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mylen [45]
3 years ago
11

A 1200-kg car is travelling east at a rate of 9 m/s. A 1600-kg truck is travelling south at a rate of 13 m/s. The truck accident

ally runs a stop sign and collides with the car in a completely inelastic collision. What is the speed of the combined mass after the collision
Physics
1 answer:
weeeeeb [17]3 years ago
3 0

Answer:

Around 3.57m/s

Explanation:

p=mv

Let's denote the momentum, mass, and velocity of the car with the subscript 1, and for the truck use 2. After the collision, the combined momentum can be denoted with the subscript 3.

p_1=1200\cdot 9=10800 \\\\p_2=1600\cdot 13=20800 \\\\20800-10800=10000 \\\\1200kg+1600kg=2800kg \\\\10000=2800v_3 \\\\v_3\approx 3.57m/s

Hope this helps!

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5) Find the initial velocity for a 700 kg car that
Serhud [2]

Answer:

Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.

Maximum initial speed is 40.6 m/s

Minimum initial speed is 16.6 m/s

Explanation:

Assume this is a NET impulse so we can ignore friction.

An impulse results in a change of momentum

The impulse applied was

p = Ft = 1400(6.0) = 8400 N•s

p = mΔv

Δv = 8400 / 700 = 12 m/s

If the impulse was applied in the direction the car was already moving, the initial velocity was

vi = 28.6 - 12 = 16.6 m/s

if the impulse was applied in the direction opposite of the original velocity, the initial velocity was

vi = 28.6 + 12 = 40.6 m/s

Other angles of Net force would result in various initial velocities.

5 0
2 years ago
This should be correct
sweet-ann [11.9K]

Maybe it is, maybe it isn't.  We can't tell, until we see what "this" is. 

Show us a drawing, an equation, an expression, a statement ... something !

5 0
3 years ago
How wind Form?<br> Explain please
snow_lady [41]
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6 0
3 years ago
Read 2 more answers
A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is f
Vinil7 [7]

Answer:

Solution is given in the attachments,below.

7 0
3 years ago
A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
Bingel [31]

Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
3 years ago
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