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2 years ago

Peristalsis is the action caused by contractions of muscles in this structure in the digestive system...

Physics

1 answer:

ollegr [7]2 years ago

8 0

Answer:

I believe the answer is esophagus

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2. What is the percent composition of sulfur in H2SO4?

allochka39001 [22]

Answer:

C. 32.7%

Explanation:

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 % pls mark brainliest

6 0

2 years ago

The two cyclists travel at the same speed on level ground. They approach a low hill and decide to coast up instead of hard pedal

Sergeeva-Olga [200]

Answer:

<u>Option-(E): </u>Bike B; its wheels have a smaller moment of inertia.

Explanation:

The two cyclists travel at the same speed on level ground, as they approach a low hill and decide to coast up instead of hard pedaling. At the top of the hill, as bike "B" will have a larger speed, as compared to the bike "A".Due, to the effect of having smaller moment of inertia, I=m×r².

Having no friction nor air resistance, and all the wheels roll on the ground without slipping. While, the B have larger speed,v due to the fact that bike "B" will have a larger speed,"v" as compared to the bike "A".

3 0

2 years ago

The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s

Bumek [7]

Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: $m_{(t)} =m_{0} e^{-kt}$ , where $m_{0}$ is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

$m_{(1)} =m_{0} e^{-k(1)}$ , but $m_{(1)}=\frac{99m_{0} }{100}$ , so we have $\frac{99m_{0} }{100}=m_{0}e^{-k}$ , then $k=-ln(\frac{99}{100})=0.01$

Now using k found we must to find $m_{(5)}$ .

$m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg$

6 0

2 years ago

Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite

nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

$\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s$

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

$J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s$

Hence, this is the required solution.

7 0

2 years ago

Plss help me While playing at the park, Katie and her sister take turns going down the slide and playing on the swings.

Stels [109]

Wait it’s B I’m pretty sure

8 0

2 years ago