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olya-2409 [2.1K]

4 years ago

10

The power dissipated in each of two resistors is the same. The current across resistor A is triple that across resistor B. If th

e resistance of resistor B is R, what is the resistance of A?

1 answer:

Svetllana [295]4 years ago

5 0

Answer:

Explanation:

this is the answer to your question

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Who are the scientists given credit for discovering electrons, protons, and neutrons?

nydimaria [60]

J.J. Thompson is the scientist who recieved credit for discovering them.

7 0

3 years ago

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0

3 years ago

Some homes that use baseboard heating use copper tubing. hot water runs through and heats the copper tubing, which in turn heats

AlekseyPX

When you heat a certain substance with a difference of temperature $\Delta T$ the heat (energy) you must give to it is
$E(=Q) =mc\Delta T$
where $c$ is the specific heat of that substance (given in J/(g*Celsius))
In this case
$E=645*0.3850*(28.22-13.20) =3729.8 (Joule)$

Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a difference of 1 degree Celsius = 1 degree Kelvin

7 0

3 years ago

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em> after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

$x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}$

$x_{cm} = \dfrac{8 \times0 + 16 \times 30}{8 + 16} = 20$

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

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4 0

3 years ago

How much heat is needed to raise the temperature of 50.0 g of water by 25.0°C

love history [14]

Answer:

Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

4.18

J

g

∘

C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

1 g

of that substance by

1

∘

C

.

In water's case, you need to provide

4.18 J

of heat per gram of water to increase its temperature by

1

∘

C

.

What if you wanted to increase the temperature of

1 g

of water by

2

∘

C

? You'd need to provide it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

=

increase by 2

∘

C



2

×

4.18 J

To increase the temperature of

1 g

of water by

n

∘

C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

To cause a

1

∘

C

increase in the temperature of

m

grams of water, you'd need to supply it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

for m g of water



m

×

4.18 J

This means that in order to increase the temperature of

m

grams of water by

n

∘

C

, you need to provide it with

heat

=

m

×

n

×

specific heat

This will account for increasing the temperature of the first gram of the sample by

n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

, and so on until you reach

m

grams of water.

And there you have it. The equation that describes all this will thus be

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q

=

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

50.0

−

25.0

)

∘

C

q

=

10,450 J

Rounded to three sig figs and expressed in kilojoules, t

Explanation:

3 0

3 years ago

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