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3 years ago

how much heat must be added to 30.0 g of solid iron to melt it, assuming the iron is already at it's melting point?

Physics

1 answer:

Dmitry [639]3 years ago

6 0

Answer:

2,800°F (1,538°C)

Explanation:

30g of iron will melt at the same temperature as 1g of iron.

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Which bright object is in shadow

IRINA_888 [86]

My best guess would be sun because it is bright but is surrounded by shadows on all sides.

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3 years ago

Identify the chemical change below

lesya [120]

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4 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

Greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but

maw [93]

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

${\large{\bold{\rm{\underline{To\; find}}}}}$

★ The speed of the hound and the hare

${\large{\bold{\rm{\underline{Solution}}}}}$

★ The speed of the hound and the hare = 25:18

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

$\dashrightarrow$ As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

(5/3)a × 5

25/3a metres

Now the distance travelled by the hare in it's 6 leaps..!

6a metres

Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

25/3a = 6a

25/3 = 6

25:18

5 0

3 years ago

Hello, I want to ask. . anyone knows the answer.

stealth61 [152]

I would say D. because you round to the nearest whole number and 0.04 is way less than 0.5 which is a good rounding up number.

5 0

3 years ago

how much heat must be added to 30.0 g of solid iron to melt it, assuming the iron is already at it's melting point?​

how much heat must be added to 30.0 g of solid iron to melt it, assuming the iron is already at it's melting point?