, V at the point given the data from the question is –4458.87 V

The electric potential or electromotive force (EMF) is defined as the energy supplied by a battery per unit charge. Mathematically, it can be expressed as:

Electromotive force (EMF) = Work (W) / charge (Q)

V = EMF = W / Q

V = W / Q

V = –3.09 / 6.93×10⁻⁴

V = –4458.87 V

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Answer:

Explanation:

During an earthquake, there are both transverse and longitudinal wave. The answer is c.

Answer:

Explanation:

ans is c. Both transverse and longitudinal

Answer:

Option B

Explanation:

The reliability of an experiment can be increased by conducting experiment sufficient number of times keeping other things constant. Hence, the 10 p coin needs to be flipped again to measure reliability for more than one times.

Hence, option B is correct

Answer:

Electromotive force or EMF is equal to the terminal potential difference when no current flows. EMF (ϵ) is the amount of energy (E) provided by the battery to each coulomb of charge (Q) passing through.

Answer:

k = 933.88 N/m

Explanation:

Given that,

The mass of a load, m = 8.1 kg

The displacement in the spring, x = 8.5 cm

We need to find the force constant of the spring. We know that Hooke's law is given by :

F = -kx

Also,

F = mg (Force due to its weight)

So,

[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{8.1\times 9.8}{0.085}\\\\k=933.88\ N/m[/tex]

So, the force constant of the spring is equal to 933.88 N/m.

Answer:

Explanation:

To find the direction of this vector we need o find the angle that has a tangent of the y-component over the x-component:

[tex]tan^{-1}(\frac{11.1}{-43.0})=-14.5[/tex] but since we are in Q2 we have to add 180 degrees to that angle giving us 165.5 degrees

Answer:

Capacitance is a derived physical quantity measured in farad

Answer:

"Farad" is another term for the coulomb/volt measurement of capacitance, so both of those options are the correct answer. Amperes and coulomb/second are measurements of electric flow, in other words, how strong a current is.

Answer:

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

Explanation:

La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.

Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:

a ⇒ b

c ⇒ x

Entonces [tex]x=\frac{c*b}{a}[/tex]

La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?

1 km ⇒ 3280.84 pies

12 km ⇒ x

[tex]x=\frac{12 km*3280.84 pies}{1 km}[/tex]

x= 39370.08 pies

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

Answer:

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Explanation:

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Answer:

so 9/3=3 current is 3 amperes

Explanation:

The fomula to calculate resistance is:

voltage/cutrent

9 V/3 A= 3 ohms

Answer:

a. Acceleration, a = 0.28 m/s²

b. Distance, S = 156 meters

Explanation:

Given the following data;

Initial velocity = 30 km/h

Final velocity = 45 km/h

Time = 15 seconds

a. To find the acceleration;

Conversion:

30 km/h to m/s = 30*1000/3600 = 8.33 m/s

45 km/h to m/s = 45*1000/3600 = 12.5 m/s

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation;

[tex]a = \frac{12.5 - 8.3}{15}[/tex]

[tex]a = \frac{4.2}{15}[/tex]

Acceleration, a = 0.28 m/s²

b. To find the distance travelled, we would use the second equation of motion given by the formula;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

[tex] S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}[/tex]

[tex] S = 124.5 + 0.14*225[/tex]

[tex] S = 124.5 + 31.5 [/tex]

S = 156 meters

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

Answer: 400 m

Explanation:

Vf= 20 + (4*10)

Vf= 60 [m/s]

x= (60^2 - 20^2) / (2*4)

x= 400 m

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

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Answer:

Explanation:

Im going to be using the rules for significant digits properly so I hope you're quite familiar with them. The equation we need for this is

F - f = ma where F is the applied force (our unknown), f is the frictional force, m is the mass, and a is the acceleration. Filling in:

F - 18.2 = 2.25(1.50) and

F = 2.25(1.50) + 18.2  Do the multiplication first and round to get

F = 3.38 + 18.2   The addition rules tell us that we will be rounding to the tenths place after we add to get

F = 21.6 N

Answer:

Explanation:

F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?

We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that

15 = 5a so

a = 3

Then the F is cut into thirds, so

5 = 5a so

a = 1

The second acceleration is one-third of the first one, where the Force is 3 times greater.

[tex]\huge{ \mathcal{  \underline{ Answer }: }}[/tex]

[tex] \boxed{ \mathrm{Work done = Force × displacement}}[/tex]

Therefore the distance covered :

[tex] \large \boxed{50 \: m}[/tex]

_____________________________

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Answer:

magnification is 4

Explanation:

m= image height / object height

m= 12/3

m= 4

Answer:

Explanation:

1 mole of gas contains 6.02x10^23 molecules. That is the Avogadro's number. To find the number of moles, you divide the number of molecules by the Avogadro's number.

The correct answer is C.

Answer:

Explanation:

ans is C. divided by Avogadro no.

Answer:

20kgm/s

Explanation:

p=mv

p=5*4

p=20kgm/s

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

Answer:

Solution given:

strength of electric field [E]:?

distance[d]:0.2m

charge[Q]=[tex]1.56×10^{-6} C[/tex]

we have

E =K [tex] \frac{Q}{r²}[/tex]

E=[tex]9×10^{9} C[/tex][tex] \frac{1.56×10^-6 C}{0.2²} [/tex]

E=[tex] 3.51×10^{5}[/tex]V/m

strength of electric field : [tex] 3.51×10^{5}[/tex]V/m

Answer:

D.)

Explanation:

the current separates on each branch according to the resistance it experience.

Answer:D

Explanation:

Answer:

I am not sure if this is the answer

(B) what is the height of the building if it hits the ground after those 5 seconds.