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3 years ago

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Um corpo de massa 50g recebe 300 calorias e sua temperatura sobe de -10°C até 20°C. Determine a capacidade térmica do corpo e o

calor especifico da substancia que o constitui.

1 answer:

kykrilka [37]3 years ago

7 0

The specific heat capacity of the substance is $0.963 J/g^{\circ}C$

Explanation:

When an object of mass m is supplied with a certain amount of energy Q, its temperature increases according to the equation:

$Q=mC_s \Delta T$

where

m is the mass of the object

$C_s$ is its specific heat capacity

$\Delta T$ is the increase in temperature of the object

In this problem, we have

$Q=300 cal \cdot 4.814 = 1444.2 J$

m = 50 g

$\Delta T = 20C-(-10C)=30^{\circ}C$

Therefore, we can solve for $C_s$ to find its specific heat capacity:

$C_s = \frac{Q}{m\Delta T}=\frac{1444.2}{(50)(30)}=0.963 J/gC$

Learn more about specific heat capacity:

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List two industrial uses for nonmetallic mineral resources

Anni [7]

Answer:

To make useful chemicals and abrasives

Explanation:

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2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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3 years ago

A lacrosse ball that is thrown straight upwards reaches a maximum height of 4.5 m. At what initial velocity was it thrown? (note

shtirl [24]

Answer:

The initial velocity was 9.39 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

The ball is thrown straight upward with initial velocity u

The ball reaches a maximum height of 4.5 m

At the maximum height velocity v = 0

The acceleration of gravity is -9.8 m/s²

We need to find the initial velocity

The best rule to find the initial velocity is <em>v² = u² + 2ah</em>, where v is

the final velocity, u is the initial velocity, a is the acceleration of

gravity and h is the height

⇒ v = 0 , h = 4.5 m , a = -9.8 m/s²

⇒ 0 = u² + 2(-9.8)(4.5)

⇒ 0 = u² - 88.2

Add 88.2 to both sides

⇒ 88.2 = u²

Take square root for both sides

⇒ u = 9.39 m/s

<em>The initial velocity was 9.39 m/s</em>

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3 years ago

A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?

Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

A force of $55\text{ }N$ is acting on a wagon along the road. The wagon weights $7.5\text{ }kg$ . Acceleration of the wagon is given as $5.3\text{ }\frac{m}{s^{2}}$ .

Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

Gravitational Force and Normal Reaction cancel out each other.

Net External Force = Mass of system/wagon $\times$ Acceleration of wagon

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$F_{friction}$ has a negative sign because it opposes the motion of the wagon.

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Anything that has mass

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