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3 years ago

5

Scientific theories are deductive in nature.?

1 answer:

dolphi86 [110]3 years ago

8 0

Answer:

deductive reasoning usually follows steps .

That is, how we predict what the observations should be if the theory were correct

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9. A 30 cm ruler is found to have a center of mass of 15.6 cm. The percent error of the center of mass is _____, if the ruler is

grigory [225]

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

4 0

3 years ago

If you are driving your car at 45.0 miles per hour, and you have been driving 2.50 hours, how far has your car gone?

Greeley [361]

Answer:

113 miles

Explanation:

45.00 x 2.50= 1.12.5 so 113 miles in 2.50 hours

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3 years ago

Which term describe the substance that forms when two or more types of atoms join?

xz_007 [3.2K]

C i’m pretty sure but don’t take me as your first answer i just need to answer a question to ask one.

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3 years ago

Read 2 more answers

A series rlc circuit is in resonance with 600 hz. by what factor must the capacitance be multiplied change the resonance frequen

quester [9]

Fvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

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3 years ago

Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that

Greeley [361]

Answer:

The value is $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

So here heat gained water is mathematically represented as i.e

$Q_w = m_w * c_w * (T_w - T )$

substituting 0.150 kg for $m_w$ , 4200 J/kg.°C for $c_w$ , 100°C for $T_w$ and 75°C for T

We have

$Q_w = 0.150 * 4200 * (100 - 75 )$

$Q_w =15750 \ J$

The Heat loss by the ice is mathematically represented as

$Q_i = Q_1 + Q_2 + Q_3$

Here $Q_1$ is the energy to move the ice to its melting point which is evaluated as

$Q_1 = m_i * c_i * ( T_o -T_i)$

Here $m_i$ is the mass of ice

$c_i$ is the specific heat of ice with value $2.05 * 10^3 J/kg.°C$

$T_o$ temperature of ice at melting point with value 0°C

$T_i$ is the temperature of ice with value -12°C

$Q_2$ is the energy to move the ice from its its melting point to liquid which is evaluated as

$Q_2 = m_i * L$

Here L is the Latent heat of melting of ice with value $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as

$Q_1 = m_i * c_w * ( T -T_o)$

So

$Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=> $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

We have that

$m_i * 673600 =15750$

=> $m_i = \frac{15750}{673600}$

=> $m_i = 0.0234 \ kg$

8 0

4 years ago