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4 years ago

List two reasons for why a non-physicist might be interested in electrostatic interactions.

Physics

1 answer:

-Dominant- [34]4 years ago

3 0

Answer:

all areas of knowledge that wish to understand the physical, chemical and biological process must know electrostatics

Explanation:

Electrostatic interactions, have many rare manifestations in nature, which causes many reasons to study them.

- Lightning is a very striking form of electricity

- The biological processes are governed by currents of inanes and potential differences

- The transfer of nutrients and fertilizers to plants is with ion exchange, electrostatic forces

- all modern electronics is based on electricity

- the electric charge in very dry places, creates high currents that can create fires or kill people

In summary all areas of knowledge that wish to understand the physical, chemical and biological process must know electrostatics

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Which non-mineral nutrients are essential in photosynthesis? (Select all that apply.) oxygen calcium hydrogen carbon

Lunna [17]

Answer:

carbón, hydrogen and oxygen

4 0

3 years ago

Two objects made of the same material are heated to 60 oC and 90 oC. According to

nlexa [21]

Assuming that there is in a vacuum, the two object will cool at the same rate, because the objects are made of the same material they will have the same cooling rate, assuming the surrounding temperature is the same.

4 0

3 years ago

Read 2 more answers

Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially

statuscvo [17]

Answer:

Final speed is 900.06 m/s at $0.2215^{\circ}$

Solution:

As per the question:

Mass of the first asteroid, m = $15\times 10^{3}\kg$

Mass of the second asteroid, m' = $20\times 10^{3}\kg$

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, $\theta = 20^{\circ}$

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

$\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}$

Thus applying vector addition and momentum conservation, the final velocity is given by:

$(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}$ (1)

Now,

$(m +m')v_{final} = (35\times 10^{3})v_{final}$

$(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}$

$(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}$

$2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}$

Now, substituting the suitable values in eqn (1), we get:

$v_{final} = 900.06\ m/s$

Now, the direction for the two vectors is given by:

$\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}$

$\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}$

5 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

4 years ago

Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with

Firdavs [7]

Answer:

wavelength = 0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 * 0.220 s/2

x= 162.8 m

8 0

3 years ago