What gases are exchanged in the respiratory system

Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

8 0

3 years ago

HELP ASAP TIMED TEST

balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

$U_1 = mgH$

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

$\displaystyle U_2=\frac{1}{2}mv^2$

Since the energy is conserved, U1=U2

$\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]$

For the speed to be double, we need to drop the snowball from a height H', and:

$\displaystyle mgH'=\frac{1}{2}m(2v)^2$

Operating:

$\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]$

Dividing [2] by [1]

$\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}$

Simplifying:

$\displaystyle \frac{H'}{H}=4$

Thus:

H' = 4H

Correct choice: b 4H

4 0

3 years ago

What are the best units for DENSITY of a solid like a rock or a lemon? *

Ulleksa [173]

You would use answer 2O g/cm3

4 0

3 years ago

IM DESPERATE PLS HELP ME I DONT UNDERSTAND THIS! PLEASE PLEASE PLEASE ANSWER BACK

OlgaM077 [116]

Answer:

The speed with which the man flies forward is 5.5 m/s

Explanation:

The mass of the man = 100 kg

The mass of the scooter = 10 kg

The speed with which the man was traveling on the scooter = 5 m/s

The speed of the scooter after it hits the rock = 0 m/s

Let v represent the speed with which the man flies forward

The formula for momentum, P, is P = Mass × Velocity

The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;

The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s

The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v

When the momentum is conserved, we have;

550 kg·m/s = 100 kg × v

∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.

The speed with which the man flies forward = v = 5.5 m/s

8 0

3 years ago

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0

3 years ago