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WARRIOR [948]
3 years ago
10

what is the acceleration of a bowling ball that starts at rest and moves 300m down the gutter in 22.4 sec

Physics
1 answer:
exis [7]3 years ago
7 0
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
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Savatey [412]

Answer:

The wavelengths of C1 is 10.4m, A6 is 0.193m and B7 is 0.0861m

Explanation:

Using the formula V = f×λ . Then substitute the following values into the formula:

a) v=340m/s

f=32.7 Hz

λ=V ÷ f

= 340 ÷ 32.7

= 10.4m (3s.f)

b) λ=340 ÷ 1760

= 0.193m (3s.f)

c) λ=340÷3951.1

= 0.0861m (3s.f)

(Correct me if I am wrong)

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The driver of a 1000 kg car traveling on a highway at 35 m-s'applies brake to avoid hitting a van in front of him, which had com
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2 years ago
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Two identical charges, each -8.00 E-5 C, are separated by a distance of 20.0 cm (100 cm = 1 m). What is the force of repulsion?
Rom4ik [11]

F = 1440 N. The repulsion force between two identical charges, each -8.00x10⁻⁵C separated by a distance of 20.0 cm is 1440 N.

The easiest way to solve this problem is using Coulomb's Law given by the equation F=k\frac{|q_{1}*q_{2}|}{r^{2} }, where k is the constant of proportionality or Coulomb's constant, q₁ and q₂ are the charges magnitude, and r is the distance between them.

We have to identical charges of -8.00x10⁻⁵C, are separated by a distance of 20.0 cm, and we need to know the force of repulsion between the charges.

First, we have to convert 20.0 cm to meters.

(20.0 cm x 1m)/100cm = 0.20 m

Using the Coulomb's Law equation:

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}} \frac{|-8.00x10^{-5}C*-8.00x10^{-5}C|}{(0.20m)^{2} }

F = 9.00x10^{9}\frac{Nm^{2}}{C^{2}}(1.6x10^-7\frac{C^{2} }{m^{2} } })\\F = 1440N

5 0
3 years ago
A 2-inch, f/4 reflector has a focal length of:
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I’m pretty sure it’s 8 sorry if I’m wrong
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