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Alexeev081 [22]
3 years ago
6

Is the SI unit of work newton?

Physics
2 answers:
Naya [18.7K]3 years ago
7 0

Answer:

Newton is the SI unit for force . Newton is kg m2

notsponge [240]3 years ago
6 0

Answer:

No

Explanation:

The s.i. unit of work is Joules. Work is measured in Joules. It is the product of force and the distance moved in the direction of the force. Force is in Newton while distance is measured in metres.

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ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
4 0
3 years ago
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
Gekata [30.6K]

Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

<u>Now the difference in the column is :</u>

\delta h=h_w-h_m

\delta h=25-1.838

\delta h=23.162\ cm

7 0
3 years ago
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
3 years ago
A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35m x 0.55m. The magnetic field has a
Cloud [144]

Answer:

E = 0.38V

Explanation:

See attachment below.

4 0
3 years ago
Which of these forms of radiation passes most easily through the disk of the Milky Way?
Rama09 [41]
<h2>Answer: 3 - infrared light</h2>

Explanation:

<u>There are certain areas of the Milky Way that cannot be observed using the visible range of the electromagnetic spectrum</u> (this includes blue light and red light). This is because these areas are covered or hidden behind columns of interstellar dust and dark matter.

However, using infrared light and sometimes radio waves, it is possible to observe the galaxy better, because this light manages to pass through all that interstellar dust.

4 0
3 years ago
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