Hypoventilation dramatically increases carbonic acid concentration and involves_______

If the volume of a container increases what effect will this have on pressure? Why?

frozen [14]

As the volume of the container increases the pressure inside will decrease because the atoms have more room to move around in.

5 0

2 years ago

Read 2 more answers

A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

pentagon [3]

Answer:

The block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

$v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}$

Now, plug in the given values and solve for 'a'. This gives,

$a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2$

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

$v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}$

Now, plug in the given values and solve for 'S'. This gives,

$S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m$

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0

3 years ago

Please helo me to 1st question

k0ka [10]

Explanation:

6 min in seconds:

6×60

<h2>360 seconds </h2><h2 /><h2 />

6 min in hours:

<h2>0.1 hours</h2>

6 0

3 years ago

Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant

evablogger [386]

(a) $2.56\cdot 10^{-5} C$

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

$F=(0.021)(1.1)=0.0231 N$

The electrostatic force between the two balloons can be also written as

$F=k\frac{Q^2}{r^2}$

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

$Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C$

(b) $1.6\cdot 10^{14}$ electrons

The magnitude of the charge of one electron is

$e=1.6\cdot 10^{-19}C$

While the magnitude of the charge on one balloon is

$Q=2.56\cdot 10^{-5} C$

This charge can be written as

$Q=Ne$

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

$N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}$

5 0

3 years ago

If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil

Hoochie [10]

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second and closest to the source charge. The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source when Furthest away the bond with charge will be less effective 

5 0

3 years ago

Hypoventilation dramatically increases carbonic acid concentration and involves________?