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2 years ago

When an ideal diatomic gas is heated at constant

Physics

1 answer:

ira [324]2 years ago

7 0

Answer:

Q=∆U+W

Explanation:

work done+ change in internal energy = heat supplied to change the internal energy

(1st law of thermodynamics)

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You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too

AleksandrR [38]

Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²

The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)

3 0

2 years ago

A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

3 0

2 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

2 years ago

Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge b

Elena L [17]

Answer:

a) 0.167 μC/m^2

b) 1.887 * 10^4 V/m

Explanation:

Hello!

First let's find the surface charge density:

Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:

σ = Q/A

Where A is the area of the satellite, which is:

A=4πr^2 = πd^2 = π(1.9m)^2

Therefore:

σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2

Now let's calculate the electric field

Just outside the surface of the satellite the elctric field will be:

E = σ/ε0

Where ε0=8.85×10^−12 C/Vm

Therefore:

E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m

E = 1.887 * 10^4 V/m

5 0

2 years ago

In a chemical reaction, the total charge of the reactants must be

MA_775_DIABLO [31]

the same with that of products

Explanation:

In a chemical reaction, the total charge of the reactants must be the same with that of products.

Charges must be conserved or balanced in chemical reactions.

In both acidic and basic/neutral medium electrons are used to balance the charge.
The appropriate number of electrons is added to the side with a larger charge.
One electron is used to balance each positive charge.
This ensures that the sum of charges on both sides the same.

Learn more:

Balanced equation brainly.com/question/5297242

#learnwithBrainly

4 0

2 years ago