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andreyandreev [35.5K]

3 years ago

14

When you hold a frozen banana treat in your hands, which statement BEST describes what is taking place? A. Heat is flowing from

you to the banana. B. Heat is flowing from the banana to you. C. Cold is flowing from you to the banana. D. Cold is flowing from the banana to you.

1 answer:

Andrej [43]3 years ago

3 0

Answer:

The answer is A

Explanation:

It is A because your body heat is warmer than the banana and when you hold it the heat is transferring over.

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If the equation on the board had shown 3 atoms of carbon on the reactants side, how many atoms of carbon would need to be repres

Sholpan [36]

3

Explanation:

If the equation on the board had shown 3 atoms of carbon on the reactants side, it would need 3 atoms of carbon on the product side to balance it.

This is in accordance to the law of conservation of mass which states that "matter is neither created nor destroyed during the course of chemical reaction, they are only transformed from one form to another".

We expect the same number of atoms on both sides of the equation. For example:

Fe + S → FeS

1 iron atom combines with 1 sulfur atom to produce FeS which contains 1 iron and sulfur atoms.

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6 0

2 years ago

Read 2 more answers

A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cros

EleoNora [17]

Answer:

The ratio of the young's modulus of steel and copper is $1.79\times10^{-2}$

(c) is correct option.

Explanation:

Given that,

Length of steel wire = 4.7 m

Cross section $A = 3\times10^{-3}\ m^2$

Length of copper wire = 3.5 m

Cross section $A = 4\times10^{-5}\ m^2$

We need to calculate the ratio of young's modulus of steel and copper

Using formula of young's modulus for steel wire

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$

$Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}$ ....(I)

The young's modulus for copper wire

$Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}$ ....(II)

From equation (I) and (II)

The ratio of the young's modulus of steel and copper

$\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}$

$\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}$

$\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}$

$\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}$

Hence, The ratio of the young's modulus of steel and copper is $1.79\times10^{-2}$

5 0

3 years ago

A Carnot engine performs 2.5 * 104 J of work in each cycle and has an efficiency of 66%. (a) How much heat does the engine extra

Natalka [10]

a.

The heat the engine <em>extracts </em>from its heat source in each cycle is 3.79 × 10⁴ J

The efficiency of the Carnot engine, ε = W/Q where W = work done per cycle = 2.5 × 10⁴ J and Q = heat extracted in each cycle.

Making Q subject of the formula, we have

Q = W/ε

Since ε = 66% = 0.66, we substitute the values of the variables into the equation for Q.

So, Q = W/ε

Q = 2.5 × 10⁴ J/0.66

Q = 3.79 × 10⁴ J

So, the heat the engine <em>extracts </em>from its heat source in each cycle is 3.79 × 10⁴ J

b.

The temperature of its heat source is 588.76 °C

Also, the efficiency of the Carnot engine, ε = 1 - T/T' where T = temperature of exhaust heat = 20° C = 273 + 20 = 293 K and T' = temperature of heat source.

Making T' subject of the formula, we have

T' = T/(1 - ε)

Since ε = 66 % = 0.66

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

T' = T/(1 - ε)

T' = 293 K/(1 - 0.66)

T = 293 K/0.34

T = 861.76 K

We convert this to Celsius.

T(K) = T(°C) + 273

T(°C) = T(K) - 273

T(°C) = 861.76 - 273

T(°C) = 588.76 °C

So, the temperature of its heat source is 588.76 °C

Learn more about Carnot engine here:

brainly.com/question/13170743

5 0

2 years ago

A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.

Bond [772]

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

4 0

3 years ago

Calculate the wavelength in centimeters of radar energy at a frequency of 10 GHz. What is the frequency in gigahertz of radar en

ddd [48]

Answer:

Energy will be equal to $6.6\times 10^{-27}J$

Frequency will be equal to $12\times 10^8Hz$

Explanation:

We have given frequency of the radar f = 10 GHz $=10\times 10^6Hz$

Speed of light $c=3\times 10^8m/sec$

Plank's constant $h=6.6\times 10^{-34}js$

So energy $E=h\nu$ ,here h is plank's constant and $\nu$ is frequency

So energy $E=6.6\times 10^{-34}\times 10^{7}=6.6\times 10^{-27}J$

In second case we have given wavelength = 25 cm = 0.25 m

Wavelength is equal to $\lambda =\frac{c}{f}$

So $f=\frac{c}{\lambda }=\frac{3\times 10^8}{0.25}=12\times 10^8Hz$

So frequency will be equal to $12\times 10^8Hz$

8 0

3 years ago