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4 years ago

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Calculate the gravitational force on the washers using the formula F = ma where a = 10 m/s2 and m is the mass of the washers. Us

e the mass of the washers in kilograms. Record your calculations, to no more than three decimal places, in Table D. Note that the force of gravity on the washers is the same as the applied force of the washers on the car. The force applied to the car by one washer is kg · m/s2. The force applied to the car by two washers is kg · m/s2. The force applied to the car by three washers is kg · m/s2. The force applied to the car by four washers is kg · m/s2.

2 answers:

Radda [10]4 years ago

6 0

0.049

0.098

0.147

0.196

Neko [114]4 years ago

5 0

Answer:

0.049

0.098

0.147

0.196

Explanation:

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Answer and Explanation:

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3 years ago

How does rocks and fossils determine Earth's age?

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3 years ago

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A mother pushes a baby stroller 10 meters by applying 40 newtons of force. How much work was done?

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A small bulb is rated at 7.5 W when operated at 125 V. The tungsten filament has a temperature coefficient of resistivity α = 0.

alisha [4.7K]

To solve this problem we will apply the concepts related to resistance as a function of temperature, product of the relationship between the squared voltage and the power. Mathematically this is,

$R = \frac{v^2}{P}$

Here,

R = Resistance (At function of temperature)

v = Voltage

P = Power

Then we have,

R at 140°C (7 times room temperature),

$R(140\°C) = \frac{125^2}{7.5}$

$R(140\°C) = 2083.33\Omega$

The relationship between normal temperature and increased temperature would then be given by,

$R(140\°C) = R(20\°C)(1 +\alpha (\Delta T))$

$R(140\°C) = R(20\°C)(1+(4.5*10^{-3})(140-20))$

$R(20\°C) = \frac{2083.33}{1.54}$

$R(20\°C) = 1352.81\Omega$

Therefore the correct value of the group of answer is 1350

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3 years ago

One ball of mass 0.600kg travelling 9.00m/s to the right collides head on elastically with a second ball of mass 0.300kg travell

Alina [70]

Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

where v₁' and v₂' are the balls' respective velocities after the collision.

Kinetic energy is also conserved, so

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²

or

m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²

From the momentum equation, we have

(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'

which simplifies to

10.0 m/s = 2 v₁' + v₂'

so that

v₂' = 10.0 m/s - 2 v₁'

From the energy equation, we have

(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

which simplifies to

67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

or

226 m²/s² = 2 (v₁')² + (v₂')²

Substituting v₂' yields

226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²

which simplifies to

3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0

Solving for v₁' using the quadratic formula gives two solutions,

v₁' ≈ -2.33 m/s or v₁' = 9.00 m/s

but the second solution corresponds to the initial conditions, so we omit that one.

Then the second ball has velocity

v₂' = 10.0 m/s - 2 (-2.33 m/s)

v₂' ≈ 14.7 m/s

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2 years ago