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Pani-rosa [81]
2 years ago
12

A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea

ches the lowest point on its path, what is the maximum height of the pendulum?
Physics
2 answers:
Scrat [10]2 years ago
4 0

Answer : Maximum height of the pendulum is 0.43 m

Explanation :

Given that,

Mass of the pendulum, m = 2 kg

Speed of the pendulum, v = 2.9 m/s

When it reaches the lowest point on its path, the potential energy at the highest point is equal to the kinetic energy at the lowest point.

i.e.

KE=PE

\dfrac{1}{2}mv^2=mgh

h is the height of the pendulum.

h=\dfrac{v^2}{2g}

h=\dfrac{(2.9\ m/s)^2}{2\times 9.8\ m/s^2}

h=0.43\ m

Hence, this is the required solution.                              

Vanyuwa [196]2 years ago
3 0
You first us 1/2(mv^2) to solve for the potential energy and then put that in to PE=m*g*h and solve for hight

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Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

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igor_vitrenko [27]

A lady bug moves 10 cm forward and 5 cm backwards

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so the average speed is given as

v = \frac{d}{t}

v = \frac{15}{20}

v = 0.75 cm/s

so its average speed is 0.75 cm/s

5 0
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How far could a rabbit run if it ran 36km/h for 5.0min?
Korolek [52]

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A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
2 years ago
6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord
kiruha [24]

Answer:

Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)

Explanation:

4 0
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