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4 years ago

An iron cube has edge 15cm long at 20c .What will be? The new surface area of a face when the temperature rises to 80c

Physics

1 answer:

grigory [225]4 years ago

3 0

Answer:

271cm^2

Explanation:

volume 1= 15^3 =3375

temp. 1. = 20+273 = 293

temp. 2. = 50+273 = 353

volume 2 =?

According to Charles law

volume is proportional to temperature

v2 = v1 * t2 / t1

v2 = 3375 * 353 / 293

v2 = 4066cm^3

v = area * length

4066 = area * 15

area = 4066/15 = 271cm^2

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3 years ago

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A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?

Ber [7]

U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:

v^2=u^2+2as
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7 0

4 years ago

Se deja caer una pelota inicialmente en reposo desde una altura de 50m sobre el nivel del suelo. ¿cuanto tiempo requiere para ll

umka2103 [35]

Answer:

a) t = 3.2 s

b) $v_{f} = -32 m/s$

Explanation:

a) El tiempo requerido para llegar al suelo se puede calcular usando la siguiente fórmula:

$t = \sqrt{\frac{2y_{0}}{g}}$

En donde:

$y_{0}$ : es la altura inicial = 50 m

g: es la gravedad = 10 m/s²

$t = \sqrt{\frac{2y_{0}}{g}} = \sqrt{\frac{2*50 m}{10 m/s^{2}}} = 3.2 s$

Entonces, el tiempo requerido para llegar al suelo es 3.2 s.

b) La rapidez de la pelota justo antes del choque es el siguiente:

$v_{f} = v_{0} - gt$

En donde:

$v_{0}$ : es la velocidad inicial = 0 (dado que se deja caer en resposo)

$v_{f} = v_{0} - gt = 0 - 10 m/s^{2}*3.2 s = -32 m/s$

Por lo tanto, la rapidez de la pelota justo en el momento anterior del choque es -32 m/s (el signo negativo es porque la pelota está cayendo).

Espero que te sea de utilidad!

6 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago