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gregori [183]
3 years ago
12

How does air resistance affect the velocity of a falling object?

Physics
1 answer:
Rashid [163]3 years ago
4 0
A falling object (directly downward) is slowed down by air resistance. In turn it would take longer to fall.
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assume that in an unpolished floor a lady us pushing a 40 kg box to the left with an applied force of 192 N. find the sum of for
son4ous [18]
The sum of force is 7,680...............
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The object of badminton is to hit the birdie or "shuttlecock" over the net and onto to court to score how many points to win gam
xxTIMURxx [149]

Answer:

either 7 or 21

Explanation:

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3 years ago
Can someone plz help me
Ymorist [56]

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6 0
3 years ago
A cosmic ray proton moving toward the Earth at 5.00 10 m/s 7 × experiences a magnetic force of 1.70 10 N −16 × . What is the str
EastWind [94]

Answer:

Magnetic field will be equal to B=3\times 10^{-5}T

Explanation:

We have given velocity of proton 5\times 10^7m/sec

Magnetic force experienced by proton F=1.7\times 10^{-16}N

Charge on proton q=1.6\times 10^{-19}C

Angle between field and velocity \Theta =45^{\circ}

Force in magnetic field is equal to F=qBVsin\Theta

So 1.7\times 10^{-16}=1.6\times 10^{-19}\times 5\times 10^7\times B\times sin45^{\circ}

B=3\times 10^{-5}T

So magnetic field will be equal to B=3\times 10^{-5}T

7 0
3 years ago
An object at 20∘c absorbs 25.0 j of heat. what is the change in entropy δs of the object?
anastassius [24]
From the definition of entropy, the entropy change of an object is
\delta S =  \frac{Q}{T}
where
Q is the heat absorbed
T is the absolute temperature

in our problem, we have Q=25.0 J, while the absolute temperature is (converting in Kelvin)
T=20 ^{\circ}C + 273 = 293 K

and so the entropy change is
\delta S=  \frac{25.0 J}{293 K}=0.085 JK^{-1}
5 0
4 years ago
Read 2 more answers
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