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Mumz [18]
3 years ago
11

The amount of pressure exerted by a solid is based on

Physics
2 answers:
Veronika [31]3 years ago
5 0

Answer:

pressure= force/area

A solid resting on a horizontal surfaceexerts a normal contactforce equals to its weight. The pressure of the solid on the surface depends on the area of contact. (b) the area of contact between the two surfaces. The greater the force or the smaller the area the greater the pressure.

mart [117]3 years ago
4 0

Answer:force and area

Explanation:

The greater the force the greater the pressure,the smaller the area the greater the pressure

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How are frequency, wavelength, and energy related? Choose all that apply.
marta [7]

Explanation:

The energy of a wave is given by :

E=\dfrac{hc}{\lambda}

Where

h is Planck's constant

c is the speed of light

\lambda is wavelength

Energy is inversely proportional to wavelength. Also, the relation between frequency and wavelength is inverse.

If the frequency is high, the wavelength will be shorter.

Hence, the correct options are :

Higher frequencies have shorter wavelengths.

Shorter wavelengths have lower energy.

Lower frequencies have lower energy.

7 0
2 years ago
If you wanted to know how fast a tiger can run, which measurements would you use?
sp2606 [1]
<span>D is the correct answer. If you wanted to know how fast a tiger can run, you would use meters and seconds. Meters will tell you the distance that the tiger ran and seconds will tell you how quickly the tiger ran the distance.</span>
6 0
2 years ago
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This passage best illustrates which reaction to multiculturalism?
Luden [163]

Answer:

D

Explanation:

by having different ethnic groups

7 0
2 years ago
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Two resistors of resistances R1 and R2, with R2&gt;R1, are connected to a voltage source with voltage V0. When the resistors are
Contact [7]

Answer:

r=0.127

Explanation:

When  connected in series

Current = I

When connected in parallel

Current = 10 I

We know that equivalent resistance

In series  R = R₁+R₂

in parallel  R= R₁R₂/(R₂+ R₁)

Given that voltage is constant (Vo)

V = I R

Vo = I (R₁+R₂)  ------------1

Vo = 10 I (R₁R₂/(R₂+ R₁)) -------2

From above equations

10 I (R₁R₂/(R₂+ R₁)) = I (R₁+R₂)  

10  R₁R₂ =  (R₁+R₂) (R₂+ R₁)

10  R₁R₂  = 2  R₁R₂  + R₁² + R₂²

8 R₁R₂  =     R₁² + R₂²

Given that

r =  R₁/R₂

Divides by R₂²

8R₁/R₂  = ( R₁/R₂)²+ 1

8 r = r ² + 1

r ² - 8 r+ 1 =0  

r= 0.127 and r= 7.87

But given that R₂>R₁  It means that r<1 only.

So the answer is r=0.127

8 0
3 years ago
La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.
KATRIN_1 [288]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

3 0
2 years ago
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