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Lynna [10]
3 years ago
9

What variable is represented on the y-axis?

Physics
1 answer:
gladu [14]3 years ago
8 0

Distance/ Time which means Distance is on horizontal and time is on vertical

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What is the string theory.
Leno4ka [110]
String the·o·ry
noun
a cosmological theory based on the existence of cosmic strings.
7 0
3 years ago
Read 2 more answers
Is it possible to have a net torque when all of the forces sum to zero? Explain.
nexus9112 [7]

Answer:

Yes it is possible

Explanation:

When two equal magnitude forces are acting on the rod in opposite direction

Then the net force on the system is always zero in that case

so we will have

F - F = 0

now for the system net torque due to these forces is given by

\tau = F(r_1) + F(r_2)

here we know that

r_1, r_2 = distance of the forces from reference about which torque is measured

so here we can say that net force is zero on the system while torque is not zero

in all such case object will rotate about a fixed position with change angular speed

3 0
3 years ago
Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

8 0
3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
• How were historical solar system models different from our current model?
horrorfan [7]
One of the most famous historical model was the geocentric model thought of by plato. This model says that the earth is in the center of the universe and that each of the planets follow complicated paths that go backwards sometimes. This model was used until a new one was accepted. This model is the one we have today called the heliocentric model. It was bought of by Copernicus. Hope that answers your question!
4 0
3 years ago
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