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Lunna [17]
3 years ago
11

I have 6 questions from newtons first law, Second law and third law topic.

Physics
1 answer:
iogann1982 [59]3 years ago
7 0
Rolling friction is less than sliding friction

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If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll
faltersainse [42]
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

    s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    Displacement in 4 seconds = 332 - 76 = 256 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{256}{4}=64ft/s

    Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

    Displacement in 4 seconds = 78 - 332 = -254 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{-254}{4}=-63.5ft/s

    Average velocity in second 4 seconds is 63.5 ft/s downward

3 0
3 years ago
Noble gases, such as argon and neon, are known for being extremely non-reactive. Neon and argon are non-reactive because they A.
Greeley [361]
B. have eight electrons in their outer shell
3 0
3 years ago
Read 2 more answers
Unit 12 Exam
lapo4ka [179]

Answer:

You increase the acceleration of the car.

6 0
3 years ago
The octopus’s tentacle keeps _ right after it is bitten off ? a. Moving b. Breathing c. Growing
konstantin123 [22]

Answer:

The answer is A

Explanation:

The octopus’s tentacle keeps moving right after it is bitten off

5 0
2 years ago
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = \left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0
3 years ago
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