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WINSTONCH [101]
3 years ago
6

A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?

Physics
2 answers:
riadik2000 [5.3K]3 years ago
3 0

=F×s×cosa=2×g×0,8×cos90°= 0

natulia [17]3 years ago
3 0

Answer:

work done=force×distance

=(2×10)×0.8

=16J

I hope this helps

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By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp
strojnjashka [21]

Answer:

A'=2A

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}

When the spring is in its equilibrium position, that is x=0, the object speed its maximum. So, we have:

\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}

In order to double its maximum speed, that is v'{max}=2v_{max}. We have:

A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A

6 0
3 years ago
After a projectile is fired into the air, what is the magnitude of the acceleration
sergiy2304 [10]

Answer: Option A;  9.8 m/s^2

Explanation:

When an object is in the air, and there is no air resistance acting on the object, the only force that will act on the object is the gravitational force (on the vertical axis).

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We know that the gravitational acceleration is equal to:

g = 9.8m/s^2

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The correct option is the first one:

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7 0
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or Rt = 1.69 ohms

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I = V/Rt = 2.37 A

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