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Mrac [35]
3 years ago
11

When a 0.1-kilogram pendulum bob reaches the top of its swing, how much kinetic energy does it have?

Physics
1 answer:
zvonat [6]3 years ago
7 0
When ANY pendulum bob reaches the top of its swing,
it stops momentarily and then begins to drop.  At the instant
when it stops, its kinetic energy is zero.
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If the velocity of an object is doubled, its kinetic energy is ______
swat32
Increased by a factor of 4
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A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

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The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point
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To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.
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3 years ago
Two important economic activities in Norden are _____.
tensa zangetsu [6.8K]
Fishing and Agriculture.
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Read 2 more answers
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
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