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Paul [167]
2 years ago
11

4. A plane accelerates down a runway due to the constant resultant force of

Physics
1 answer:
Doss [256]2 years ago
4 0

Answer:

See below

Explanation:

Net force acting on the plane after overcoming frictional forces is

2.7 x 10^6   -  6.8 x 10^4   -  8.5 x 10^5 = 1 782 000 N

Work = F x d

        = 1 782 000 x 1400 m = 2.5 x 10^9   J       or  2.5 x 10^6  kJ

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What is limiting friction​
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Answer:

Explanation:

As you start increasing the force on an object from 0 N it the object will stay at rest and at a certain magnitude of force the object will start moving. The friction acting on that object is known as limiting frictional force. The magnitude of this frictional force is equal to the force that we apply on the object as it just start to move.

It's important to note that the limiting frictional force is the largest frictional force act on that object in the above explained process. The dynamic frictional force ( the frictional force acting on an moving object.) is always less than the limiting frictional force.

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3 years ago
Your friend, Dayana, says air is not matter. How could you convince her that air is matter?
nikitadnepr [17]

Answer:

Air is a mixture of gases. Gas is a state of matter. Hence, air is matter.

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2 years ago
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Answer:

I would like aliens if they exist

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4 years ago
What types of waves make up the electromagnetic spectrum?
vova2212 [387]
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4 0
3 years ago
Read 2 more answers
A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa
gavmur [86]

Answer : The correct option is, (d) 535^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of copper = 0.10cal/g^oC

c_2 = specific heat of water = 1.00cal/g^oC

m_1 = mass of copper = 120 g

m_2 = mass of water = 300 g

T_f = final temperature of mixture = 35^oC

T_1 = initial temperature of copper = ?

T_2 = initial temperature of water = 15^oC  

Now put all the given values in the above formula, we get:

120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC

T_1=535^oC

Therefore, the temperature of the kiln was, 535^oC

7 0
4 years ago
Read 2 more answers
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