Help ;-;

May some one help a poor MHA fan out with science

guapka [62]

Answer:

nm

Explanation:

4 0

3 years ago

Which of the following statements is true about earth's magnetic poles

Dahasolnce [82]

Answer: b) they are the areas where Earth's magnetic field is weakest

Explanation:

According to classical physics, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This is because for <em>classical physics</em>, naturally, magnetic monopoles can not exist.

In this context, Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet. Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.

Being the magnetic poles the places where the Earth's magnetic field is weakest. And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).

4 0

3 years ago

You travel at a speed of 24 m/s. How much time will it take you to travel 127m (round to two decimal places if needed)

shtirl [24]

Answer: 5.29 seconds

Explanation:

if the speed is 24 meters per second than it would take you 5.29 seconds or 127meters/24m/s to get 5.29s

4 0

3 years ago

8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an

77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = $\frac{F}{\frac{\pi }{4}(d^2) }$

To find force (F) =

120 = $\frac{F}{\frac{\pi }{4}(100^{2} )}$

F = 120 * (π/4) * (100)

F = 9425N

Calculate the true strain (ε) = In (l₀ / l₁)

where

l₀ = initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

= In 1.1667

= 0.154cm/cm

Calculate the true stress (σ) at fracture point

= $\frac{F}{\frac{\pi }{4}(d^2) }}$

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = $\frac{9425}{\frac{\pi }{4}(9.26^2) }}$

= 140MPa

To find the strength coefficient (K) of the material bar

K = $\frac{140}{\sqrt{0.154} }$

K = $\frac{140}{0.3925}$

= 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K = 356.75MPa

σ = $356.75(0.05)^0^.^5$

= $356.75 ( 0.2236)$

= 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

6 0

3 years ago

A player passes a basketball to another player who catches it at the same level from which ot was thrown. The initial speed of t

34kurt

(a) The initial direction of the ball is horizontal.

(b) The time of flight of the ball is 0.65 s.

<h3>Initial direction of ball</h3>

The initial direction of the ball is horizontal since the player catches the ball at the same level which implies that the speed of the of the ball was constant through out the motion.

<h3>Time of flight of the ball</h3>

The time of flight of the ball is calculated using the following kinematic equation.

$t = \frac{X}{v_x} \\\\t= \frac{4.6}{7.1} \\\\t = 0.65 \ s$

Learn more about time of flight here: brainly.com/question/4441382

7 0

2 years ago