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2 years ago

How long was a 60 W light bulb turned on if it used a total of 580 J of energy?

1 answer:

7 0

Here's the tool you need. You can't answer the question without this:

   "1 watt"
means
   "1 joule of energy, generated, used, or moved, every second".

So 60 watts = 60 joules per second

   Total energy generated,
      used, or moved = (power) x (time).

   580 joules = (60 watts) x (time)

Divide each side
by (60 watts): Time = (580 joules) / (60 joules/sec)

   = (9 and 2/3) seconds .

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A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

$v_i = 5.10 m/s$

displacement of the package as it is dropped on ground

$d = -105 m$

acceleration is due to gravity

$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

7 0

3 years ago

What energy is directly dependent upon velocity and mass?

erastova [34]

Answer:

Kinetic energy

Explanation:

3 0

2 years ago

Which equation is used to calculate the magnetic force on a charge moving in a magnetic field?

nadya68 [22]

Answer:

Explanation:

6 0

3 years ago

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

2 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

   Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:   1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

   T² = (1600 m) / (0.65 m/s²)

   T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

   T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)

7 0

2 years ago