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expeople1 [14]
3 years ago
6

Which formula describes acceleration? m/s^2 m/s s/m m2

Physics
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

m/s^2

Explanation:

Force = mass × acceleration

kgm/s^2 = kg × acceleration

where acceleration = Force ÷ mass

= kg m/s^2 ÷ kg

:Acceleration = m/s^2

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Six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then
Nataly_w [17]

Answer:

The option (b) is correct.

Explanation:

The expression for the power in terms of work done is as follows;

P=\frac{W}{t}

Here, W is the work done, t is the time taken and P is the power.

According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.

The expression for the equivalent resistance in the series combination is as follows;

R_{eq}=R_{1}+R_{2}......+R_{6}

Put R_{1},R_{2}......,R_{6}=R.

R_{eq}=R+R+R+R+R+R

R_{eq}=6R

It is given in the problem that a seventh resistor is added (in series).

R_{eq}=R_{1}+R_{2}......+R_{7}

Put R_{1},R_{2}......,R_{7}=R.

R_{eq}=R+R+R+R+R+R+R

R_{eq}=7R

The expression for the power in terms of voltage and resistance is as follows;

P=\frac{V^{2}}{R}

Here, R is the resistance.

From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.

If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.

Therefore, the option (b) is correct.

5 0
3 years ago
During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=
Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

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6 0
2 years ago
The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ
olya-2409 [2.1K]

Answer: 8400 J

Explanation:

The formula referenced in the question is:

Q=m. c. \Delta T  

Where:

Q  is the thermal energy

m=100g \frac{1 kg}{1000 g}=0.1 kg is the mass  of the water sample

c=4200 \frac{J}{kg\°C}  is the specific heat capacity of  water

\Delta T=20\°C  is the variation in temperature

Solving:

Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)  

Q=8400 J  This is the thermal energy released

8 0
2 years ago
What is the definition of psychology
Kazeer [188]
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5 0
3 years ago
Read 2 more answers
How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied
adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

Next, we shall determine the charge. This can be obtained as follow:

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0
2 years ago
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