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2 years ago

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A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the

gauge pressure on the bottom surface is 1133 Pa. What is the density of the fluid

1 answer:

Zolol [24]2 years ago

4 0

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

$\Delta P=P_{b}-P_{t}$

Where, $P_{b}$ = Pressure at bottom

$P_{t}$ = Pressure at top

put the value into the formula

$\Delta P=1133-594$

$\Delta P=539\ Pa$

Using formula of pressure for density

$\Delta P = \rho g h$

$\rho =\dfrac{\Delta P}{gh}$

Where, $\rho$ = density

P = pressure

h = height

Put the value in to the formula

$\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}$

$\rho =1100\ kg/m^3$

Hence, The density of the fluid is 1100 kg/m³.

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For metalloids on the periodic table, how do the group number and the period number relate?

lapo4ka [179]

Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right

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2 years ago

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A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal

ASHA 777 [7]

Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
$F_{centripetal\:force} = \frac{m*s^2}{R}$

Solving:
$F_{centripetal\:force} = \frac{m*s^2}{R}$
$F_{centripetal\:force} = \frac{68*3.9^2}{6.5}$
$F_{centripetal\:force} = \frac{68*15.21}{6.5}$
$F_{centripetal\:force} = \frac{1034.28}{6.5}$
$\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}$
Answer:
<span>B.159 N</span>

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2 years ago

A potential difference of 10.0 volts exists between two points, A and B, within an electric field. What is the

Viefleur [7K]

Answer:

1. 5.0 x 10^2 C

Explanation:

V=W/Q

10 = 2.0 x 10^-2/Q

Q = 2.0 x 10^-2/ 10

Q = 5.0 x 10^2 C

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2 years ago

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

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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.

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2 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

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2 years ago