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solniwko [45]
3 years ago
6

Kenny wants to get to Washington DC within 4 hours. Washington DC is 133 miles away from where he is. What is the avg speed he m

ust go in order to meet his time frame ?
Physics
1 answer:
umka21 [38]3 years ago
4 0
He must travel 35 mph
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Please answer this.
lilavasa [31]

Answer:

the answer is c

Explanation:

it's c because the moon has to be a full moon to be a solar eclipse when the sun moon and earth line up

6 0
3 years ago
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The mass of an object is a measure of its
valentina_108 [34]
Hope it helped you.

A. Acceleration is a wrong answer. Acceleration is the rate of change to in velocity. It involves a change in speed and direction. 

B.  Force is a wrong answer. Force is a directions from push/or pull.

C. Inertia is a correct. Inertia is a tendency of an object to stay at the rest or preserve its state of motion.

D. Velocity is a incorrect answer. Velocity is the placement of an object during a specific unit of time. It has two measurements are needed to determine velocity.

-Charlie

-Thank you so much!

Have a great day!
4 0
3 years ago
Which of the following is a chemical change?
klemol [59]

a). Water is still H₂O after it freezes.

b). Ice is still H₂O after it melts.

c). Wire is still Cu when it's bent.

d). Paper combines with the O₂ in the air, and turns into
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3 0
3 years ago
Which of the following stars is likely to be the coldest? Select one: a. The white-colored star b. The orange-colored star c. Th
worty [1.4K]
The White colored star is likely to be the coldest
6 0
3 years ago
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A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
3 years ago
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