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Firlakuza [10]
2 years ago
13

The table below shows the measurements you took in an experiment. Trial Length ( miles) 1.9 4.2 N 3 5.9 4 What is the longest me

asurement you made in units 1mi = 1.61 km . O A. 5.2 km B. 8.7 km C. 3.7 km D. 9.5 km

Physics
1 answer:
spin [16.1K]2 years ago
7 0

Answer:

9.5km

Explanation: just got it right

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After their loss to U of A, the ASU sumdevils traveled back to Tempe which is 187
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The average speed is 20.8 m/s

Explanation:

The average speed for the trip is given by:

v=\frac{d}{t}

where

d is the distance covered

t is the time elapsed

For the trip in this problem, we have:

d = 187 km = 187,000 m is the distance travelled

The initial  time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

t = 2.5 h \cdot (60)(60)=9000 s

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v=\frac{187,000}{9000}=20.8 m/s

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"The greater the height of an object, the *BLANK* its gravitational potential energy."
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A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

7 0
3 years ago
Read 2 more answers
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