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3 years ago

Can you change how much power you use while exercising?

1 answer:

8 0

Answer:Most treadmills and stationary bikes use electricity, but what if you could produce electricity while exercising? With a pedal-power generator, you can! And you can use the electricity immediately to power a television, computer, stereo or other electronics — or store it in batteries to use it later.

Explanation:

You might be interested in

Which ball (if either) has the greatest speed at the moment of impact

Ierofanga [76]

Are there any options??

I would have to say metal of course but without options I can't assume anything

5 0

3 years ago

Read 2 more answers

The particles that are found in the nucleus of an atom are

kvv77 [185]

Answer: Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom

Explanation: hope this helps

8 0

3 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

2 years ago

A 1.00-kg mass is attatched to a string 1.0m long and completes a horizontal circle in 0.25s. What is the centripetal accelerati

liraira [26]

-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .

-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .

-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²

-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =

64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .

</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.

You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.

3 0

3 years ago