4. A car travels 3 km due North, then 5 km East Represent

two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s

solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3) of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>

The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

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6 0

1 year ago

Wagon wheel. While working on your latest novel about settlers crossing the Great Plains in a wagon train, you get into an argum

OverLord2011 [107]

Answer:

I = 16.7kgm²

Explanation:

Since, Torque is given by,

\tau = F*r = I*\alpha

here, I = Moment of inertia = ??

\alpha = angular acceleration of wheel = a/r

F = tangential tension acting on the wheel = T

a = acceleration of bag of sand = 2.95 m/s^2

r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m

from force balance on sand bag,

mg - T = m*a

T = m*(g-a)

m = mass of sand bag = 20 kg

So, I = T*r/\alpha = m*(g-a)*r/(a/r)

Using known values:

I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74

I = 16.7 kgm² = Moment of inertia of wheel experimentally

also, Moment of inertia of wheel theoretically(I') = M*r²

given, M = mass of wheel = 70 kg

I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically

7 0

3 years ago

In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion

Leno4ka [110]

Answer:

4 m/s²

Explanation:

When the elevator is 1 m below point of contact , compression will be 1 m.

Restoring force in the spring will be 10600 N. Friction force of 17000N will also act in upward direction . The weight of 2000 x 9.8 N will act downwards

Force in down ward direction = 2000 x 9.8

= 19600 N

Force in upward direction

= 10600 + 17000

= 27600 N

Net force in upward direction

= 27600 - 19600

= 8000 N

Acceleration in upward direction

= 8000 / 2000

= 4 m/s²

5 0

3 years ago

A person walks 4.0 kilometers north, then 4.0 kilometers west. His displacement is closest to

larisa86 [58]

Use the Pythagoras for the magnitude and the tan^-1 x = -1 for the angle

displacement = 4^2 + 4^2 = 32 = 4 sqrt(2) = 5.65 km

angle is 135 degrees.

7 0

3 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

2 years ago