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3 years ago

How does the diaphragm remove air from the lungs?

Physics

1 answer:

LekaFEV [45]3 years ago

4 0

Answer:

d) When the diaphragm pushes upward, it decreases the volume of the lungs, pushing air out.

Explanation:

the relaxed diaphragm muscle is dome-shaped, the contracted muscle is horizontal.

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What is the final velocity of a car that is originally traveling 12 m/s and then undergoes an acceleration of 2.3m/s squared for

Katarina [22]

To solve this problem we use the general kinetic equations.

We need to know the time it takes for the car to reach 130 meters.

In this way we have to:

$x(t) = x_0 + v_0t + 0.5at ^ 2$

Where

$x_0$ = initial position

$v_0$ = initial velocity

$a$ = acceleration

$t$ = time

$x(t)$ = position as a function of time

$130 = 0 + 12(t) + 0.5(2.3)t ^ 2$

$1.15t ^ 2 + 12t - 130$ .

We use the quadratic formula to solve the equation.

$t = \frac{-12 \± \sqrt {(12) ^ 2-4(1.15)(- 130)}}{2 (1.15)}$

t = 6.63 s and t = -17.1 s

We take the positive solution. This means that the car takes 6.63 s to reach 130 meters.

Then we use the following equation to find the final velocity:

$v_f = v_0 + at$

Where:

$v_f$ = final speed

$v_f = 12 +2.23(6.63)$

The final speed of the car is 27.25 m/s

3 0

4 years ago

A 700 kg racecar slowed from 30 m/s to 15 m/s. What was the change in momentum?

xxMikexx [17]

Momentum equation is

change in momentum = mass•initial velocity•final velocity

so....

p=700(15) because your initial is 30, and your final in 15, so you subtract! hope that helped!

8 0

3 years ago

A small marble attached to a massless thread is hung from a horizontal support. When the marble is pulled back a small distance

s2008m [1.1K]

Answer:

f' = 2 f

Explanation:

The frequency of the pendulum that swings in simple harmonic motion is given by :

$f={2\pi}\sqrt{\dfrac{l}{g}}$

Where

l is the length of pendulum

g is the acceleration due to gravity

If the length of the thread is increased by a factor of 4, such that, l' = 4 l, let f' is the new frequency such that,

$f'={2\pi}\sqrt{\dfrac{l' }{g}}$

$f'={2\pi}\sqrt{\dfrac{4l}{g}}$

$f'=2\times {2\pi}\sqrt{\dfrac{l}{g}}$

f' = 2 f

So, the new frequency of the pendulum will become 2 time of initial frequency. Hence, the correct option is (b) "2f"

3 0

3 years ago

The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With

tangare [24]

Answer:

0.7 mJ

Explanation:

Identify the unknown:

The work required to turn the dial from 180° to 0°

List the Knowns:

Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F

Voltage of the battery: V = 130 V

Set Up the Problem:

Energy stored in a capacitor:

U_c=1/2*V^2*C

=1/2*Q^2/C

When the dial is set at 180°:

U_c=1/2*(130)^2*350*10^-12=10^-4

Q=√2*U_c*C=4*10^-7

When the dial is set at 0°:

U_c=1/2*(4*10^-7)^2/100*10^-12

=8*10^-4 J

Solve the Problem:

ΔU_c=7*10^-4 J

=0.7 mJ

note:

there maybe error in calculation but method is correct

4 0

3 years ago

Read 2 more answers

Sprinting near the end of a race, a runner with a mass 63 kg accelerates from a speed of 25 m/s to a speed of 26 m/s in 54 s. To

gavmur [86]

Answer:

1.1655 N

Explanation:

Given that,

Initial speed, u = 25 m/s

Final speed, v = 26 m/s

Time taken, t = 54 s

So, Applying equation of motion as:

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{26-25}{54}\ m/s^2\\\Rightarrow a=0.0185\ m/s^2$

According to the Newton's second law of motion:-

$Force=Mass\times Acceleration$

Mass = 63 kg

So,

$Force=63\times 0.0185\ kgm/s^2$

Force = 1.1655 N

5 0

4 years ago