All categories

4 years ago

If a ball is tossed straight up into the air, what position is its potential energy the greatest

1 answer:

4 0

Potential energy is the greatest when the the kinetic energy is the least. This occurs when the ball loses the energy to fly higher, and begins to fall back to the ground.

You might be interested in

you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll

-BARSIC- [3]

Technically, it should roll forever.

6 0

3 years ago

Sean drew this diagram and labeled an unknown gas as gas A. Which gas does A represent?

EleoNora [17]

Using the information given, gas-A is still completely unknown.

3 0

3 years ago

The use of a beat or a rhythm to remember something is

Mamont248 [21]

<span>The answer is Music Mnemonics. Mnemonic is originally coming from the Ancient Greek word means “memory”. It is the study and development of systems for improving and assisting the memory. Mnemonic techniques are more specific memory aids. Music Mnemonic is a way we used to remember song lyrics and rhythmic movement.</span>

5 0

3 years ago

Read 2 more answers

We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goe

o-na [289]

Answer:

See answer

Explanation:

Given quantities:

$\eta = 0.05\\ W=90[W]\\r=0.0285[m]$

where $\eta$ is the efficiency of the lightbulb (visible light is 5% of the total power), $W$ is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

$I =\frac{P}{A}$

a) Now the effective power is $\eta*W$ , therefore:

$I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]$

b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:

$S_{average}= \frac{EB}{2\mu_{0}}[W/m]$

now $E$ and $B$ are related:

$E=cB\\ B=\frac{E}{c}$ and $c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}$

replace in $S_{average}$

$S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]$

we replace the values and we get:

$E= \sqrt{\frac{2I}{\epsilon_{0}c}}$

$E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]$

therefore

$B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]$

8 0

3 years ago

Suppose you were marooned on a tropical island and had to use seawater (density 1.10 g·cm3) to make a primitive barometer. what

Paul [167]

We know that P = hρg

Where:

P - pressure Pa,

h - height in meter,

ρ – would be the density in kg / m^3; and

g - acceleration due to gravity is m / s^2

p = hρg if h = 0.735 meter, ρ = 13600 kg / meter^3, g = 10 meter/ sec^2

p = 0.735*13600*10 = 99960 Pa or

P = 1 x 10^5 Pa

Now with sea water if we have to make a barometer:

ρ = 1100 kg / meter^3 (given)

P = hρg if we put the value of P calculated and the value of ρ = 1100 kg / meter^3 given, we will

get, 1 x 10^5 = h*1100*10

therefore, h = 9.09 meter or 29.82 feet of water.

3 0

4 years ago