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3 years ago

5. What is an excess of electric charge on an object?

Physics

1 answer:

ValentinkaMS [17]3 years ago

8 0

Answer:

Explanation:. What is an excess of electric charge on an

object?

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A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = $I\alpha$

$\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2$

Now

$\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians$

Here t = 1 minutes = 60 seconds

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3 years ago

Jay bought a toy car. To make the car move, he must turn a key attached to a spring. Turning the key winds the spring tight. The

lapo4ka [179]

The chemical energy in Jay's body, to kinetic energy in the car

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3 years ago

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An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength

Vladimir [108]

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

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3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

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4 years ago

Why is it important for scientist to do research?

sweet-ann [11.9K]

Answer:

Because if they dont research first they will be unprepared

Explanation:

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3 years ago

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