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2 years ago

14

In this lab, you will use a dynamics track to generate collisions between two carts. If momentum is conserved, what variable cha

nge would result in a velocity change after a collision? In the space below, write a scientific question that you will answer by doing this experiment.

2 answers:

BartSMP [9]2 years ago

8 0

In collision type of problems since momentum is always conserved

we can say

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

So here along with this equation we also required one more equation for the restitution coefficient

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

so above two equations are required to find the velocity after collision

here the change in velocity occurs due to the contact force while they contact in each other

so this is the impulse of collision while they are in contact with each other while in collision which changes the velocity of two colliding objects

Alborosie2 years ago

5 0

how does changing mass affect colliding objects?

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Answer:

The decelerating force is $3\times 10^{- 11}\ N$

Solution:

As per the question:

Frontal Area, A = $10\ m^{2}$

Speed of the spaceship, v = $1\times 10^{6}\ m/s$

Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

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t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

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1 year ago

A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

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To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

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$\Delta T$ = Temperature Raise

Our values are given as,

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$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

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Replacing we have,

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$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

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2 years ago

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

Vedmedyk [2.9K]

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2 years ago

Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

ioda

Answer:

$6.1875\times 10^{-8}$

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies $I=\frac {k}{r^{2}}$ where r is the distance

Since intensity is given then when at 3 m

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Since we have the constant then at 4m

Intensity, $I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}$

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1 year ago

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Answer:

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f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

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2 years ago