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algol13
4 years ago
7

How does the range change as the angle increases?

Physics
1 answer:
RSB [31]4 years ago
5 0
It first increases then it decreases give me brainlest please
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Explain how the earth is round and not flat
frosja888 [35]
It is a conglomeration of rock that collected together over time, and it rotating in a spinning ball that is orbiting a sun that is in the middle of our solar system. 

You should look at a picture of the earth from a satellite.<span />
5 0
3 years ago
Read 2 more answers
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
3 years ago
Function of a simple pendulum​
Misha Larkins [42]

Answer:

A pendulum is a mechanical machine that creates a repeating, oscillating motion. A pendulum of fixed length and mass (neglecting loss mechanisms like friction and assuming only small angles of oscillation) has a single, constant frequency. This can be useful for a great many things.

From a historical point of view, pendulums became important for time measurement. Simply counting the oscillations of the pendulum, or attaching the pendulum to a clockwork can help you track time. Making the pendulum in such a way that it holds its shape and dimensions (in changing temperature etc.) and using mechanisms that counteract damping due to friction led to the creation of some of the first very accurate all-weather clocks.

Pendulums were/are also important for musicians, where mechanical metronomes are used to provide a notion of rhythm by clicking at a set frequency.

The Foucault pendulum demonstrated that the Earth is, indeed, spinning around its axis. It is a pendulum that is free to swing in any planar angle. The initial swing impacts an angular momentum in a given angle to the pendulum. Due to the conservation of angular momentum, even though the Earth is spinning underneath the pendulum during the day-night cycle, the pendulum will keep its original plane of oscillation. For us, observers on Earth, it will appear that the plane of oscillation of the pendulum slowly revolves during the day.

Apart from that, in physics a pendulum is one of the most, if not the most important physical system. The reason is this - a mathematical pendulum, when swung under small angles, can be reasonably well approximated by a harmonic oscillator. A harmonic oscillator is a physical system with a returning force present that scales linearly with the displacement. Or, in other words, it is a physical system that exhibits a parabolic potential energy.

A physical system will always try to minimize its potential energy (you can accept this as a definition, or think about it and arrive at the same conclusion). So, in the low-energy world around us, nearly everything is very close to the local minimum of the potential energy. Given any shape of the potential energy ‘landscape’, close to the minima we can use Taylor expansion to approximate the real potential energy by a sum of polynomial functions or powers of the displacement. The 0th power of anything is a constant and due to the free choice of zero point energy it doesn’t affect the physical evolution of the system. The 1st power term is, near the minimum, zero from definition. Imagine a marble in a bowl. It doesn’t matter if the bowl is on the ground or on the table, or even on top of a building (0th term of the Taylor expansion is irrelevant). The 1st order term corresponds to a slanted plane. The bottom of the bowl is symmetric, though. If you could find a slanted plane at the bottom of the bowl that would approximate the shape of the bowl well, then simply moving in the direction of the slanted plane down would lead you even deeper, which would mean that the true bottom of the bowl is in that direction, which is a contradiction since we started at the bottom of the bowl already. In other words, in the vicinity of the minimum we can set the linear, 1st order term to be equal to zero. The next term in the expansion is the 2nd order or harmonic term, a quadratic polynomial. This is the harmonic potential. Every higher term will be smaller than this quadratic term, since we are very close to the minimum and thus the displacement is a small number and taking increasingly higher powers of a small number leads to an even smaller number.

This means that most of the physical phenomena around us can be, reasonable well, described by using the same approach as is needed to describe a pendulum! And if this is not enough, we simply need to look at the next term in the expansion of the potential of a pendulum and use that! That’s why each and every physics students solves dozens of variations of pendulums, oscillators, oscillating circuits, vibrating strings, quantum harmonic oscillators, etc.; and why most of undergraduate physics revolves in one way or another around pendulums.

Explanation:

7 0
3 years ago
PLEASE HEELP!!!
Ann [662]

Answer:

The "pressure" of the electricity is electric potential. Electric potential is the amount of energy available to push each unit of charge through an electric circuit. The unit of electric potential is the volt. ... A volt is the force needed to move one amp through a conductor that has 1 ohm of resistance

8 0
3 years ago
Can anyone help me with this question please​
JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

  1. v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0
3 years ago
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