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Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

8 0

3 years ago

An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

4 0

3 years ago

if the charge of each two particles is doubled and the seperation between them is also doubled. the force between the two partic

jok3333 [9.3K]

Explanation is in this link

https://doubtnut.app/physics-question/when-the-distance-between-two-charged-particles-is-doubled-the-force-between-them-becomes-422318076

4 0

3 years ago

An electric current transports of charge in seconds. Calculate the size of the electric current. Be sure your answer has the cor

aleksklad [387]

Answer:

The size of the electric current is 154.74 A.

Explanation:

It is assumed that,

Charge, q = 93 C

Time, $t=601\ ms=601\times 10^{-3}\ s$