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Lapatulllka [165]
2 years ago
13

A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?​

Physics
1 answer:
ELEN [110]2 years ago
5 0

Answer:

\boxed {\boxed {\sf 50 \ kg}}

Explanation:

We are asked to find the mass of a cabinet, given the force and acceleration. According to Newton's Second Law of Motion, force is the product of mass and acceleration. The formula for this is:

F= m \times a

The force is 200 Newtons, but we should convert the units to make unit cancellation easier. 1 Newton is equal to 1 kilogram meter per second squared, so the force of 200 Newtons is 200 kilogram meters per second squared.

The mass is unknown and the acceleration is 4 meters per second per second or 4 meters per second squared.

  • F= 200 kg*m/s²
  • a= 4 m/s²

Substitute the values into the formula.

200 \ kg *m/s^2 = m \times 4 \ m/s^2

We are solving for the mass, m, so we must isolate the variable. It is being multiplied by 4 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 4 m/s²

\frac {200 \ kg *m/s^2}{4 \ m/s^2}= \frac{m \times 4\ m/s^2}{4 \ m/s^2}

\frac {200 \ kg *m/s^2}{4 \ m/s^2} =m

The units of meters per second squared cancel.

\frac {200 \ kg }{4 }=m

50 \ kg =m

The mass of the cabinet is <u>50 kilograms.</u>

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a motorcycle is capable of accelerating at 5.1 m/s starting from rest how far can it travell in 1.5 sec
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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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