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Gekata [30.6K]
3 years ago
14

Characteristics we use to tell the differences between kinds of matter pare called _ properties

Physics
1 answer:
lorasvet [3.4K]3 years ago
6 0

The answer to the blank is physical.

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The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl
Aleks04 [339]

The capacitance of the capacitor is 1.8\cdot 10^{-9}F

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k is the dielectric constant of the medium

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

d=3.5\cdot 10^{-5}m is the separation between the plates

A=0.063m \cdot 0.054m = 0.0034 m^2 (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the  capacitor is

C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

5 0
3 years ago
The gravitational potential energy of an object is equal to its weight multiplied by its
Elina [12.6K]

Answer:

Height above a surface

Explanation:

Gravitational potential energy is the energy which an object possesses due to its position above a surface.

It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.

It is given mathematically as:

P. E. = m*g*h

where m = mass of the body

g = acceleration due to gravity

h = height above a surface

m*g represents the weight of the object.

Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.

3 0
3 years ago
1) A plane lands on a runway with a speed of 125 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m
valentina_108 [34]

Answer: 1) a = 9.61m/s² pointing to west.

             2) (a) Δv = - 37.9km/s

                  (b) a = - 6.10⁷km/years

Explanation: Aceleration is the change in velocity over change in time.

1) For the plane:

a=\frac{\Delta v}{\Delta t}

a=\frac{125}{13}

a = 9.61m/s²

The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.

2) Total change of velocity:

\Delta v = v_{f}-v_{i}

\Delta v = -17.1-(+20.8)

\Delta v= -37.9km/s

The interval is in years, so transforming seconds in years:

v = \frac{-37.9}{3.15.10^{-7}}

v=-12.03.10^{7}km/years

Calculating acceleration:

a=\frac{-12.03.10^{-7}}{2.01}

a=-6.10^{7}

Acceleration of an asteroid is a = -6.10⁷km/years .

5 0
2 years ago
Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ
Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

v=v_o+at    (1)

x=v_ot+\frac{1}{2}at^2   (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s

Next, you solve the equation (1) for the acceleration a:

a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0
3 years ago
During an earthquake you should.
White raven [17]

Well, you should TAKE COVER UNDER A STUDY DESK, TABLE OR AGAINST AN INSIDE WALL. Which is C. This is because, doing this protects you from the falling objects. If you do A or B, you will probably get crushed by falling objects because you are exposed. Hope I helped.
3 0
2 years ago
Read 2 more answers
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