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zepelin [54]
3 years ago
9

If you planned to bike to a park that was five miles away, what average speed would you have to maintain to arrive in about 15 m

in? (
Physics
1 answer:
AfilCa [17]3 years ago
5 0

The average speed you need to maintain is 0.33 miles per minute.

The average speed is the speed that you maintain throughout the journey. It is obtained as; Total distance covered/Total time taken

If we have the following information from the question;

Distance covered = 5 miles

Time taken = 15 mins

Then it follows that;

Average speed = Distance/time

Average speed= 5 miles/15mins

Average speed = 0.33 miles per minute.

Learn more: brainly.com/question/17661499

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Lena [83]
The answer would be B
7 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
Two negative charges are both -3.0C push each other apart with a force of 19.2N. How far apart are the two charges?
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06758 m

Explanation:

it is correct because that is what

5 0
3 years ago
A very long thin wire carries a uniformly distributed charge, which creates an electric field. The electric field is (2300 N/C ,
Leto [7]

Answer:

λ= 5.24 × 10 ⁻² nC/cm

Explanation:

Given:

distance r = 4.10 cm = 0.041 m

Electric field intensity E = 2300 N/C

K = 9 x 10 ⁹ Nm²/C

To find λ = linear charge density = ?

Sol:

we know that E= 2Kλ / r

⇒ λ = -E r/2K         (-ve sign show the direction toward the wire)

λ = (- 2300 N/C × 0.041 m) / 2 ×  9 x 10 ⁹ Nm²/C

λ = 5.24 × 10 ⁻⁹ C/m

λ = 5.24 nC/m = 5.24 nC/100 cm

λ= 5.24 × 10 ⁻² nC/cm

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Answer:

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Explanation:

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8 0
3 years ago
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