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3 years ago

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Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. Elect

rical outlets. Which type of current is described by these statements? A) AC B) AC/DC C) DC D) Nuclear

2 answers:

LenaWriter [7]3 years ago

7 0

Answer:DC

Explanation:

Karolina [17]3 years ago

3 0

Answer:

C. D.C.

Explanation:

The current that is being described here is D.C. or direct current. It is the D.C. that moves in one direction from negative to positive. May be created by a battery. It is different from the A.C.( alternating current whose polarity changes regularly). It is A.C. that is used in electrical outlets and not D.C.So, the current option is C.

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An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago

How does 'g' vary from place to place?

r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

8 0

3 years ago

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

3 years ago

Please help me asap

ddd [48]

Answer:

if I aint wrong it would 2nd one

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3 years ago

Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck

artcher [175]

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

8 0

3 years ago