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marissa [1.9K]
2 years ago
7

Question 2 of 20

Physics
1 answer:
alukav5142 [94]2 years ago
8 0

Answer: a

Explanation: because i said so

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If you travel at 3 m/s for 12 seconds, how far did you travel?
ella [17]

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

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2 years ago
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When will he love me 
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Johannes Keller’s theory was based upon analysis of planet observations taken by tycoon Brahe
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3 years ago
At the lowest point in a vertical dive (radius = 0.58 km), an airplane has a speed of 300 km/h which is not changing. Determine
murzikaleks [220]

Answer:

The centripetal acceleration is a = 11.97 \ m/s^2

Explanation:

From the question we are told that

     The radius  is r =  0.58 \ km =  0.58 * 1000  =  580 \ m

      The speed is v  = 300\ km /hr =  \frac{300 *1000}{1 * 3600 }  =  83.33 \ m/s

The centripetal acceleration of the pilot is mathematically represented as

       a =  \frac{v^2 }{r}

substituting  values

      a =  \frac{(83.33)^2 }{580}

     a = 11.97 \ m/s^2

7 0
2 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
iren2701 [21]
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
3 0
3 years ago
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