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solong [7]
3 years ago
11

With mechanical waves, what is moving and what stays in roughly the same place?

Physics
1 answer:
Zielflug [23.3K]3 years ago
6 0
The part that moves are called anti-nodes. The stationary pars are nodes
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If you touch the two probes together while the DMM is set to resistance, what will happen?
rosijanka [135]

If the meter's battery is good and the meter is working OK, it should read zero ohms.

Most meters have a little adjustment wheel on the side to SET the reading to exactly zero when the probes are touched together.

8 0
3 years ago
Determine the number of revolutions through which a typical automobile tire turns in 1 yr. Suppose the automobile travels 13500
RSB [31]

Answer:

Number of revolution made by tire is 1.57 x 10⁷

Explanation:

Radius of tire, r = 0.220 m

Circumference of tire, C = 2πr

Substitute the value of r in the above equation.

C = 2 x π x 0.220 m = 1.38 m

Total distance covered by tire in a year, D = 13500 miles

But 1 mile = 1609.34 m

So, D = 13500 x 1609.34 m

Number of revolutions take by tire, N = \frac{D}{C}

N=\frac{13500\times1609.34}{1.38}

N = 15743543

5 0
3 years ago
At which location could you place the north pole of a bar magnet so that it would be pushed away from the magnet shown?
Gala2k [10]
It’s a because b makes more since shown as the magnet
7 0
3 years ago
Read 2 more answers
An aluminum rod of length 3.3 m and crosssectional area 3.8 cm2 is inserted vertically into a thermally insulated vessel contain
Sveta_85 [38]

Answer:

V_vap = 161.2 L

Explanation:

The total mass of the aluminum rod is given as;

m = ρ∙V = ρ∙L∙A

Where;

ρ is density = 2700 kg/m³

L is length = 3.3m

A is cross sectional area = 3.8 cm² = 3.8 x 10⁻⁴ m²

Thus;

m = 2700kg/m³•3.3m•3.8 × 10⁻⁴m²

= 3.3858kg

By cooling down the submerged half of the aluminum rod releases an heat amount of

Q = (1/2)∙m∙cp∙∆T

Where;

cp is specific heat of aluminum aluminum = 900 J/kg

∆T is change in temperature = 274 - 4.2 = 269.8 K

Thus;

Q = (1/2)•3.3858•900•(269.8)

= 411069.978 J

The liquid absorbs this heat and vaporizes partially, such that the heat equals vaporized mass times latent heat of vaporization:

Q = m_vap•∆h_vap

Making m_vap the subject;

m_vap∙ = Q/∆h_vap

Where ∆h_vap is latent heat of vaporization given as 20900J/kg

Thus,

m_vap∙ = 411069.978/20900

= 19.668 kg

Let's divide this mass by the density of liquid helium and we get the liquid volume which has vaporized:

V_vap∙= m_vap/ρ

V_vap∙ = 19.668/122

V_vap∙ = 0.1612 m³

Converting to litres;

V_vap = 0.1612 x 1000

V_vap = 161.2 L

3 0
3 years ago
How is Uranus colder than Neptune
STatiana [176]
With Uranus at an average distance of 2.88 billion kilometres from the Sun and Neptune at an average distance of 4.5 billion kilometres it would be very easy to point out which of the gas giants is the coldest, but if you were you were to say that Neptune was the coldest, you’d be wrong.<span>Given that we expect planets further from the Sun to be colder than those closer, this does make Neptune and Uranus quite a mysterious pair. Uranus and Neptune are brimming with volatiles such as water, methane and ammonia and due to their composition in comparison to Jupiter and Saturn, which are comprised mainly of hydrogen and helium, are labelled the ice giants. Scientists have measured how hot Uranus and Neptune should be and have found that Uranus is very cold and very dim</span>
4 0
3 years ago
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