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pav-90 [236]
3 years ago
15

Dos láminas, una de cobre y otra de hierro, se encuentran soldadas y empotradas enuna pared como lo muestra la figura 13. Si las

láminas se encuentra a 20 ºC ysabiendo que el coeficiente de dilatación térmica del cobre es mayor que el hierro,entonces se podría predecir que a una temperatura de 100 ºC
A) El extremo libre se doblará hacia A.
B) el extremo libre se doblará hacia B.
C) las láminas se dilatarán sin doblarse.
D) las láminas se contraerán sin doblarse.
E)la lámina de coeficiente de dilatación térmica menorimpedirá la dilatación de la otra.
Physics
1 answer:
PolarNik [594]3 years ago
7 0

Answer:

umm i gotta put this in English!!!!!

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A 10 kg mass rests on a table. What acceleration will be generated when a force of 5 N is applied? a 0.5 m/s2 b 3.5 m/s2 c 5 m/s
rosijanka [135]

Answer:

a= 0.5m/s^2

Explanation:

Force applied on an object is known as

F=m.a  (Newton's second law states it)

a=F/m

a=5/10=0.5m/s^2

3 0
3 years ago
The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of
adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = ma_a (0.35)

1.8v^2(1.1) - 1200(9.8)(1.25) = 1200a(0.35)

1.8v^2(1.1) - 14700 = 420 a   ------- equation (1)

F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a             --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

{1.1 * 1200 \ a} - 14700 = 420 a

1320 a - 14700 = 420 a

1320 a -  420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is  16.33 m/s²

5 0
3 years ago
Melissa's favorite exercise equipment at the gym consists of various springs. in one exercise, she pulls a handle grip attached
konstantin123 [22]

The formula for force exerted on/by a spring is 

F = k*e where k is the spring constant and x is the distance stretched from unstrained position. This should allow you to find what you need.

 

Using F = k x e, 


where k is the spring constant, 


and e is the extension, 


The F is her weight = 45 X 0.80 


= 36 N

6 0
3 years ago
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

8 0
3 years ago
Consider this statement: Air is matter. Which facts best support the statement?
mihalych1998 [28]
The answers is A and C hope this helps :)
8 0
3 years ago
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