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Verizon [17]
3 years ago
9

2) How much work is required to pull a sled 15 meters if you use 30N of force?

Physics
1 answer:
geniusboy [140]3 years ago
7 0

2 people

Explanation:

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2,4,5

Air pressure is created by the weight of the atmosphere pushing on Earth’s surface.

Denser air is heavier than less dense air.

Air is less dense at higher altitudes.

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A primary succession does not include any invasive species <br><br> True or false
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What is the change in internal energy if 60 J of heat is released from a system and 30 J of work is done on the system? Use U =
k0ka [10]

The change in internal energy of the system is +30 J

Explanation:

We can solve this problem by using the first law of thermodynamics, which states that the change in internal energy of a system is given by the equation:

\Delta U = Q -W

where

\Delta U is the change in internal energy

Q is the heat absorbed by the system (positive if it is absorbed, negative if it is released)

W is the work done by the system (positive if it is done by the system, negative if it is done by the surroundings on the system)

Therefore, in this problem, we have

Q=-60 J (heat released by the system)

W=-30 J (work done on the system)

Therefore, the change in internal energy is

\Delta U = -60 - (-30) = +30 J

Learn more about thermodynamics:

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3 years ago
Which statement is true regarding the vectors shown.
kap26 [50]

Answer:

Y-> + F-> +G ->=E->

Explanation:

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3 years ago
Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure ne
ivann1987 [24]

Answer:

Energy = 7.83 x 10⁻¹⁹ J

Energy = 6.63 x 10⁻¹⁹ J

Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

Energy = \frac{hc}{\lambda}\\

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ Js

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light

Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\

<u>Energy = 7.83 x 10⁻¹⁹ J</u>

<u></u>

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\

<u>Energy = 6.63 x 10⁻¹⁹ J</u>

7 0
3 years ago
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