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2 years ago

Specific heat refers to the amount of heat required to change 1 gram of a substance by _______ degree(s) Celsius.

Physics

2 answers:

Vesnalui [34]2 years ago

8 0

ok I got your answer

it is 1 degree(s) Celsius

yuradex [85]2 years ago

5 0

Answer:

1 degree C

Explanation:

According to the definition of specific heat, the amount of heat required to raise the temperature of unit mass by unit degree Celsius.

C = Q / m × ◇T

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What is the wavelength of an earthquake wave if it has a speed of 3 km/s and a frequency of 6 Hz?

disa [49]

Wavelength = c/f.
Wavelength =0.5km

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2 years ago

Match the speed to the section that describes.

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Answer:

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2 years ago

A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor

Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

7 0

2 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago

HELP... 3.00 amu = _____ Mev. 3.22 x 10-3 2.79 x 103 3.10 x 102

DanielleElmas [232]

The correct answer is:
$2.79 \cdot 10^3 MeV$
Let's see why.

1 amu corresponds to the mass of the proton, which is:
$m_p = 1.66 \cdot 10^{-27} kg$
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
$E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J$
Now we can convert it into electronvolts:
$E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV$

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
$3 amu = 3 \cdot 934 MeV \sim 2790 MeV = 2.79 \cdot 10^3 MeV$

5 0

3 years ago

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