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puteri [66]
3 years ago
9

Melting Ice is what kind of reaction

Physics
1 answer:
Vesnalui [34]3 years ago
7 0

Answer:

Answer is Endothermic Reaction

Explanation:

Basically, melting ice is an endothermic reaction because the ice absorbs (heat) energy, which causes a change to occur.

I hope it's helpful!!

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g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resi
Anettt [7]
<h2>Question:</h2>

In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Answer:

9.1Ω

Explanation:

The circuit diagram has been attached to this response.

(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e

\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}

=> R_{X} = \frac{R_1 * R_2}{R_1 + R_2}             ------------(i)

From the question;

R1 = 3Ω,

R2 = 7Ω

Substitute these values into equation (i) as follows;

R_{X} = \frac{3 * 7}{3 + 7}

R_{X} = \frac{21}{10}

R_{X} = 2.1Ω

(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.

Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e

R = Rₓ + R3

Rₓ = 2.1Ω

R3 = 7Ω

=> R = 2.1Ω + 7Ω = 9.1Ω

Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω

4 0
3 years ago
As galáxias são maiores do que os enxames de estrelas?
GuDViN [60]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Responder \;\;a}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

As galáxias são geralmente maiores do que os aglomerados de estrelas. Como disse Geller ~ As galáxias são como as cidades em que vivem os aglomerados de estrelas. As galáxias podem ter cerca de milhares ou mais aglomerados de estrelas ~

I hope it helps ~

8 0
3 years ago
What is meant by the term blue moon
Andreyy89
“Once in a blue moon” is a common expression that has been used for a long time, and which means 'not very often,' or 'very rarely. ' It often refers to an extra full moon; however, it has been used to describe the way the moon actually looked, when for different reasons it had turned a blueish color.
www.loc.gov › astronomy › item
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4 0
3 years ago
A charged particle is injected at 211 m/s into a 0.0633‑T uniform magnetic magnetic field perpendicularly to the field. The diam
lakkis [162]

Answer:

85.68*10^3C/kg

Explanation:

For this problem we need the concept about Force in a Magnetic field,

For definition we know that

F=m\frac{v^2}{r}

Where v is the velocity, m the mass and r the radius or distance between the two points.

We know as well that

F = qvB

where q is the charge of a proton

v the velocity and B the magnetic field, then matching the two equation,

qvB=m\frac{v^2}{r}

Re-arrange for q/m (charge to mass ratio)

\frac{q}{m} = \frac{v}{Br}

Our values are,

v=211m/s

B= 0.0633T

r=0.0389m

Substituting,

\frac{q}{m} = \frac{211}{0.0633*0.0389}

\frac{q}{m} = 85689.8 C/kg = 85.68*10^3C/kg

7 0
3 years ago
What would your estimate be for the age of the universe if you measured hubble's constant to be 11 kilometers per second per mil
vredina [299]

By using Hubble's constant H₀, the estimated age of the universe would be 27,269,816,110  years.

As we know that the age of the universe is something near to the time the galaxies needed to reach their current distance:

T = D/V

where T is the time the galaxies needed, D is distance and V is speed.

By using Hubble's law we can use the equation

V = H₀*D

where H₀ is Hubble's constant

By combining these 2 equations, we get

T = D/(H₀*D) = 1/H₀ .............(A)

In conclusion, the age of the universe is something near the inverse of Hubble's constant.

From the question above, we know that:

H₀ = 11 km/s/Mly

By using equation A we get

T = 1/(11 km/s*Mly)

T = (1/11)  s*Mly/km

Convert 1 million light-years to km by (1 ly = 9.461*10^12 km)

T = (1/11) s*Mly/km

T = (1/11)*9.461*10^18 s

T = 8.6009*10^17 s

Convert to years by (1 year = 3.154*10^7 s)

T = 8.6009*10^17 s

T = (8.6009*10^17 s/3.154*10^7 s)* 1 yea

T = 27,269,816,110 years.

Learn more about Hubble's constant at: brainly.com/question/27051032

#SPJ4

6 0
2 years ago
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