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Rudik [331]
3 years ago
12

The high-speed police chase ends at an intersection as a 2,150-kg Ford Explorer (driven by Robin) traveling south at 35 m/s coll

ides with a 18,250-kg garbage truck moving east at 1.55 m/s. The Explorer and the garbage truck entangle together in the middle of the intersection and move as a single object. Determine the post-collision speed two entangled vehicles.
Physics
1 answer:
defon3 years ago
4 0

Answer:

The post-collision speed of the two entangled vehicles is 5.08m/s

Explanation:

According to the law of conservation of momentum which states that the sum of momentum of a body before collision is equal to the sum of the collision of the bodies after collision. The bodies moves with the same velovity after collision.

Mathematically, mu + MU = (m+M)v

m and M are the masses of the bodies

u and U are their respective velocities

v is their common velocity

Momentum = mass*velocity

BEFORE COLLISION;

Momentum of Ford Explorer = 2150*35

= 75,250kgm/s

Momentum of the garbage truck = 18, 250 * 1.55

= 28,287.5kgm/s

AFTER COLLISION;

Note that the body will move with the same velocity after collision.

Momentum of the bodies after entangling = (2,150+18,250)v

= 20,400v kgm/s (v is their common velocity)

According to thr law;

75,250+28,287.5 = 20,400v

103,537.5 = 20,400v

v = 103,537.5/20,400

v = 5.08m/s

The post-collision speed of the two entangled vehicles is 5.08m/s

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
To the nearest square foot, how many square feet are there in an area of 4.4 square meters?
vichka [17]

Answer:

    4.4 square meters = 47 square foot

Explanation:

We have

    1 meter = 3.28084 foot

    1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot

    4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot

    4.4 square meters = 47 square foot

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Have wavelengths that are longer than normal.
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Because we are measuring rain, a liquid, we would want to use a tool that would allow us to collect the rain for measuring. Therefore, the tool e would use to measure the amount of rainfall would be A. a graduated cylinder.

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