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qwelly [4]
3 years ago
15

What are some examples of pendulums?

Physics
1 answer:
Hatshy [7]3 years ago
8 0
Like a grandfather clock,

A pendulum = a string with a weight(can be anything from a coin to a ring to a ball) is on the end, the pendulum swings back and forth losing a little bit of energy every swing until it comes to a complete stop
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What is the total amount of kinetic and potential energy in a system ?
Solnce55 [7]

Answer:

Its the sum of the potential energy and the kinetic energy

7 0
3 years ago
A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude
finlep [7]
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
                   = (5 * 10) Kg m/s
                   = 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
5 0
3 years ago
Which stage of sleep usually comes before REM sleep
kumpel [21]
Stage 4





Hope this helps
7 0
4 years ago
Read 2 more answers
A substance has density ρ, mass m, and volume V. If the volume is tripled, what is the new mass?
Igoryamba

Since the density of the substance is ρ (rho),

==> every cm³ of this substance has ρ grams of mass.

Then

==> V cm³ of it has ρV grams of mass.  That's ' m '.

and

==> 3V cm³ of it has 3ρV grams of mass.  That's ' <em>3m</em> '.

6 0
4 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
3 years ago
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