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sertanlavr [38]

2 years ago

10

13. A diver swims to a depth of 3.2 m in a freshwater lake. What is the increase in the force pushing in on her eardrum, compare

d to what it was at the lake surface? The area of the eardrum is 0.60 cm².

1 answer:

Gnoma [55]2 years ago

6 0

Answer:

1.88 N

Explanation:

h = 3.2 m, A = 0.6 cm^2 = 0.6 x 10^-4 m^2

density of water, d = 1000 kg/m^3, g = 9.8 m/s^2

Pressure at depth h, P = h x d x g = 3.2 x 1000 x 9.8 = 31.36 x 10^3 Pa

Force = pressure x Area = 31.36 x 10^3 x 0.6 x 10^-4 = 1.88 N

Thus, the force on ear drum is 1.88 N.

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What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL

pogonyaev

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of object = $6.0g/ml$

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

$6.0g/ml=\frac{3.0g}{\text{Volume of substance}}$

${\text{Volume of substance}}=0.50ml$

Thus the volume of an irregularly shaped object is 0.50 ml

4 0

2 years ago

Write a hypothesis about the effect of the fan speed on the acceleration of the cart. Use the "if . . . then . . . because . . .

iragen [17]

As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
$s = ut + \frac{1}{2} at {}^{2}$
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time

Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)

8 0

2 years ago

Read 2 more answers

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. If a rock weighs 75.0 N on earth, compu

IgorLugansk [536]

Answer:

g = 0.905 gE

W = 67.9 N

Explanation:

given data

mass of Venus mv = 81.5% = 0.815

radius Rv = 94.9% = 0.949

weighs W = 75.0 N

solution

we apply here acceleration due to gravity at earth surface that is

g = $\frac{Gm}{R^2}$ = 9.80 m/s² ............1

so

g = $\frac{G(0.815)}{0.949R^2}$

g = 0.905 gE

and

W = m gv

W = 0.905 m gE

W = 0.905 × 75

W = 67.9 N

7 0

2 years ago

A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate

tiny-mole [99]

Answer:

a. Time = 16.11 s

b. Gauge Pressure = 1009400 Pa = 1 MPa

c. Absolute Pressure = 1110725 Pa + 1.11 MPa

d. Force = 2.22 MN

Explanation:

a.

For the accelerated part of motion of submarine we can use equations of motion.

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

where,

t₁ = time taken during accelerated motion = ?

Vf = final velocity = 4 m/s

Vi = Initial Velocity = 0 m/s (Since, it starts from rest)

a = acceleration = 0.3 m/s²

Therefore,

t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)

t₁ = 13.33 s

Now, using 2nd equation of motion:

d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

where,

d₁ = the depth covered during accelerated motion

Therefore,

d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²

d₁ = 88.89 m

Hence,

d₂ = d - d₁

where,

d₂ = depth covered during constant speed motion

d = total depth = 100 m

Therefoe,

d₂ = 100 m - 88.89 m

d₂ = 11.11 m

So, for uniform motion:

s₂ = vt₂

where,

v = constant speed = 4 m/s

t₂ = time taken during constant speed motion

11.11 m = (4 m/s)t₂

t₂ = 2.78 s

Therefore, total time taken by submarine to move down 100 m is:

t = t₁ + t₂

t = 13.33 s + 2.78 s

<u>t = 16.11 s</u>

<u></u>

b.

The gauge pressure on submarine can be calculated by the formula:

Pg = ρgh

where,

Pg = Gauge Pressure = ?

ρ = density of salt water = 1030 kg/m³

g = 9.8 m/s²

h = depth = 100 m

Therefore,

Pg = (1030 kg/m³)(9.8 m/s²)(100 m)

<u>Pg = 1009400 Pa = 1 MPa</u>

<u></u>

c.

The absolute pressure is given as:

P = Pg + Atmospheric Pressure

where,

P = Absolute Pressure = ?

Atmospheric Pressure = 101325 Pa

Therefore,

P = 1009400 Pa + 101325 Pa

<u>P = 1110725 Pa + 1.11 MPa</u>

<u></u>

d.

Since, the force to open the door must be equal to the force applied to the door by pressure externally.

Therefore, the force required to open the door can be found out by the formula of pressure:

P = F/A

F = PA

where,

P = Absolute Pressure on Door = 1110725 Pa

A = Area of door = 2 m²

F = Force Required to Open the Door = ?

Therefore,

F = (1.11 MPa)(2 m²)

<u>F = 2.22 MN</u>

8 0

2 years ago