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3 years ago

If a suspended object A is attracted to a charged object B, can we conclude that A is charged?

1 answer:

8 0

Not necessarily, object A could also be neutral, and becoming a dipole due to object B's charge. A charged object can induce a dipole in a neutral object, and that object would then become attracted without being charged.

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Science

Rama09 [41]

Answer:

Explanation:

6 0

4 years ago

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The ultraviolet region of the electromagnetic spectrum extends beyond violet in the visible region. Knowing the relationship bet

madreJ [45]

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7 0

3 years ago

I was playing goalie in a soccer league and made a save. The ball was traveling at 14 m/s and has a mass of 0.45 kg. How muck ki

Zigmanuir [339]

Answer:

Kinetic energy = 44.1 J

Explanation:

Given:

Velocity of ball = 14 m/s

Mass of ball = 0.45 kg

Find:

Kinetic energy

Computation:

An item's kinetic energy is force it has as a result of its motion. It is the amount of work necessary to move a body of a known volume from rest to a specified velocity.

Ke = (1/2)(m)(v²)

Ke = (1/2)(0.45)(14²)

Ke = (1/2)(0.45)(196)

Kinetic energy = 44.1 J

6 0

3 years ago

The most energy-efficient way to send a spacecraft to the Moon is to boost its speed while it is in circular orbit about the Ear

jeyben [28]

Answer:

98.33 %

Explanation:

On an elliptical orbit, angular momentum will be conserved .

Angular momentum = I ω = mvR

So mv₁R₁ = mv₂R₂

= v₁R₁ = v₂R₂

where v₁ is velocity and R₁ radius in low orbit (perigee)and v₂ and R₂ is velocity and radius in high orbit ( apogee ).

Here R₁ = Radius of the earth , R₂ is distance between moon and earth.

R₁ / R₂ = 1/60

v₁ /v₂ = R₂ / R₁ = 60

v₂ / v₁ = 1 / 60

1 - (v₂ / v₁ ) = 1 -( 1 / 60)

(v₁ -v₂)/v₁ = ( 60-1 )/60

(v₁ -v₂)/v₁ x 100 = 5900/60 = 98.33 %

5 0

4 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago