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snow_tiger [21]
3 years ago
14

A(n) 1.3 kg mass sliding on a frictionless surface has a velocity of 7.1 m/s east when it undergoes a one-dimensional elastic co

llision with an unknown mass that is initially at rest. After the collision the velocity of the 1.3 kg is 1.7 m/s west. What is the unknown mass? Answer in units of kg
Physics
1 answer:
Oxana [17]3 years ago
3 0

Answer: 2.12 kg

Explanation:

Since the 1.3 kg object moves to the west after the collision, the other object will move to the east after the collision.

In an elastic collision, the relative velocity after the collision is the opposite of the relative velocity before the collision. Since the 1.3 kg object’s velocity before the collision is 6.7 m/s greater than the other object, after the collision, its velocity will be 6.7 m/s less than the other object. To determine the other object’s velocity, use the following equation.

v = 1.7 – 7.1 = -5.4 m/s

The negative sign means it is moving eastward. Let’s use this number is a momentum equation to determine its mass.

Initial momentum = 1.3 * 7.1 = 9.23 east

For the 1.3 object, final momentum = 1.3 * 1.7 = 2.21 west

To determine the final momentum of the other object, add these two numbers.

Final momentum = 11.44 east

To determine its mass, use the following equation.

m * 5.4 = 11.44

m = 11.44 ÷ 5.4 = 2.12 kg

To make sure that kinetic energy is conserved, let’s round this number to 2 kg and determine the final kinetic energies.

For the 1.3 object, KE = 1/2 * 1/3* 1.7^2 = 0.48

For the 2 kg object, KE = 1/2* 2 * 5.4^2 = 29.64

Total final KE = 29.64

Initial KE = 0.5* 1.3 * 7.1^2 = 32.77

Since I rounded the mass up to 2kg, this proves that kinetic energy is conserved and the mass is correct!

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To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

P = h\rho g

Where,

h = Height

\rho = Density

g = Gravity

Our values are

\rho = 1000kg/m^3 \rightarow density of water at normal conditions

h = 7.3m

g = 9.8m/s^2

PART A) Replacing these values to find the total pressure difference we have to

P_1 = h_1 \rho g

P_1 = (7.3)(1000)(9.8)

P_1 = 71540Pa

In this way the pressure change would be subject to

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\Delta = 2.1*10^5Pa- 0.7154*10^5Pa

\Delta = 138460Pa

\Delta = 0.135Mpa

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

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2.1*10^5 = H (1000)(9.8)

H = 21.42m

Therefore the high which could a faucet be before no water would flow from it is 21.42m

5 0
3 years ago
A driver travels 4.1 km west, 17.3 km north, and finally 1.2 km at an angle of 65.4 degree north of west. What is the driver’s d
jek_recluse [69]

Answer:

Explanation:

We shall represent each displacement in vector form .

i will represent east , j will represent north .

D₁ = 4.1 west = - 4.1 i

D₂ = 17.3 north = 17.3 j

D₃ = - 1.2 cos65.4 i + 1.2 sin65.4 j

= - .5 i + 1.09 j

Total displacement

= D₁ + D₂ + D₃

=  - 4.1 i + 17.3 j - .5 i + 1.09 j

D = - 4.6 i + 18.39 j

magnitude of D

= √ ( 4.6² + 18.39² )

= √ (21.16 + 338.2 )

= √359.36

= 18.95 km .

Final displacement = 18.95 km .

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anyanavicka [17]

Answer:

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Explanation:

If:

\frac{sin\theta -\theta}{\theta } \leq 0.1o/o\\\frac{sin\theta -\theta}{\theta }\leq 10^{-3}

According the Taylor`s series:

sin\theta =\theta -\frac{\theta ^{3}  }{3!}  ....

\frac{\theta ^{2} }{6} \leq 10^{-3} \\\theta ^{2} \leq 6x10^{-3} \\\theta \leq \sqrt{6x10^{-3} } \\\theta \leq 0.0775\\\theta = 0.0775rad=4.44

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The answer is donate
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