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Deffense [45]
2 years ago
6

engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi

on between two train cars traveling toward each other. Car one is traveling south at 18 M/S and has a full load, giving it a total mass of 14,650 kg. Kartoo is traveling north at 11 m/s and has a mass of 3825 kg. After the collision car one has a final velocity of 6 M/S south what is the final velocity of car 2?
Physics
2 answers:
IRISSAK [1]2 years ago
8 0

Answer:

35 m/s south

Explanation:

took the test

Archy [21]2 years ago
4 0

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

where ;

m_1=mass of object 1\\v_1i=initial  velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2

Given in the question that;

m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?

Apply the formula as;

m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

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Luigi jumps straight upwards 15.0m/s. How high is he when he is jumping at :<br>8.0 m/s upwards
tensa zangetsu [6.8K]

Given: Initial velocity v_i= 15.0m/s

           Final Velocity v_f=8.0 m/s

To find : Height when jumpled 8.0 m/s upwards.

Solution: We already have values of initial velocity and final velocity.

We know, accereration due to gravity is given by a= -9.8 m/s^2.

It's negative because when jump it's in opposite direction.

We know formula,  v_f ^2 = v_i ^2+2ah.

Where h is the height when jumpled 8.0 m/s upwards.

Plugging values of v_i, \ v_f \ and \ a \ in \ formula \ above.

(8.0)^2 = (15.0)^2 +2(-9.8)*h

64= 225 -19.6h

Subtracting both sides by 225.

64-225= 225 -19.6h-225.

We get,

-161 = -19.6h

Dividing both sides by -19.6, we get

\frac{-161 }{-19.6} =\frac{ -19.6h}{ -19.6}

h= 8.2143

Rounding to nearest tenth, we get

h= 8.2 meter.

His height is 8.2 meter when he is jumping 8.0 m/s upwards.

8 0
2 years ago
If you are stopped, then get on a bike and accelerate 2 m/s/s south, what us your final velocity after 3 seconds? Please Explain
olga_2 [115]
Vf = Final velocity.
Vi = initial velocity
a = acceleration. 
t = time

Vf = Vi + at

Vf = 0 + (2 m/s^2)(3s)

Vf = 6 m/s south
4 0
3 years ago
A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat
Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

m g =\dfrac{mv^2}{r}

9.81 =\dfrac{v^2}{0.675}

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)

where, I =\dfrac{2}{5}mr^2

\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)

0.7 u^2 + g H = 0.7 v^2 + g(2R)

0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial  speed is 3.35 m/s

8 0
2 years ago
True or false kinetic energy increases if the speed or the mass of an object increases
Olin [163]
It would be trueeeeeee
4 0
3 years ago
Read 2 more answers
7. Small rockets are used to make tiny adjustments in the speeds of satellites. One such rocket has a
patriot [66]

Answer:

222.3 s

Explanation:

From Newton’s law, F=ma hence a=\frac {F}{m}

a=\frac {31}{65000}=0.000477

Time, t=\frac {\triangle v}{acceleration}

Where \triangle v is change in velocity

The change in velocity is equivalent to 0.106 m/s

t=\frac {0.106}{0.000477}=222.2581

4 0
2 years ago
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