8) PLZ HELP ME PLZ HELP ME ASAP ASAP ASAP

What type of weather would be in the middle of the picture, where the barometric pressure is 1000 millibars?

UkoKoshka [18]

It would be D. Hope that helped :)

4 0

2 years ago

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

$\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r$

where,

$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

$\tau _{max}=\frac{4}{3}\times \frac{V}{A}$

Applying values we get

$\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa$

3 0

2 years ago

A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met

shtirl [24]

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

$\epsilon=BvL$

where

B is the strength of the magnetic field

v is the speed of the wire

L is the length of the wire

For the wire in this problem, we have:

$\epsilon=12 V$ (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving for B, we find the strength of the magnetic field:

$B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T$

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

6 0

2 years ago

A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonis

icang [17]

Answer:

t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

$y = u t + \dfrac{1}{2}at^2$

$-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2$

$4.9 t^2 - 2.51 t - 3.16 = 0$

using quadratic formula to solve the equation

$t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}$

t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

8 0

2 years ago

The mass of the Earth is 6 × 1024 kg, the mass of the Moon is 7 × 1022 kg, and the center-to-center distance is 4 × 108 m. How f

Karolina [17]

Answer:

The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

Explanation:

Let the reference point be the center of the Earth

$X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m}$

Where;

Xcm is the distance from center of the Earth =?

Me is the mass of the Earth = 6 × 10²⁴ kg

Xe is the center mass of the Earth = 0

Mm is the mass of the moon = 7 × 10²² kg

Xm is the center mass of the moon = 4 × 10⁸ m

$X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m} = \frac{M_e(0) +M_mX_m}{M_e+M_m} = \frac{ M_mX_m}{M_e+M_m}}\\\\X_C_m = \frac{7 X 10^{22}*4X10^8}{6X10^{24}+7X10^{22}} =\frac{28 X10^{30}}{607X10^{22}}\\\\ X_C_m = 4.613 X10^6 m$

Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

8 0

2 years ago