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zhenek [66]
4 years ago
15

Part A:

Physics
2 answers:
kobusy [5.1K]4 years ago
7 0
In a perfectly ELASTIC collision between two perfectly rigid objects <span>both the momentum and the kinetic energy of the system are conserved.hope it helps</span>
skad [1K]4 years ago
6 0

Answer:

both the momentum and the kinetic energy of the system are conserved.hope it helps

Explanation:

You might be interested in
Calculate the longest wavelength visible to the human eye 164 kj mol
const2013 [10]
The working equation to be used here is the Planck's equation. This was derived using the wave behavior theory of the light and electromagnetic waves. According to this equation, electron transfer from orbital to orbital in discrete packets of energy called quanta. When an electron moves to a higher energy level, it absorbs energy. On the other hand, when it lowers to an energy level, it releases energy by emitting light. Hence, the wavelength of the light or magnetic wave can be determined.

E = hν = hc/λ, where ν is the frequency, λ is the wavelength, h is the Planck's constant equal to 6.626×10⁻³⁴ J-s and c is the speed of light equal to 3×10⁸ m/s.

Knowing the energy to be 164 kJ or 164,000 J, the wavelength is equal to

164,000 = (6.626×10⁻³⁴)(3×10⁸ m/s)/λ
λ = 1.212×10⁻³⁰ meters

8 0
4 years ago
In the example given below, Aaron applies a force of 300N and Bob applies a force of 450N :
garri49 [273]

Answer:

Explanation:

This problem is all about torque. The "rules" are that in order for a system to be in rotational equilibrium, the sum of the torques on the system have to equal 0 (in other words, they have to equal each other {cancel each other out}). The equation for torque is

τ = F⊥r where τ is torque, F⊥ is the perpendicular force, and r is the lever arm length in meters. We also have to understand that in general Forces moving clockwise are negative and Forces moving counterclockwise are positive. Now we're ready for the problem:

A. The counterclockwise torque:

τ = 300(3) so

τ = 900N*m

B. The clockwise torque:

τ = -450(2.5) so

τ = -1100N*m

C. Obviously the system is not in roational equilibrium because one side is experiencing a greater torque than the other. This system will move clockwise as it currently exists.

D. In order for the system to be in rotational equilibrium, we have to move Bob's location from the fulcrum. Let's see to where.

The torques have to be the same on both sides of the fulcrum; mathematically, that looks like this:

F⊥r = F⊥r  Filling in:

300(3) = 450r and

900 = 450r so

2 = r. This means that Bob will have to move closer to the fulcrum by a half of a meter to 2 meters from the fulcrum in order for the system to be in balance.

Isn't this so much fun?!

3 0
3 years ago
Sea and land breezes over a large region that change direction with the seasons are called_____________.
Nataliya [291]
Monsoon. Sea and land breezes over a large region that change direction with the seasons are called. Global winds.
8 0
3 years ago
the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i
Elanso [62]
Gravitational potential energy can be calculated using the formula:

PE_{grav} =mgh

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know. 
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it. 
PE_{grav} =mgh
PE_{grav} =(10kg)(3.7m/s^{2})(1m)
PE_{grav} =37J


7 0
4 years ago
In figure 1, charge q2 experiences no net electric force. What is q1?
lukranit [14]

By using Coulomb's law, we want to find the value of q₁ given that q₂ experiences no net electric force. We will find that q₁ = 8nC

<h3>Working with Coulomb's law.</h3>

Coulomb's law says that for two charges q₁ and q₂ separated by a distance r, the force that each one experiences is:

F = k\frac{q_1*q_2}{r^2}

Where k is a constant

Here we can see that q₂ interacts with two charges, then the total force on q₂ will be:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2}

And we know that it must be equal to zero, so we can write it as:

F = k\frac{q_1*q_2}{(20cm)^2} + k\frac{-2nC*q_2}{(10cm)^2} = 0\\\\k*q_2*(\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2}) = 0\\

The parenthesis must be equal to zero, so we can write:

\frac{q_1}{(20cm)^2} + \frac{-2nC}{(10cm)^2} = 0

And now we can solve this for q₁ to get:

q_1  = 2nC*(\frac{(20cm)^2}{(10cm)^2} ) = 8nC

If you want to learn more about Coulomb's law, you can read:

brainly.com/question/24743340

3 0
3 years ago
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