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KIM [24]
3 years ago
7

Can the absolute energy states of the reactants (E) and products (E2) be measured in a chemical reaction?

Physics
1 answer:
ASHA 777 [7]3 years ago
8 0

When the products of a reaction have a lower enthalpy than the reactants: The reaction is endothermic. ... Can the absolute energy states of the reactants (E1) and products (E2) be measured in a chemical reaction?

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If a car skids to a stop in 3.8s while undergoing a uniform acceleration of -9.55 m/s2, what is the car’s velocity?
Lapatulllka [165]

Answer:

36.29 m/s

Explanation:

We know from theory that v= v_0 +a\Delta t. Let's replace the value we know in our equation and solve for the only value left.

0 = v_0 - 9.55 \cdot 3.8 \\ v_0 = 9.55 \cdot 3.8 = 36.29 m/s

3 0
2 years ago
A period of one year on Earth is the time it takes Earth to
Kazeer [188]

D make one revolution around the sun hope this helps

7 0
3 years ago
A car accelerates at a rate of 5ft/s/s for a time of 9 seconds. How far does the car go?
vlada-n [284]

The car will move in a speed of 45 meter per second

4 0
3 years ago
Read 2 more answers
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
2 years ago
The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2
Artyom0805 [142]

Right, as you mentioned in the comments, you find d by plugging in the different values of t.

For t=1\,\mathrm s, we have

d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1\,\mathrm s)^2

d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(1\,\mathrm s^2\right)

d=4.905\,\mathrm m

Similarly, for t=2\,\mathrm s, you get

d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(2\,\mathrm s\right)

d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(4\,\mathrm s^2\right)

d=19.62\,\mathrm m

8 0
3 years ago
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