All categories

2 years ago

The two upper chambers of the human heart are known as:

Physics

2 answers:

Juli2301 [7.4K]2 years ago

7 0

I would guess the correct answer is choice A. Atriums

True [87]2 years ago

3 0

The answer is atriums

You might be interested in

Which of the following is calculated by dividing work by time?

Svet_ta [14]

A is the answer i think

4 0

3 years ago

Why do I need to use the Unit Circle in Physics? And how do I use it?

stiks02 [169]

Answer:

The unit circle helps in making so many calculations and equations easy.

Explanation:

You cannot separate the knowledge of trigonometry to application in equations to physics. The unit circle is known to have a radius of one. This means that the distance from the centre of the circle, regardless of the unit of measurement, to any point of the edge of the circle is 1. Since the unit circle is very helpful in trigonometry, and trigonometry in turn is the projection of triangles and angles that is very crucial in the calculation of momentum, velocity and other factors of physics, the importance of the unit circle cannot be overemphasized in physics.

6 0

2 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

28. A stone is projected at a cliff of height h with an initial speed of 42.0 mls directed at angle θ0 = 60.0° above the horizon

Julli [10]

Answer:

a) 51.8 m, b) 27.4 m/s, c) 142 m

Explanation:

Given:

v₀ = 42.0 m/s

θ = 60.0°

t = 5.50 s

Find:

h, v, and H

a) y = y₀ + v₀ᵧ t + ½ gt²

0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²

h = 51.8 m

b) vᵧ = gt + v₀ᵧ

vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)

vᵧ = -17.5 m/s

vₓ = 42.0 cos 60.0

vₓ = 21.0 m/s

v² = vₓ² + vᵧ²

v = 27.4 m/s

c) vᵧ² = v₀ᵧ² + 2g(y - y₀)

0² = (42.0)² + 2(-9.8)(H - 51.8)

H = 142 m

4 0

3 years ago

Kelli drew a diagram to compare cast and imprint fossils. Which label belongs in the area marked Y? Involves minerals replacing

STatiana [176]

Answer:

the answer is c

Explanation:

an imprint gives evidence of an organisms activity they don't normally fill with sediments

8 0

2 years ago

Read 2 more answers