All categories

3 years ago

The depth of a pond is 1.5m. Calculate the pressure caused by the water at the bottom of the pond.

Physics

2 answers:

Fantom [35]3 years ago

8 0

Area=1.5(1.5)=2.25m^2
Force of gravity=10N

$\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}$

$\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}$

$\\ \sf\longmapsto Pressure=4.4Pa$

aleksley [76]3 years ago

7 0

Answer:

Area=1.5(1.5)=2.25m^2

Force of gravity=10N

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}

⟼Pressure=

Area

Force

\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}

⟼Pressure=

2.25

\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}

⟼Pressure=4.4Pa

You might be interested in

A scientist records the motion of an object. She writes down that the

Phantasy [73]

The answer is Velocity

3 0

3 years ago

A student blows in a tube with one closed in that is 0.49 m long. If the frequency of the sound is 358 Hz what is the speed wave

BaLLatris [955]

Answer:175.42m/s

Explanation:

Wavelength=0.49m

Frequency=358Hz

Velocity=frequency x wavelength

Velocity=358x0.49

Velocity=175.42m/s

6 0

3 years ago

Calculate the mass of an object that has a momentum of 100kg x m/sec and velocity of 4 m/sec

lidiya [134]

P= mv

where p is momentum
m is mass
v is velocity

so it's given p= 100kgm/sec
v= 4m/s

so putting in the formula

100= m × 4

m = 25kg

3 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

How is the electrical conductivity of a metal explained by metallic bonds.

love history [14]

Answer:

Metallic bonding is the bond that exist between the atoms. The electrons do not only flow at its respective atoms but instead, contribute to a sea of delocalised electrons. As a result, metals can conduct electricity as the delocalised electrons are able to carry charges.

Explanation:

8 0

2 years ago

The depth of a pond is 1.5m. Calculate the pressure caused by the water at the bottom of the pond.​

The depth of a pond is 1.5m. Calculate the pressure caused by the water at the bottom of the pond.