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laiz [17]
3 years ago
6

A train slows its speed from 52 kilometers per hour to 46 kilometers per hour in 0.04 hour. What is the acceleration o the train

during this time?​
Physics
1 answer:
BigorU [14]3 years ago
3 0

Answer: here you go i have to put 20 letters in so just ignore this and look at the link.

Download pdf
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A force of 1 N is equal to which combination of SI units?
zavuch27 [327]

Answer:

F = M a = kg * m/s^2

1 N = 1 kg * m/s^2

(B) is correct

4 0
2 years ago
Ball is rolling for 12 seconds and a distance of 6 meters what is its speed
o-na [289]

Speed=Distance/Time

So S=6/12=.5 m/s

5 0
3 years ago
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What 2 kinds of substances does convection occur ?
ch4aika [34]

Liquids and Gases

Explanation:

Convection is a form of heat transfer that is predominant in liquids and gaseous substances. This form of heat transfer is driven by density differences between gases and liquids.

  • Convection involves the actual movement of the particles of a medium.
  • Boiling of food is clinical example of convection in liquids. Hot part of the food in contact with the heat source becomes less dense and more buoyant. They rise to the top of the medium and are replace by the denser and colder part of the food.
  • Land and sea breeze is an example of convection in gases. The land warms the air around it during the day. It is hotter and less dense. The air mass moves to replace the ones on the sea where the air is relatively cold due to high specific heat capacity of water.
  • The reverse process occurs at night.

Learn more:

Heat transfer from the sun brainly.com/question/1140127

#learnwithBrainly

4 0
3 years ago
HERES THE ANSWER AND QUESTION
tatyana61 [14]

Answer:

agree

Explanation:

5 0
3 years ago
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At what angle two forces P + Q and (P - Q) act so that their resultant is :
stiv31 [10]

Use resultant formula

\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}

So

#1

A be p+q and B be p-q

\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2

\\ \rm\Rrightarrow 2cos\alpha=1

\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}

\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}

#2

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)

\\ \rm\Rrightarrow 2cos\beta=0

\\ \rm\Rrightarrow cos\beta=0

\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}

#3

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2

\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)

\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}

\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)

8 0
2 years ago
Read 2 more answers
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