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hammer [34]
3 years ago
6

Mount st Helens is a volcano that erupted with great force in 1980 which phrase

Physics
2 answers:
Rashid [163]3 years ago
5 0

Answer:

high-viscosity magma and lava broken into fragments

Explanation:

ololo11 [35]3 years ago
3 0
Which phrase is there more information I’m confused sorry
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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th
Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L  = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

Explanation:

Given that :

The internal shear force V  =  80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress  \tau can be calculated by using the equation:

\tau = \dfrac{VQ}{Ib}

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10  mm

From the centroid ;

Q = Q_1 + Q_{2}

Q = A_1y_1 + A_{2}y_{2}  

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5)  ) mm³

Q = ( 38500 +  307125 ) mm³

Q = 345625 mm³

\tau_H = \dfrac{VQ}{Ib}

\tau_H = \dfrac{80*10^3  * 345625}{64900000*10 }

\tau_H = \dfrac{2.765*10^{10}}{649000000 }

\tau_H = 42.60400616 \ N/mm^2

\mathbf{\tau_H \approx  42.604 \ N/mm^2}

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

7 0
3 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 times 10^6 m and a mass of 7.35 times
Eduardwww [97]

Answer:

aaaaaaaaaaaaaaaaaaaaaahh sorry

6 0
3 years ago
Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field
morpeh [17]

To develop the problem it is necessary to apply the concepts related to Magnetic Field.

The magnetic field is defined as

B = \frac{\mu_0 I}{2\pi r}

Where,

\mu_0 = Permeability constant in free space

r = Radius

I = Current

Our values are given as,

B = 0.1T

d = 4.5mm

r = 2.25mm

If the maximum current that the wire can carry is I, then

B = \frac{\mu_0 2I}{4\pi r}

I = \frac{Br}{2\frac{\mu_0}{4\pi}}

I = \frac{(0.1T)(2.25*10^{-3}m)}{2(1*10^{-7}N/A^2)}}

I = 1125A

Therefore the maximum current is 1125A

4 0
4 years ago
What is the surface area if the shape?
Oliga [24]
=112yd²
Because 2x4=8
8x2=16
8x4=32
32x2=64
2x8=16
16x2=32

16+64+32=112
4 0
2 years ago
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