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Anna35 [415]
3 years ago
13

Differences between horrseand horse​

Physics
2 answers:
liq [111]3 years ago
6 0
The difference is that they are spelled differently
kenny6666 [7]3 years ago
5 0
The proper difference between hor Rse and horse is both shows the same thing that is the horse
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Please i need answer for part d
matrenka [14]
It is c that’s what it is that’s the answer
4 0
2 years ago
What is the magnetic field at the center of a circular loop
Elan Coil [88]

Answer:

The magnetic field at the center of a circular loop is 3.14\times10^{-5}\ T.

Explanation:

Given that,

Radius = 4.0 cm

Current = 2.0 A

We need to calculate the magnetic field at the center of a circular loop

Using formula of magnetic field

B = \dfrac{I\mu_{0}}{2r}

Where, I = current

r = radius

Put the value into the formula

B =\dfrac{2.0\times4\pi\times10^{-7}}{2\times4.0\times10^{-2}}

B =0.00003141\ T

B=3.14\times10^{-5}\ T

Hence, The magnetic field at the center of a circular loop is 3.14\times10^{-5}\ T.

7 0
3 years ago
A plumbing branch is composed of 68' of 1" diameter Type L copper pipe, three 90-degree elbows (wrought), and a gate valve. The
valina [46]

Answer:

7.404 psi

Explanation:

See attachment please

5 0
3 years ago
A small 0.14 kg metal ball is tied to a very light (essentially massless) string that is 0.9 m long. The string is attached to t
Vsevolod [243]

Answer:

a) v=2.743m/s

b) a_c = 8.363m/s^2

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

y' = 0.9cos(55)

y' = 0.516

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

E_p=E_k\\mgh=\frac{mv^2}{2}

Solving for v:

v=\sqrt{2gh}

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

a_c=\frac{v^2}{r}

a_c = 8.363m/s^2

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N

4 0
3 years ago
At temperatures near absolute zero, Bc approaches 0.142 T for vanadium, a type-I superconductor. The normal phase of vanadium ha
ioda

Answer:

b) field is zero,  c) the magnetic field does not change in intensity or direction

e) M = -H = Bo /μ₀ ,  g)  M = 0

Explanation:

Part b

superconductors are formed by so-called Coper pairs that are electrons linked through a distortion in the network, this creates that they must be treated as an entity so we have an even number of charge carriers and the material must behave with diamagnetic , Meissner effect, consequently the magnetic field inside its superconductor is zero

the correct answer is Zero

Part c

 outside the superconducting cylinder the magnetic field does not change in intensity or direction

Part E

Magnetization is defined by the equation

       B = μ₀ (H + M)

with field B it is zero inside the superconductors

        M = -H = Bo /μ₀

         

where Bo is the magnetic induction in the normal state

Part g

 As outside the cylinder there is no material zero magnetization

        M = 0

6 0
3 years ago
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