How is energy wasted by the radio

What are the potential out comes of force

AysviL [449]

Answer:

F in the definition of potential energy is the force exerted by the force field, e.g., gravity, spring force, etc. The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r.

Explanation:

5 0

3 years ago

a 905 - g meteor impacts the earth at a speed of 1623 m/s. if all of its energy is entirely converted to heat in the meteor, wha

nlexa [21]

Answer: 2859.78 k

Explanation: By using the law of conservation of energy, the kinetic energy of the meteor equals the heat energy.

Kinetic energy = 1/2mv^2

Heat energy = mcΔθ

Where m = mass of meteor , v = velocity of meteor = 1623 m/s

c = specific heat capacity of meteor (iron) = 460.548 j/kg/k

Δθ = change in temperature of meteor = ?

From law ofconservation of energy, we have that

1/2mv^2 = mcΔθ

By cancelling "m" on both sides, we have that

v^2/2 = cΔθ

v^2 = 2cΔθ

(1623)^2 = 2× 460.548 × Δθ

2634129 = 921.096 × Δθ

Δθ = 2634129 / 921.096

Δθ = 2859.78 k

6 0

3 years ago

SOMEONE PLEASE HELP ME

Leviafan [203]

Answer:

The same.

Explanation:

Temperature is a measure of the average kinetic energy of a substance; if two pieces of iron have the same temperature, then their particles will have the same amount of average kinetic energy.

5 0

3 years ago

Read 2 more answers

Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radi

dlinn [17]

Answer:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

$\omega = 75.8 \frac{rev}{min}\cdot \frac{2\pi rad}{1rev}\cdot \frac{1min}{60s} = 7.94rad/s$

A 6.84cm diameter top spinning at 375 degrees per second:

$\omega = 375 \frac{^\circ}{s}\cdot \frac{2\pi rad}{360^ \circ} = 6.54rad/s$

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

$\omega = \frac{v}{r} = \frac{v}{\frac{b}{2}\sqrt 2} = \frac{2.51 m/s}{\frac{0.458 m}{2} \sqrt 2} = 7.75rad/s$

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

$\omega = \sqrt \frac{a}{r} = \sqrt \frac{25.4 m/s^{2}}{0.521m} = 6.98rad/s$

<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

I hope it helps you!

4 0

3 years ago

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

3 years ago