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belka [17]
3 years ago
7

What is linear momentum?

Physics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Linear momentum is a vector quantity defined as the product of an object’s mass and it's velocity

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Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 20 mph and half the d
Gnoma [55]

Answer:

On the way to grandmother´s, the average speed was 30 mph. On the way back, the average speed was 40 mph.

Explanation:

The average speed is given by the variation of the position over time.

Mathematically:

ΔX / Δt = v

where:

ΔX = distance (final position - initial position)

Δt = time (final time - initial time)

v = speed

On the way to Grandmother´s, we can calculate how much time Julie drove at each speed:

ΔX / Δt = v

ΔX / v = Δt

50 mi / 20 mph = 2.5 h

In the same way, we can calculate how much time she drove at 60 mph:

50 mi / 60 mph = 0.83 h

In total, she drove a distance of 100 mi in (2.5 h + 0.83 h) 3.33 h. Then, the average speed on the way to Grandmother´s was:

<u>ΔX / Δt = v = 100 mi / 3.33 h = 30 mph</u>

In the return trip, we do not know the distance nor the time that she drove at each speed, but we know that for each part of the trip, the time is the same (Δt)  and we also know that the total distance is 100 mi and the total time is 2Δt:

v1 = ΔX1 / Δt

v2 = ΔX2 / Δt

ΔX2 + ΔX1  = 100

where

v1 = speed during the first part of the trip (20 mph)

v2 = speed during the second part of the trip (60 mph)

ΔX1 = distance driven at 20 mph

ΔX2 = distance driven at 60 mph

Δt = time

If we divide v2/v1, we will get:

v2/v1 = (ΔX2 / Δt) / (ΔX1 / Δt)

60 mph / 20 mph = ΔX2 / ΔX1

3 = ΔX2 / ΔX1

3ΔX1 = ΔX2

Then we can replace ΔX2 for 3ΔX1 in this equation:

ΔX2 + ΔX1  = 100 mi

3ΔX1 + ΔX1 = 100 mi

4ΔX1 = 100 mi

ΔX1 = 25 mi

And now, we can solve Δt from the equation of v1:

v1 = ΔX1 / Δt

Δt = ΔX1 / v1 = 25 mi / 20 mph = 1.25 h

The average speed on the return trip is then:

<u>v = 100 mi / 2Δt = 100 mi / 2.5 h = 40mph</u>

8 0
4 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
I need help me with my question
lisov135 [29]

The tilt of the moon's axis does not allow for monthly alignment, so the lunar and solar eclipse do not happen every month.

<h3>How do the lunar and solar eclipse occur?</h3>
  • For the occurrence of lunar and solar eclipse, the sun, moon and the earth must remain in a plan and along a straight line.
  • When the earth appears in between the sun and the moon, lunar eclipse occurs.
  • When the moon appears in between the sun and the earth, solar eclipse occurs.
  • The moon and earth are rotating not only around the sun, but also around the black hole of Milky way galaxy.
  • So they are not present in a plan as well as in a straight line in every full moon and new moon time.

Thus, we can conclude that the option D is correct.

Learn more about the lunar eclipse and solar eclipse here:

brainly.com/question/8643

#SPJ1

4 0
2 years ago
A uniform horizontal beam of weight 481 N and length 3.32 m has two weights hanging from it. One weight of 381 N is located 0.87
Andre45 [30]

Answer:

1143 N at 1.59 m from the left end

Explanation:

For the system to produce equilibrium, the total force and moment must be 0. Since the total weight downward is

481 + 381 + 281 = 1143 N

Therefore the magnitude of the force acting upward to balance this system must be the same of 1143 N

That alone is not enough, we also need the position of the force for the total moment to be 0.

Let x be the length from the this upward force to the left side. And let the left point be the point of reference for moment arm:

481 * 3.32/2 + 381 * 0.8798 + 281*(3.32 - 0.8798) - 1143*x = 0

x = (481*1.66 + 381 * 0.8798 + 281*2.4402)/1143 = 1.59m

5 0
4 years ago
What is the highest and lowest frequency a human being can hear​
nata0808 [166]

Every person is different.  But for a planet-wide overall average that roughly represents all human beings on Earth, the figures usually used are:

from  20 Hz  to  20,000 Hz .

6 0
3 years ago
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