What statement best describes the kind of change that could have taken place

1 answer:

5 0

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AND what was there after. Is there a reason why you're concealing that
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A 0.850 kg mass is placed on a

densk [106]

Answer:

Picture

Explanation:

I am trying to solve this Q

but not sure if this a correct answer

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5 0

3 years ago

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped

lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$

6 0

3 years ago

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$