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Ulleksa [173]
3 years ago
10

A car of mass 1000 kg travelling at a velocity of 25 m/s collides with another car of mass 1500kg which is at rest. The two cars

stick and move off together. What is the velocity of the two cars after the collision?​
Physics
1 answer:
Svetach [21]3 years ago
4 0

Answer:

<em>The velocity of the two cars is 10 m/s after the collision.</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is

P=m.v

If we have a system of bodies, then the total momentum is the sum of them all

P=m_1v_1+m_2v_2+...+m_nv_n

If some collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum  takes the form:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).

Knowing both cars stick and move together after the collision, their velocity is found solving for v':

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

\displaystyle v'=\frac{1000*25+1500*0}{1000+1500}

\displaystyle v'=\frac{25000}{2500}

v' = 10 m/s

The velocity of the two cars is 10 m/s after the collision.

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3 0
3 years ago
A student drops a ball off the top of building and records that the ball takes 3.82s to reach the ground. Determine all unknowns
klemol [59]

Explanation:

Given parameters:

Time of drop  = 3.82s

Unknown:

Final velocity of the ball  = ?

Height of fall  = ?

Solution:

To solve this problem, we apply the appropriate motion equation.

  To find the final velocity;

   v  = u + gt

v is the unknown final velocity

u is the initial velocity  = 0m/s

g is the acceleration due to gravity  = 9.8m/s²

t is the time

 v  = 0  + 9.8 x 3.82  = 37.4m/s

Height of drop;

       H  = ut + \frac{1}{2}gt²  

So;

        H  = (0 x 3.82) + ( \frac{1}{2} x 9.8 x 3.82²) = 71.5m

4 0
3 years ago
A hammer taps on the end of a 3.8-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pu
Elan Coil [88]

Answer:

the speed of sound in this metal is 1668 m/s

Explanation:

given information:

x = 3.8 m

t = 8.80 ms = 0.0088 s

the speed of sound in the metal

v = x/Δt

to find the Δt of the metal, we have to calculate the interval time sound in the air

t_{air} = x/v_{air}, v_{air} = 343 m/s

so,

t_{air}  = 3.8/343 = 0.01108 s

Δt = t_{air}  - t

    = 0.011079 - 0.0088

    = 0.002279

v = x/Δt

  = 3.8/ 0.002279

  = 1668 m/s

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photoshop1234 [79]

Wanning gibbous

hope it helps u

<em>Plz </em><em>mrk </em><em>me </em><em>brainlest</em>

3 0
3 years ago
¿Con qué cualidad del sonido está relacionada la Amplitud de onda?
Keith_Richards [23]

Answer:

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Explanation:

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