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Keith_Richards [23]
3 years ago
6

If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th

e tangential acceleration?
Physics
1 answer:
Rufina [12.5K]3 years ago
8 0
Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a). 
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
<span>Radius = 150 cramwells </span>
<span>t = sqrt(150/1.6) = 9.68 s.

I hope this helps.</span>
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<h2><u>Answer:</u></h2>

Cynophobia

<h3><u>Explanation:</u></h3>

Cynophobia originates from the Greek words that signify "dog" (cyno) and "fear" (phobis). An individual who has cynophobia encounters a dread of mutts that is both unreasonable and tenacious. It's something beyond feeling of scaredness whether a dog is barking or an individual is around dogs.

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3 years ago
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Answer:

Part a)

a= 0.32 m/s^2

Part b)

F_c = 3.6 N

Part c)

F_c = 5.5 N

Explanation:

Part a)

As we know that the friction force on two boxes is given as

F_f = \mu m_a g + \mu m_b g

F_f = 0.02(10.6 + 7)9.81

F_f = 3.45 N

Now we know by Newton's II law

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9.1 - 3.45 = (10.6 + 7) a

a = \frac{5.65}{17.6}

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Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

F_c - F_f = m_b a

F_c = \mu m_b g + m_b a

F_c = 0.02(7)(9.8) + 7(0.32)

F_c = 3.6 N

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

F_p - F_f - F_c = m_b a

9.1 - 0.02(7)(9.8) - F_c = 7(0.32)

F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)

F_c = 5.5 N

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3 years ago
A woman and her dog are out for a morning run to the river, which is located 4.0 KM away. The woman runs at 2.5 M/S in a straigh
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River shore is located at distance

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for the same time interval the dog will run to and fro with speed 4.5 m/s

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Answer:

Explanation:

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