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mote1985 [20]
3 years ago
14

A chemical reaction occurs. Which of the following would indicate that energy is transformed during the reaction?

Physics
2 answers:
tresset_1 [31]3 years ago
6 0

all of these ;) HOPE IT HELPS


weqwewe [10]3 years ago
4 0
The answer would be 4. All of the above

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You used the right-hand rule to determine the z component of the angular momentum, but as a check, calculate in terms of positio
KengaRu [80]

Answer:

case x py   L is in the positive z direction

case y px   L the negative z direction

Explanation:

The angular amount is defined by the relation

         L = r x p

the bold are vectors, where r is the position vector and p is the linear amount vector.

The module of this vector can be concentrated by the relation

         L = r p sin θ

the direction of the vector L can be found by the right-hand rule where the thumb points in the direction of the displacement vector, the fingers extended in the direction of the moment p which is the same direction of speed and the palm points in the direction of the angular momentum L

in the case x py

the thumb is in the x direction, the fingers are extended in the direction and the palm is in the positive z direction

In the case y px

the thumb is in the y direction, the fingers are in the x direction, the palm is in the negative z direction

7 0
3 years ago
A block slides down an incline plane that makes a 30 degree angle with the
Alona [7]

Hi there!

We know that:

Force due to gravity = Mgsinθ

Force due to friction = μMgcosθ

Let the positive direction be directed in the direction of the block's acceleration, which is downward.

Thus:

ΣF = Mgsinθ - μMgcosθ

Solving for acceleration requires diving all terms by the mass, so:

a = gsinθ - μgcosθ

Substitute in given values. (g = 9.8 m/s²)

a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²

7 0
3 years ago
A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp
Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

4 0
4 years ago
A student thinks that any real vibration must be damped. Is the student correct? If so, give convincing reasoning. If not, give
Stels [109]

Answer:

Yes the student is correct

Explanation:

The first law of thermodynamics states that energy can neither be created nor destroyed

The second law of thermodynamics states that the entropy (disorderliness) of an isolated system always increases

Therefore, whereby energy is not supplied to maintain the orderly oscillatory motion with constant amplitude, the amplitude of the system is bound to reduce with time that is the vibration of the system must be damped

6 0
3 years ago
in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's
GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}

Where

  • I₁ = Intensity at distance 1 (W/m²)
  • I₂ = Intensity at distance 2 (W/m²)
  • d₁ = distance 1 from a light source (m)
  • d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}

\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}

\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}

I_{2} = \frac{200} {36}

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

Learn more about intensity of light here:

brainly.com/question/13155277

#SPJ4

3 0
1 year ago
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