Ask question

Ask question

All categories

3 years ago

9

The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary t

o increase the object's temperature from 6°C to 12°C? 4Q 2Q Q 3Q

2 answers:

Tresset [83]3 years ago

5 0

Answer:

The quantity of heat required to increase the temperature of the object from 6°C to 12°C is 3Q

Explanation:

Heat capacity is the quantity of heat required to increase the temperature of an object.

Q = mcΔθ

where;

Q is the quantity of heat

m is the mass of the object

c is specific heat capacity of the object

Δθ is change in temperature = T₂ - T₁

For the first sentence of this question;

Q = mc(6-4)

Q = mc(2)

Q = 2mc

For the second sentence of this question;

Let Q₂ be the quantity of heat required to increase the temperature of the object from 6°C to 12°C

Q₂ = mcΔθ

Q₂ = mc(12-6)

Q₂ = mc(6)

Q₂ = 6mc

Q₂ = 3(2mc)

Recall, Q = 2mc

Thus, Q₂ = 3Q

bixtya [17]3 years ago

4 0

Answer:

The answer is 3Q

Explanation:

The metal temperature increases in a linear way, we could get a difference between final and initial temperature

$DT=FinalTemperature-InitialTemperature$

We get a temperature difference of 2 degrees per each heat addition. If we add the same heat 3 times more, it will increase to 12 degrees

You might be interested in

PLEASE ANSWER QUICKLY

Rom4ik [11]

Answer:

a luminous ball of plasma

6 0

3 years ago

A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme

Shkiper50 [21]

Answer: The density of this piece of jewelry is $8.90g/cm^3$

Explanation:

To calculate the density, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = $14.7cm^3$ $1cm^3=1ml$

Putting values in above equation, we get:

$\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3$

Thus density of this piece of jewelry is $8.90g/cm^3$

8 0

3 years ago

The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

3 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

$F = \dfrac{d^4G\delta}{8D^3N}$

$\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm$

$F = \dfrac{d^4G}{8D^3}\times 0.004$

$F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004$

F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

$\tau = \dfrac{8FDk_s}{\pi d^3}$

$\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}$

= 442.6 Mpa

The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa

now,

$\tau_s \leq 0.45 S_u$

$\tau_s \leq 0.45 \times 1300$

$\tau_s \leq 585\ Mpa$

since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

6 0

3 years ago

Read 2 more answers

When a charge of 8 C flows past any point along a circuit in 2 seconds, the current is ________ A?

RoseWind [281]

The answer would be 4 A.

7 0

3 years ago