Imagine that you’re observing a collision. Which action would allow you to determine whether the collision is inelastic?A.
Measure the force of impact.
B.
Observe whether the colliding objects change their shape.
C.
Measure the acceleration of the colliding objects.
D.
Measure the mass of the colliding objects.
Answers
Answer:
Its B, observe weather the colliding objects change their shape .
Explanation:
I hope this will help you, if not so advance sorry :)
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Answers
2. Calculate the pressure in atmospheres using the van der Waals equation. For N2 , a=1.35 (L2⋅atm)/mol2 , and b=0.0387 L/mol
Answers
Answer:
1) 16.88 atm
2) 34.47 atm
Explanation:
Data:
Volume=0.700L
Temperature = 300K
Number of moles=0.480 mol
Ideal gas constant=0.082057 L*atm/K·mol
1) The ideal gas law is:
(1)
with P the pressure, T the temperature, n the number of moles, V the volume and R the ideal gas constant , so solvig (1) for P:
2) The vander Walls equation is:
solving for P
The pressure in atmospheres is 0.974 atm using the ideal gas law and 0.962 atm using the van der Waals equation for N2.
1. To calculate the pressure in atmosphere using the ideal gas law, we can use the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Rearranging the equation, we have P = (nRT)/V. Plugging in the given values, we get P = (0.480 mol * 0.0821 L·atm/mol·K * 300 K) / 0.700 L = 0.974 atm.
2. To calculate the pressure in atmosphere using the van der Waals equation, we can use the equation (P + an^2/V^2)(V - nb) = nRT, where a and b are constants specific to the gas being used. Rearranging the equation, we have P = (nRT/(V - nb)) - an^2/V^2.
Plugging in the given values and the constants for N2, we get P = (0.480 mol * 0.0821 L·atm/mol·K * 300 K/(0.700 L - 0.0387 L/mol * 0.480 mol))^2 - 1.35 (L^2·atm)/mol^2 * (0.480 mol)^2/(0.700 L)^2 = 0.962 atm.
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Answers
Answer:
Explanation:
4 meters east
2. no troughs
3. long wavelengths
4. short wavelengths
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I think it #3(not Sure) I think Its because of their long wavelengths that tsunamis behave as shallow-water waves.
___m/s
What is the direction of the kayaker’s resultant velocity?
___ ° south of west
please help
Answers
The magnitude of the kayaker's resultant velocity is 4.96 m/s.
The direction of the kayaker’s resultant velocity from the triangle is 20⁰ South of west.
Resultant velocity
The magnitude of the kayaker's resultant velocity is calculated by drawing the vector representation of the velocity to form a triangle, with side lengths of 3.7 m/s and 4 m/s, with 80⁰ angle between the two sides.
Let the resultant velocity = R
The direction of the kayaker’s resultant velocity from the triangle is 20⁰ South of west (alternate angles).
Learn more about resultant velocity here: brainly.com/question/24767211
Answer:
7.2 m/s
49 south of west
Explanation:
I got it correct on Edge. :)
Answers
This question is written by a master of deception and distraction.
It's full of so many red herrings that it's tough to walk past it too closely.
In order to answer this question correctly, you don't need to know
the magnitude of the test charge, the distance it travels, or the field
strength between the plates. The whole question lies in that magic
word "equipotential".
An 'equipotential' line is a path along which the electrical potential
is the same at every point. It's analogous to a 'contour line' on a
topographic map, or an isobar on a weather map.
A charge moving along such a line uses no energy to travel that path,
and it encounters no change in voltage anywhere in its travels.