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mamaluj [8]
2 years ago
10

What is unique regarding the abdominal muscle when is compared to other muscle in the body?

Physics
1 answer:
marissa [1.9K]2 years ago
8 0
Answer:a is the answer
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The volume of the lung 0.0024m^3 following exhalation and the pressure is 101.70KPa. Calculate the volume of the lungs during in
Mekhanik [1.2K]

According to Boyle-Mariotte:

p₁V₁=p₂V₂=>V₂=p₁V₁/p₂= 0.0024*101.70/ 84.16=0.0028 m³

5 0
3 years ago
How do objects become negatively charged using the contact method
Kamila [148]

If a negative object is used to charge a neutral object, then both objects become charged negatively. In order for the neutral sphere to become negative, it must gain electrons from the negatively charged rod. A metal sphere is electrically neutral. It is touched by a positively charged metal rod.

7 0
3 years ago
You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i
Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

W_s = \frac{1}{2}kc^2

Work lost to friction:

W_f =\mu mg(c-x)

Energy:

E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2

(a) Solve for v:

v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}

(b) Solve \frac{dv}{dx}=0 for x:

\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}

\frac{dv}{dx}=0 if:

\mu g-\frac{k}{m}x = 0

x_{max} = \frac{\mu gm}{k}

v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}

6 0
3 years ago
A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu
serg [7]

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

5 0
3 years ago
How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec     time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0
11 months ago
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